From bef1cf931e89dbba06df186fb5d7f64a9a78a1e7 Mon Sep 17 00:00:00 2001 From: "W. Trevor King" Date: Mon, 29 Jun 2009 14:30:33 -0400 Subject: [PATCH] Reorganized problems by textbook and imported my previous courses' code. --- .../notes/equations/Giancoli_6/chapter_19.tex | 39 ++ .../notes/equations/Giancoli_6/chapter_20.tex | 78 ++++ .../notes/equations/Giancoli_6/chapter_21.tex | 41 ++ .../notes/equations/Giancoli_6/chapter_22.tex | 76 ++++ .../Serway_and_Jewett_4_wking/chapter_1.tex | 15 + .../Serway_and_Jewett_4_wking/chapter_10.tex | 68 ++++ .../Serway_and_Jewett_4_wking/chapter_2.tex | 36 ++ .../Serway_and_Jewett_4_wking/chapter_3.tex | 29 ++ .../Serway_and_Jewett_4_wking/chapter_4.tex | 12 + .../Serway_and_Jewett_4_wking/chapter_5.tex | 16 + .../Serway_and_Jewett_4_wking/chapter_6.tex | 49 +++ .../Serway_and_Jewett_4_wking/chapter_7.tex | 47 +++ .../Serway_and_Jewett_4_wking/chapter_8.tex | 30 ++ .../Serway_and_Jewett_4_wking/format.tex | 8 + .../introduction.tex | 3 + .../Serway_and_Jewett_4_wking/main.aux | 65 +++ .../Serway_and_Jewett_4_wking/main.log | 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wking}/problem12.T.tex (100%) rename latex/problems/{ => wking}/problem28.compton-cat.T.tex (100%) diff --git a/latex/notes/equations/Giancoli_6/chapter_19.tex b/latex/notes/equations/Giancoli_6/chapter_19.tex new file mode 100644 index 0000000..593e69a --- /dev/null +++ b/latex/notes/equations/Giancoli_6/chapter_19.tex @@ -0,0 +1,39 @@ +\section{Chapter 19: Electric Forces and Electric Fields} + +\subsection{Coulomb's law} +\begin{equation} + \vect{F}_{12} = k_e \frac{q_1 q_2}{r^2} \rhat_{12} +\end{equation} +Where $k_e \approx 8.988\E{9}\U{N$\cdot$m$^2$/C$^2$}$ is the \emph{Coulomb constant}, + $\vect{F}_{12}$ is the force on $q_1$ due to $q_2$, and + $\rhat_{12}$ is a unit vector pointing from $q_1$ to $q_2$. + +\subsection{Electric field} + +\begin{equation} + \vect{E} \equiv \frac{\vect{F}_e}{q_0} +\end{equation} +So for a point charge, the electric field at a point \vect{r} is given by +\begin{equation} + \vect{E}(\vect{r}) = k_e \frac{q}{r^2}\rhat +\end{equation} +And for a group of charges, the electric field is given by +\begin{equation} + \vect{E}(\vect{r}) = k_e \int\frac{dq}{r^2}\rhat +\end{equation} + +\subsection{Electric flux} + +The electric flux is the amount of electric field ``flowing'' through a surface $S$: +\begin{equation} + \Phi_E \equiv \int_S \vect{E} \cdot d\vect{A} +\end{equation} + +When you know something about the symmetry of the problem, you can often use \emph{Gauss's law} +\begin{equation} + \Phi_E = \oint \vect{E} \cdot d\vect{A} = \frac{q_{in}}{\epsilon_0} +\end{equation} + +Where the $\oint$ symbol reminds us that the surface must be closed, + $q_{in}$ is the enclosed charge, and + $\epsilon_0 \approx 8.854\E{-12}\U{C$^2$/N$\cdot$m$^2$}$ is the \emph{permittivity of free space}. diff --git a/latex/notes/equations/Giancoli_6/chapter_20.tex b/latex/notes/equations/Giancoli_6/chapter_20.tex new file mode 100644 index 0000000..957d57b --- /dev/null +++ b/latex/notes/equations/Giancoli_6/chapter_20.tex @@ -0,0 +1,78 @@ +\section{Chapter 20: Electric Potential and Capacitance} + +\subsection{Potential} +The change in potential $\Delta U$ of a charge $q_0$ moving from $A$ +to $B$ in an electric field $\vect{E}(\vect{r})$ is given by +\begin{equation} + \Delta U = -q_0 \int_A^B \vect{E}\cdot d\vect{s} +\end{equation} +Where $d\vect{s}$ is an infinitesimal displacement vector. + +The electric potential is defined as +\begin{equation} + \Delta V = \frac{\Delta U}{q_0} = -\int_A^B \vect{E}\cdot d\vect{s} +\end{equation} +Or, taking the derivative of both sides in the $x$ direction +\begin{equation} + E_x = -\frac{dV}{dx} +\end{equation} +With similar cases in the $y$ and $z$ directions. +For those of you with vector calculus who recognize the gradient +$\vect{\nabla}$ you can express the above formula as: +\begin{equation} + \vect{E} = - \vect{\nabla}V +\end{equation} + +The voltage a distance $r$ from a point charge $q$ is given by +\begin{equation} + V = -\int_A^B \frac{k_e q}{r^2} dr = k_e \frac{q}{r} +\end{equation} +Which we can combine with the definition of the electric potential to +find the potential energy of two point charges seperated by a distance +$r_{12}$: +\begin{equation} + U = q_1 V = k_e \frac{q_1 q_2}{r_{12}} +\end{equation} + +Just as we could integrate over a charge distribution to find the +electric field at a given point, we can find electric potential with +\begin{equation} + V = k_e \int \frac{dq}{r} +\end{equation} +Which is usually an easier integral because the integrand is a \emph{scalar}. + +\subsection{Capacitance} + +The capacitance $C$ of a pair of conductors (plates, other shapes...) is defined as +\begin{equation} + C \equiv \frac{Q}{\Delta V} +\end{equation} +Where $Q$ is the charge on one of two oppositely charged conductors, +and $\Delta V$ is the voltage difference between the conductors. + +The capacitance of a capacitor consisting of two parallel plates of +area $A$ seperated by a distance $d$ is given by +\begin{equation} + C = \frac{\epsilon_0 A}{d} +\end{equation} +where $\epsilon_0$ is the permittivity of free space. +(Which comes from using the definition of capacitance, and the +electric field between two charged plates that we calculated in +recitation 2, problem P19.62.) + +The equivalent capacitance of several capacitors in parallel is +\begin{equation} + C_{eq} = C_1 + C_2 + C_3 + \cdots +\end{equation} +and in series is +\begin{equation} + \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \cdots +\end{equation} + +The electric potential energy $U$ stored in a capacitor is +\begin{equation} + U = \frac{1}{2} C (\Delta V)^2 +\end{equation} + + + diff --git a/latex/notes/equations/Giancoli_6/chapter_21.tex b/latex/notes/equations/Giancoli_6/chapter_21.tex new file mode 100644 index 0000000..3bda9b5 --- /dev/null +++ b/latex/notes/equations/Giancoli_6/chapter_21.tex @@ -0,0 +1,41 @@ +\section{Chapter 21: Current and Direct Current Circuits} + +\subsection{Current} + +The electrical current $I$ through a conductor is defined by +\begin{equation} + I \equiv \frac{dQ}{dt} \;, +\end{equation} +the amount of charge passing through some surface per unit time. + +\subsection{Resistance} + +Resistance $R$ of a conductor is given by Ohm's law +\begin{equation} + V = I R \;, +\end{equation} +where $I$ is the current through the conductor, and $V$ is the electric potential difference across the resistor. + +\subsection{Power} + +The power dissipated by a current $I$ dropping through an electric potential difference $V$ is given by +\begin{equation} + P = IV \;, +\end{equation} +so using Ohm's law, the power dissipated by a resistor is +\begin{equation} + P_R = I (IR) = I^2 R = \p({\frac{V}{R}})^2 R = \frac{V^2}{R} \;. +\end{equation} + +\subsection{Kirchhoff's rules} + +Complicated steady-state circuits can be analyzed using Kirchhoff's two rules. +The loop rule conserves energy by forcing the electric potential change about a loop to be 0. +\begin{equation} + \sum_{\text{elements in a loop}} \Delta V = 0 \;. +\end{equation} +The junction rule conserves charge by forcing the current into a node to balance the current out of a node. +\begin{equation} + \sum_{\text{branches from a node}} I = 0 \;. +\end{equation} +By combining these two equations, we can derive equivalent capacitances and resistances for circuit elements in series or in parallel, and, more generally, find the steady-state behavior to any steady-state circuit's problem. diff --git a/latex/notes/equations/Giancoli_6/chapter_22.tex b/latex/notes/equations/Giancoli_6/chapter_22.tex new file mode 100644 index 0000000..880389a --- /dev/null +++ b/latex/notes/equations/Giancoli_6/chapter_22.tex @@ -0,0 +1,76 @@ +\section{Chapter 22: Magnetic Forces and Magnetic Fields} + +\subsection{Cross products} + +The magnitude of $\vect{A}=\vect{B}\times\vect{C}$ is given by +\begin{equation} + |\vect{A}| = |\vect{B}|\cdot|\vect{C}|\cdot\sin\theta \;, \label{eq.cross_mag} +\end{equation} +where $\theta$ is the angle between \vect{B} and \vect{C}. +In terms of $(x,y,z)$ components, \vect{A} is given by +\begin{equation} + \vect{A} = \vect{B}\times\vect{C} + = \begin{vmatrix} + \ihat & \jhat & \khat \\ + B_x & B_y & B_z \\ + C_x & C_y & C_z + \end{vmatrix} + = \ihat(B_y C_z - B_z C_y) - \jhat (B_x C_z - B_z C_x) + \khat (B_x C_y - B_y C_x) + = A_x \ihat + A_y \jhat + A_z \khat \;, +\end{equation} +so +\begin{align} + A_x &= (B_y C_z - B_z C_y) & A_y &= (B_z C_x - B_x C_z) & A_z &= (B_x C_y - B_y C_x) \label{eq.cross_comp} +\end{align} +The direction of \vect{A} is given either by solving Eqn.\ \ref{eq.cross_comp} explicitly +or by using Eqn.\ \ref{eq.cross_mag} and using a \emph{right hand rule}. + + +\subsection{Magnetic force} + +The magnetic force on a moving charged particle is given by +\begin{equation} + \vect{F}_B = q \vect{v}\times\vect{B} \;, +\end{equation} +where $q$ is the charge on the particle, \vect{v} is the particle's velocity, and \vect{B} is the external magnetic field. + +The magnetic force on a current carrying wire is given by +\begin{equation} + \vect{F}_B = I \vect{l}\times\vect{B} \;, +\end{equation} +where $I$ is the current in the wire, \vect{l} is the length of the wire in the direction of the current, and \vect{B} is the external magnetic field. + + +\subsection{Magnetic dipole moments} + +The magnetic dipole moment \vect{\mu} of a loop carrying current $I$ is +\begin{equation} + \vect{\mu} = I\vect{A} \;, +\end{equation} +where \vect{A} is the area of the loop. +The torque on such a loop placed in an external magnetic field \vect{B} is +\begin{equation} + \tau = \mu\times\vect{B} \;. +\end{equation} + + +\subsection{Generating magnetic fields} + +The Biot-Savart law determines the magnetic field generated by an infinitesimal current (much like Coulomb's law determined the electric field generated by a point charge). +\begin{equation} + d\vect{B} = \frac{\mu_0}{4\pi} \frac{I d\vect{s}\times{\rhat}}{r^2} \;, +\end{equation} +where $\mu_0 = 4\pi\E{-7}\U{T$\cdot$m/A}$ is the permeability of free space, $d\vect{s}$ is the infinitesimal length of the current $I$, and \vect{r} is the vector from the location of the current to the location of the magnetic field. + +Amp\`ere's law determines the magnetic field generated by a current, in a manner suitable for symmetric current distributions (much like Gauss's law determined the electric field for symmetric charge distributions). +\begin{equation} + \oint \vect{B}\cdot d\vect{s} = \mu_0 I_\text{in} \;, +\end{equation} +where $\oint$ signifies integration about a closed loop, and $I_\text{in}$ is the total current penetrating the loop. + +You can use Amp\`ere's law to determine the magnetic field from some standard current distributions as follows +\begin{align} + B &= \frac{\mu_0 I}{2 \pi r} && \text{Distance $r$ from a long, straight wire} \\ + B &= \frac{\mu_0 N I}{2 \pi r} && \text{Inside a torroid of radius $r$} \\ + B &= \mu_0 n I && \text{Inside a solenoind with $n$ turns per unit length} +\end{align} diff --git a/latex/notes/equations/Serway_and_Jewett_4_wking/chapter_1.tex b/latex/notes/equations/Serway_and_Jewett_4_wking/chapter_1.tex new file mode 100644 index 0000000..b001950 --- /dev/null +++ b/latex/notes/equations/Serway_and_Jewett_4_wking/chapter_1.tex @@ -0,0 +1,15 @@ +\section{Chapter 1: Introduction and vectors} + +\subsection{Trig} +sohcahtoa +\begin{align} + \sin\theta &= \frac{\mbox{opposite}}{\mbox{hypotenuse}} \\ + \cos\theta &= \frac{\mbox{adjacent}}{\mbox{hypotenuse}} \\ + \tan\theta &= \frac{\mbox{opposite}}{\mbox{adjacent}} +\end{align} + +\subsection{Vectors} +\begin{equation} + \vect{R} = \vect{A} + \vect{B} + = (A_x + B_x)\ihat + (A_y + B_y)\jhat + (A_z + B_z)\khat +\end{equation} diff --git a/latex/notes/equations/Serway_and_Jewett_4_wking/chapter_10.tex b/latex/notes/equations/Serway_and_Jewett_4_wking/chapter_10.tex new file mode 100644 index 0000000..aaad2c3 --- /dev/null +++ b/latex/notes/equations/Serway_and_Jewett_4_wking/chapter_10.tex @@ -0,0 +1,68 @@ +\section{Chapter 10: Rotational motion} + +\subsection{Definitions} + +\begin{align} + \omega &\equiv \frac{\mbox{d}\theta}{\mbox{d}t} \\ + \alpha &\equiv \frac{\mbox{d}\omega}{\mbox{d}t} +\end{align} + +\subsection{Relations to linear quantities} + +(All of the linear quanties are tangential) +\begin{align} + s_t &= r \theta \\ + v_t &= r \omega \\ + a_t &= r \alpha +\end{align} + +For constant $\alpha$ problems, all of the constant acceleration equations from Chapter 2 are still valid with the proper substitutions. + +\subsection{Analogs of other linear properties} +The rotational motion equivalent of mass is the \emph{moment of inertia} +\begin{align} + I &= \sum_i m_i r_i^2 \\ + I &= \int \rho r^2 \mbox{d}V +\end{align} +And the moments of inertia for some common shapes (Table 10.2, page 300) \\ +\begin{center} +\begin{tabular}{r|l} + Hoop/ring about axis & $MR^2$ \\ + Solid cylinder about axis & $\frac{1}{2}MR^2$ \\ + Hollow cylinder about axis & $\frac{1}{2}M(R_i^2 + R_o^2)$ \\ + Long thin rod perp to axis through center & $\frac{1}{12}ML^2$ \\ + Long thin rod perp to axis through end & $\frac{1}{3}ML^2$ \\ + Solid sphere about diameter & $\frac{2}{5}MR^2$ \\ + Thin spherical shell about diameter & $\frac{2}{3}MR^2$ \\ + Rectangular plate perp to one side through center & $\frac{1}{12}M(a^2 + b^2)$ +\end{tabular}\\ +\end{center} + +The kinetic energy of a rotating body is given by +\begin{equation} + K = \frac{1}{2} I \omega^2 +\end{equation} +(Note that the kinetic energy of any given particle should be expressed as \emph{either} a linear or a rotational kinetic energy, not as \emph{both} at once. +You \emph{can} use linear kinetic energy for one particle and rotational kinetic energy for another to find the kinetic energy at a single point in time.) + +The rotational equivalent to force is torque +\begin{equation} + \vect{\tau} \equiv \vect{r} \times \vect{F} +\end{equation} +Where the cross product between two vectors defined by +\begin{equation} + \vect{A} \times \vect{B} \equiv |\vect{A}| |\vect{B}| \sin\theta \widehat{rhr} +\end{equation} +Where $\widehat{rhr}$ is a unit vector in the direction specified by the right-hand rule. + +The analog to Newton's second law is +\begin{equation} + \sum \vect{\tau} = I \vect{\alpha} = \frac{\mbox{d}\vect{L}}{\mbox{d}t} +\end{equation} + +The angular momentum \vect{L} is given by +\begin{equation} + \vect{L} \equiv \vect{r} \times \vect{p} = I \omega +\end{equation} +Angular momentum is conserved if there are no external torques on the system. +As with rotational kinetic energy, you should be careful to avoid double counting by using both rotational and linear momentum to compute the momentum of the same particle, although here the units do not match (another reason to keep track of your units!). diff --git a/latex/notes/equations/Serway_and_Jewett_4_wking/chapter_2.tex b/latex/notes/equations/Serway_and_Jewett_4_wking/chapter_2.tex new file mode 100644 index 0000000..d08ab66 --- /dev/null +++ b/latex/notes/equations/Serway_and_Jewett_4_wking/chapter_2.tex @@ -0,0 +1,36 @@ +\section{Chapter 2: Motion in one dimension} + +\subsection{Definitions} +Average quantities +\begin{align} + \vect{v}_{avg} \equiv \frac{\Delta \vect{x}}{\Delta t} \\ + \vect{a}_{avg} \equiv \frac{\Delta \vect{v}}{\Delta t} +\end{align} +Instantaneous quantities +\begin{align} + \vect{v} \equiv \frac{\mbox{d}\vect{x}}{\mbox{d}t} \\ + \vect{a} \equiv \frac{\mbox{d}\vect{v}}{\mbox{d}t} +\end{align} + +\subsection{Equations of motion} +For constant velocity problems (integrating the instantaneous velocity definition) +\begin{equation} + x_f = x_i + v_x t +\end{equation} +And for constant acceleration problems (integrating the instantaneous acceleration definition twice, and manipulating a bit). +\begin{align} + v_{xf} &= v_{xi} + a_x t \label{v_of_t}\\ + x_f &= x_i + \frac{1}{2}(v_{xi} + v_{xf})t \\ + x_f &= x_i + v_{xi} t + \frac{1}{2} a_x t \label{x_of_t}\\ + v_{xf}^2 &= v_{xi}^2 + 2 a_x (x_f - x_i) \label{v_sqrt} +\end{align} +Where eqn. \ref{v_sqrt} comes from solving eqn \ref{x_of_t} for $t$ using the quadratic formula and plugging the result into eqn \ref{v_of_t}. + +The quadratic formula says that +\begin{equation} + x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} +\end{equation} +Yeilds two values of $x$ that solve the quadratic equation +\begin{equation} + 0 = a x^2 + b x + c +\end{equation} diff --git a/latex/notes/equations/Serway_and_Jewett_4_wking/chapter_3.tex b/latex/notes/equations/Serway_and_Jewett_4_wking/chapter_3.tex new file mode 100644 index 0000000..6d33e05 --- /dev/null +++ b/latex/notes/equations/Serway_and_Jewett_4_wking/chapter_3.tex @@ -0,0 +1,29 @@ +\section{Chapter 3: Motion in two dimensions} + +\subsection{Constant acceleration} +Applying our 1D equations of motion to each direction, we get the multidimensional formulas: +\begin{align} + \vect{v}_f &= \vect{v}_i + \vect{a} t \\ + \vect{r}_f &= \vect{r}_i + \vect{v}_i t + \frac{1}{2} \vect{a} t +\end{align} + +In the special case of projectile motion with $a_x = 0$ and $a_y = g$ these reduce to +\begin{align} + v_{xf} &= v_{xi} = \mbox{constant} \\ + x_f &= x_i + v_x t \label{proj_x} \\ + v_{yf} &= v_{yi} - gt \label{proj_vy} \\ + y_f &= y_i + v_{yi} - \frac{1}{2}gt^2 \label{proj_y} +\end{align} +We could apply our accelerations to all four of the constant 1D acceleration equations in chapter 2, but you get the idea... + +\subsection{Circular motion} +A particle moving in a circle of radius $r$ with velocity $v$ has a centerward acceleration of +\begin{equation} + a_c = \frac{v^2}{r} +\end{equation} + +\subsection{Frames of reference} +If an observer $O'$ is moving with velocity $\vect{v}_{O'O}$ with respect to observer $O$, their measurements of the velocity of a particle located at point $P$ are related according to +\begin{equation} + \vect{v}_{PO} = \vect{v}_{PO'} + \vect{v}_{OO'} \label{ref_frm} +\end{equation} diff --git a/latex/notes/equations/Serway_and_Jewett_4_wking/chapter_4.tex b/latex/notes/equations/Serway_and_Jewett_4_wking/chapter_4.tex new file mode 100644 index 0000000..61934b5 --- /dev/null +++ b/latex/notes/equations/Serway_and_Jewett_4_wking/chapter_4.tex @@ -0,0 +1,12 @@ +\section{Chapter 4: The laws of motion} + +\subsection{Newton's laws} +\begin{enumerate} +\item An object in motion will remain in motion unless acted upon by an outside force. +\item $\sum \vect{F} = m \vect{a}$ +\item For every force there is an equal and opposite reaction force. +\end{enumerate} +Note that the two forces referenced in the 3rd law belong to two \emph{different} free body diagrams (FBDs). +For example, the sun and earth attract each other gravitationally. +Let $F$ be the magnitude of the force, and \ihat\ be the direction from the earth to the sun. +The force on the earth due to the sun (showing up on the earth's FBD) is $F\ihat$ and the force on the sun due to the earth (showing up on the sun's FBD) is $-F\ihat$. diff --git a/latex/notes/equations/Serway_and_Jewett_4_wking/chapter_5.tex b/latex/notes/equations/Serway_and_Jewett_4_wking/chapter_5.tex new file mode 100644 index 0000000..a57092c --- /dev/null +++ b/latex/notes/equations/Serway_and_Jewett_4_wking/chapter_5.tex @@ -0,0 +1,16 @@ +\section{Chapter 5: More applications of Newton's laws} + +\subsection{Friction} +Let $\mu_s$ and $\mu_k$ be the static and kinetic coefficients of friction (respectively) between an object and a surface, and let $F_N$ be the normal force on the object due to the surface. +The respective forces of friction are given by +\begin{align} + F_{sf} &\le \mu_s F_N \\ + F_{kf} &= \mu_k F_N +\end{align} + +\subsection{Drag} +Objects moving through viscous materials (air, water, etc.) experience a velocity dependent resistive force +\begin{equation} + F_{drag} = -b v +\end{equation} +Where $b$ depends on the particular system under consideration. diff --git a/latex/notes/equations/Serway_and_Jewett_4_wking/chapter_6.tex b/latex/notes/equations/Serway_and_Jewett_4_wking/chapter_6.tex new file mode 100644 index 0000000..c9f8367 --- /dev/null +++ b/latex/notes/equations/Serway_and_Jewett_4_wking/chapter_6.tex @@ -0,0 +1,49 @@ +\section{Chapter 6: Energy and energy transfer} + +\subsection{Work} +In general +\begin{equation} + W \equiv \int_{\vect{r}_i}^{\vect{r}_f}\vect{F}\cdot\mbox{d}\vect{r} +\end{equation} +Or for the one dimensional case +\begin{equation} + W \equiv \int_{x_i}^{x_f}\vect{F}\cdot\ihat \mbox{d}x \label{1D_work_int} +\end{equation} +Where the dot product $\vect{A} \cdot \vect{B}$ is defined as +\begin{equation} + \vect{A} \cdot \vect{B} \equiv |\vect{A}||\vect{B}|\cos\theta \label{dot} +\end{equation} +Where $\theta$ is the angle between the vectors. +So in the 1D, constant force-and-angle case +\begin{equation} + W = \vect{F}\cdot\Delta\vect{r} \label{1D_work_const} +\end{equation} + +The force from a spring is given by Hooke's law +\begin{equation} + F_s = -kx \label{Hooke} +\end{equation} +So the work done by a spring from $x_i$ to $x_f$ is +\begin{equation} + W_s = \int_{x_i}^{x_f} (-kx)\mbox{d}x = \frac{1}{2}kx_i^2 - \frac{1}{2}kx_f^2 \label{Espring} +\end{equation} + +\subsection{Kinetic energy} + +\begin{equation} + K = \frac{1}{2} m v^2 \label{ke} +\end{equation} + +\subsection{Work-kinetic energy theorem} + +\begin{equation} + W_{net} = K_f - K_i = \Delta K \label{wk_ke_thm} +\end{equation} + +\subsection{Power} + +\begin{align} + P_{avg} &\equiv \frac{W}{\Delta t} \\ + P &\equiv \frac{\mbox{d}W}{\mbox{d}t} = \vect{F} \cdot \vect{v} + = \frac{\mbox{d}E}{\mbox{d}t} +\end{align} diff --git a/latex/notes/equations/Serway_and_Jewett_4_wking/chapter_7.tex b/latex/notes/equations/Serway_and_Jewett_4_wking/chapter_7.tex new file mode 100644 index 0000000..83e54d1 --- /dev/null +++ b/latex/notes/equations/Serway_and_Jewett_4_wking/chapter_7.tex @@ -0,0 +1,47 @@ +\section{Chapter 7: Potential energy} + +\subsection{Conservative forces} +A force is \emph{conservative} if the work it does on a particle is independent of the path the particle takes between two given points. +The potential energy change is the inverse of the work done by the force +\begin{equation} + \Delta U = U_f - U_i = - \int_{x_i}^{x_f} F_x dx +\end{equation} +Or, taking the derivative of both sides with respect to $x$ +\begin{equation} + F_x = - \frac{\mbox{d}U}{\mbox{d}x} +\end{equation} + +The gravitational potential of a particle under a gravitational force $F_g = mg$ is (relative to the energy at some height $h_r = 0$) +\begin{equation} + U_g = -\int_0^h (-mg)\mbox{d}y = mgh +\end{equation} + +The spring potential of a particle under a spring force $F_s = -kx$ is (relative to the unstretched energy at $x_r = 0$) +\begin{equation} + U_s = -\int_0^x (-kx)\mbox{d}x = \frac{1}{2}kx^2 +\end{equation} + +The gravitational potential energy of a particle under Newton's gravitational force $F_G$ is (relative to the energy at $r_r = \infty$) +\begin{equation} + U_G = -\int_\infty^r \left(\frac{-G m_1 m_2}{r^2}\right) \mbox{d}r + = G m_1 m_2 \int_\infty^r \frac{1}{r^2} \mbox{d}r + = \left. \frac{-G m_1 m_2}{r} \right|_\infty^r + = \frac{-G m_1 m_2}{r} +\end{equation} +And so on for any other conservative forces... + +\subsection{Mechanical energy} +The total mechanical energy at any moment is +\begin{equation} + E_{mech} \equiv K + U +\end{equation} +Conserving energy using this formula +\begin{align} + E_i = K_i + U_i &= E_f = K_f + U_f \\ + \Delta K &= -\Delta U +\end{align} +Which is identical to the work-kinetic energy theorem from Chapter 6 (eqn \ref{wk_ke_thm} with $W_{net} = -\Delta U$). +The only difference between these two approaches is that nonconservative forces do not have really well defined ``potential energies''. + +\subsection{Equilibria} +In a potential energy diagram, a point of \emph{stable equilibrium} is a local minimum, a point of \emph{unstable equilibrium} is a local maximum, and a region of \emph{neutral equilibrium} is a region of constant potential energy. diff --git a/latex/notes/equations/Serway_and_Jewett_4_wking/chapter_8.tex b/latex/notes/equations/Serway_and_Jewett_4_wking/chapter_8.tex new file mode 100644 index 0000000..9da6944 --- /dev/null +++ b/latex/notes/equations/Serway_and_Jewett_4_wking/chapter_8.tex @@ -0,0 +1,30 @@ +\section{Chapter 8: Momentum and collisions} + +\subsection{Linear momentum} +Momentum +\begin{equation} + \vect{p} \equiv m \vect{v} \label{p_def} +\end{equation} +is conserved +\begin{equation} + \sum\vect{p}_i = \sum\vect{p}_f +\end{equation} + +\subsection{Impulse} +The impulse-momentum theorem +\begin{equation} + \vect{I} = \int_{t_1}^{t_2} \sum \vect{F} \mbox{d}t = \Delta \vect{p} +\end{equation} + +\subsection{Types of collisions} +\begin{itemize} +\item Inelastic collision: kinetic energy is not conserved. +\item Perfectly inelastic collision: the particles stick together afterwards. +\item Elastic collision: kinetic energy is conserved. +\end{itemize} +In all types of collisions (without external forces), momentum is conserved. + +\subsection{Center of mass} +\begin{equation} + \vect{r}_{CM} = \frac{\sum_i m_i \vect{r}_i}{\sum_i m_i} +\end{equation} diff --git a/latex/notes/equations/Serway_and_Jewett_4_wking/format.tex b/latex/notes/equations/Serway_and_Jewett_4_wking/format.tex new file mode 100644 index 0000000..30cb464 --- /dev/null +++ b/latex/notes/equations/Serway_and_Jewett_4_wking/format.tex @@ -0,0 +1,8 @@ +% general layoutfor the document + +\topmargin -0.5in +\headheight 0.0in +\headsep 0.0in +\textheight 10in +\oddsidemargin -0.5in +\textwidth 7.5in diff --git a/latex/notes/equations/Serway_and_Jewett_4_wking/introduction.tex b/latex/notes/equations/Serway_and_Jewett_4_wking/introduction.tex new file mode 100644 index 0000000..1cd16d9 --- /dev/null +++ b/latex/notes/equations/Serway_and_Jewett_4_wking/introduction.tex @@ -0,0 +1,3 @@ +\section{Introduction} +I've just gone through and compiled the important equations from the ends of the various chapters, and the equations given on the quizzes so far. +Hope this helps :). diff --git a/latex/notes/equations/Serway_and_Jewett_4_wking/main.aux b/latex/notes/equations/Serway_and_Jewett_4_wking/main.aux new file mode 100644 index 0000000..b78d9c1 --- /dev/null +++ b/latex/notes/equations/Serway_and_Jewett_4_wking/main.aux @@ -0,0 +1,65 @@ +\relax +\ifx\hyper@anchor\@undefined +\global \let \oldcontentsline\contentsline +\gdef \contentsline#1#2#3#4{\oldcontentsline{#1}{#2}{#3}} +\global \let \oldnewlabel\newlabel +\gdef \newlabel#1#2{\newlabelxx{#1}#2} +\gdef \newlabelxx#1#2#3#4#5#6{\oldnewlabel{#1}{{#2}{#3}}} +\AtEndDocument{\let \contentsline\oldcontentsline +\let \newlabel\oldnewlabel} +\else +\global \let \hyper@last\relax +\fi + +\@writefile{toc}{\contentsline {section}{Introduction}{1}{section*.1}} +\@writefile{toc}{\contentsline {section}{Chapter 1: Introduction and vectors}{1}{section*.2}} +\@writefile{toc}{\contentsline {subsection}{Trig}{1}{section*.3}} +\@writefile{toc}{\contentsline {subsection}{Vectors}{1}{section*.4}} +\@writefile{toc}{\contentsline {section}{Chapter 2: Motion in one dimension}{1}{section*.5}} +\@writefile{toc}{\contentsline {subsection}{Definitions}{1}{section*.6}} +\@writefile{toc}{\contentsline {subsection}{Equations of motion}{1}{section*.7}} +\newlabel{v_of_t}{{10}{1}{Equations of motion\relax }{equation.10}{}} +\newlabel{x_of_t}{{12}{1}{Equations of motion\relax }{equation.12}{}} +\newlabel{v_sqrt}{{13}{1}{Equations of motion\relax }{equation.13}{}} +\@writefile{toc}{\contentsline {section}{Chapter 3: Motion in two dimensions}{2}{section*.8}} +\@writefile{toc}{\contentsline {subsection}{Constant acceleration}{2}{section*.9}} +\newlabel{proj_x}{{19}{2}{Constant acceleration\relax }{equation.19}{}} +\newlabel{proj_vy}{{20}{2}{Constant acceleration\relax }{equation.20}{}} +\newlabel{proj_y}{{21}{2}{Constant acceleration\relax }{equation.21}{}} +\@writefile{toc}{\contentsline {subsection}{Circular motion}{2}{section*.10}} +\@writefile{toc}{\contentsline {subsection}{Frames of reference}{2}{section*.11}} +\newlabel{ref_frm}{{23}{2}{Frames of reference\relax }{equation.23}{}} +\@writefile{toc}{\contentsline {section}{Chapter 4: The laws of motion}{2}{section*.12}} +\@writefile{toc}{\contentsline {subsection}{Newton's laws}{2}{section*.13}} +\@writefile{toc}{\contentsline {section}{Chapter 5: More applications of Newton's laws}{2}{section*.14}} +\@writefile{toc}{\contentsline {subsection}{Friction}{2}{section*.15}} +\@writefile{toc}{\contentsline {subsection}{Drag}{2}{section*.16}} +\@writefile{toc}{\contentsline {section}{Chapter 6: Energy and energy transfer}{3}{section*.17}} +\@writefile{toc}{\contentsline {subsection}{Work}{3}{section*.18}} +\newlabel{1D_work_int}{{28}{3}{Work\relax }{equation.28}{}} +\newlabel{dot}{{29}{3}{Work\relax }{equation.29}{}} +\newlabel{1D_work_const}{{30}{3}{Work\relax }{equation.30}{}} +\newlabel{Hooke}{{31}{3}{Work\relax }{equation.31}{}} +\newlabel{Espring}{{32}{3}{Work\relax }{equation.32}{}} +\@writefile{toc}{\contentsline {subsection}{Kinetic energy}{3}{section*.19}} +\newlabel{ke}{{33}{3}{Kinetic energy\relax }{equation.33}{}} +\@writefile{toc}{\contentsline {subsection}{Work-kinetic energy theorem}{3}{section*.20}} +\newlabel{wk_ke_thm}{{34}{3}{Work-kinetic energy theorem\relax }{equation.34}{}} +\@writefile{toc}{\contentsline {subsection}{Power}{3}{section*.21}} +\@writefile{toc}{\contentsline {section}{Chapter 7: Potential energy}{3}{section*.22}} +\@writefile{toc}{\contentsline {subsection}{Conservative forces}{3}{section*.23}} +\@writefile{toc}{\contentsline {subsection}{Mechanical energy}{4}{section*.24}} +\@writefile{toc}{\contentsline {subsection}{Equilibria}{4}{section*.25}} +\@writefile{toc}{\contentsline {section}{Chapter 8: Momentum and collisions}{4}{section*.26}} 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zTME2uB$>i9X3{)~>}!(zAil8Zm55bX_Dsy z8txGk?q3)lqWCDVeu&OjkkJ(#MTpi#7XyT0>a;#!s>q!xTn;98E{wa^P}NfJ!-j*^ zf>W1Y4&|FA;p5Gx{BkIiv~W39EwY`Xs;)^cr`duq5+*rYF7g}fF-dz#uoFo( z7bJKh*(zXPk@m1)Bay}ttlT7<0nXh>;|LBaNqr7S&Ln*SbCNVZ;3$zaZvc!mZ-9+N z(ib$`MCA;($Vom0fRSYd8!pjt1TGmeec-4ew}Cr&h4cyzU`cZ&CYY>); zCOBY3U|@EV^;DC*b|dqY)JL$qiNfHVk7N&lCz7=HqtUd`8MP37V4jlZIvP!rTo*P1 zB$>jtQB+p23X=6e6DA4Kb}29zL|~A|oa6(5>~ADLhzd;iBt9r$h>7CU!cQ=T`K5sK zfmDucQ^PiZWVfqmk^O{Fj with 68 dvips named colors) +%\usepackage{showkeys} % print labels in margins +\usepackage{amsmath} % environments for multiline displayed equations, and other enhancements +\usepackage{amsthm} % proof environment and extensions for the \newtheorem command +\usepackage[pdftex, + letterpaper, % sets pdf paper size to 'letter' +% backref, % adds `backlink' from bibliography to text as list of section numbers +% pdf information options + pdftitle=Equations, + pdfauthor=Trevor King, + pdfsubject=Physics 101, % pdf display options + pdfstartview=FitH % fit page to window (Fit, page; FitH, width; FitV, height) + ]{hyperref} +\usepackage[pdftex]{graphicx} % to include images +\usepackage[verbose]{wrapfig} % so text can wrap around figs. +\usepackage{subfigure} % allows sub-figures +\usepackage{pslatex} % to use PostScript fonts + +% define theorem environments +%\newtheorem{thm}{Theorem} +%\newtheorem{cor}[thm]{Corollary} +%\newtheorem{lem}[thm]{Lemma} +%\newtheorem{problem}{Problem} + +%\newtheorem*{problem13}{Problem 13} + +%\theoremstyle{definition} +%\newtheorem*{solution}{} + +\newcommand{\U}[1]{\mbox{ #1}} % units shortcut +\newcommand{\E}[1]{\ensuremath{\cdot 10 ^{#1}}} % exponent shortcut +\newcommand{\dg}{\ensuremath{^{\circ}}} % degree symbol ^o +\newcommand{\vect}[1]{\ensuremath{\mathbf{#1}}} % make vectors bold + +\newcommand{\ihat}{\vect{\hat{i}}} +\newcommand{\jhat}{\vect{\hat{j}}} +\newcommand{\khat}{\vect{\hat{k}}} + +%\newcommand{\Part}[1]{\textcolor{Red}{(#1)}} +%\newcommand{\ans}[1]{\textcolor{Red}{#1}} + +% \numberwithin{equation}{section} % number equations by section +\setcounter{secnumdepth}{0} + +%-----------------------end preamble------------------------- + +\begin{document} + +\begin{center} +{\Large Useful Equations} \\ +\end{center} + +\input{introduction} +\input{chapter_1} +\input{chapter_2} +\input{chapter_3} +\input{chapter_4} +\input{chapter_5} +\input{chapter_6} +\input{chapter_7} +\input{chapter_8} +\input{chapter_10} +\input{quiz_1} +\input{quiz_2} +\input{quiz_3} + +\end{document} diff --git a/latex/notes/equations/Serway_and_Jewett_4_wking/makefile b/latex/notes/equations/Serway_and_Jewett_4_wking/makefile new file mode 100644 index 0000000..44b6ff1 --- /dev/null +++ b/latex/notes/equations/Serway_and_Jewett_4_wking/makefile @@ -0,0 +1,15 @@ +view : pdf + xpdf main.pdf + +pdf : main.pdf + +clean : + rm *[.]pdf *[.]aux *[.]log *[.]out + +semi-clean : + rm *[.]aux *[.]log *[.]out + +main.pdf : *.tex + pdflatex main.tex + pdflatex main.tex + diff --git a/latex/notes/equations/Serway_and_Jewett_4_wking/quiz_1.tex b/latex/notes/equations/Serway_and_Jewett_4_wking/quiz_1.tex new file mode 100644 index 0000000..46a803a --- /dev/null +++ b/latex/notes/equations/Serway_and_Jewett_4_wking/quiz_1.tex @@ -0,0 +1,8 @@ +\section{Quiz 1: Vector addition and projectile motion} + +\begin{align} + v_2^2 &= v_1^2 + 2 a (x_2 - x_1) & \text{(From \ref{v_sqrt})} \\ + v_2 &= v_1 + at & \text{(From \ref{v_of_t})} \\ + x_2 &= x_1 + v_1 t + \frac{1}{2} a t^2 & \text{(From \ref{x_of_t})} +\end{align} + diff --git a/latex/notes/equations/Serway_and_Jewett_4_wking/quiz_2.tex b/latex/notes/equations/Serway_and_Jewett_4_wking/quiz_2.tex new file mode 100644 index 0000000..0a3fa8a --- /dev/null +++ b/latex/notes/equations/Serway_and_Jewett_4_wking/quiz_2.tex @@ -0,0 +1,13 @@ +\section{Quiz 2: Projectile motion, Newton's laws, and reference frames} + +\begin{align} + y &= y_0 + v_0 (\sin\theta) t - \frac{1}{2}g t^2 & \text{(From \ref{proj_y})}\\ + x &= v_0 (\cos\theta) t & \text{(From \ref{proj_x})} \\ + v_y &= v_0 (\sin\theta) - g t & \text{(From \ref{proj_vy})} \\ + v_x^2 &= v_{x0}^2 + 2 a_x \Delta x & \text{(From \ref{v_sqrt})} \\ + \sum F_{ext} &= \frac{\mbox{d}p}{\mbox{d}t} & \text{(From Newton's 2nd law and \ref{p_def})} \\ + \sum F_{ext} &= m a & \text{(Newton's 2nd law)} \\ + R &= (v_0^2 \sin 2\theta)/g & \text{(Range eqn 3.16 from book page 76)} \\ + \vect{V}_{12} &= \vect{V}_{1G} + \vect{V}_{G2} & \text{(From \ref{ref_frm})} +\end{align} +Where they have also used sohcahtoa to get $v_{xi} = v_0\cos\theta$ and $v_{yi} = v_0\sin\theta$. diff --git a/latex/notes/equations/Serway_and_Jewett_4_wking/quiz_3.tex b/latex/notes/equations/Serway_and_Jewett_4_wking/quiz_3.tex new file mode 100644 index 0000000..aff6a7c --- /dev/null +++ b/latex/notes/equations/Serway_and_Jewett_4_wking/quiz_3.tex @@ -0,0 +1,11 @@ +\section{Quiz 3: Friction, momentum, work, kinetic energy, and springs} + +\begin{align} + \Delta W &= \Delta KE & \text{(From \ref{wk_ke_thm})} \\ + KE &= 1/2 m v^2 & \text{(From \ref{ke})} \\ + \Delta W_{ext} &= 1/2 k x^2 & \text{(From \ref{Espring} with $x_f = 0$)} \\ + F &= -kx & \text{(From \ref{Hooke})} \\ + \Delta W_{12} &= \int_1^2 F.dx & \text{(From \ref{1D_work_int})} \\ + \vect{A}.\vect{B} &= AB\cos\theta & \text{(From \ref{dot})} \\ + \Delta W_{12} &= \vect{F}.\vect{X} & \text{(From \ref{1D_work_const})} +\end{align} diff --git a/latex/notes/notation/Giancoli_6/Makefile b/latex/notes/notation/Giancoli_6/Makefile new file mode 100644 index 0000000..6be3b9b --- /dev/null +++ b/latex/notes/notation/Giancoli_6/Makefile @@ -0,0 +1,20 @@ +# each MetaPost graphic has it's own file, +# so the basic definitions should get bundled out into an external file +# (encourages reuse anyway, so it's good for you :p) + +all : main.pdf + +view : all + xpdf main.pdf & + +%.pdf : %.tex *.tex + pdflatex $< + ../../tex/make_mp.sh + pdflatex $< + +semi-clean : + rm -f *.1 *.log *.mp *.mpx *.aux *.out *.cut + rm -f mp*.tex # remove metapost-generated tex + +clean : semi-clean + rm -f *.pdf diff --git a/latex/notes/notation/Giancoli_6/chapter_19.tex b/latex/notes/notation/Giancoli_6/chapter_19.tex new file mode 100644 index 0000000..ca9a95e --- /dev/null +++ b/latex/notes/notation/Giancoli_6/chapter_19.tex @@ -0,0 +1,14 @@ +\section{Chapter 19: Electric Forces and Electric Fields} + +\begin{tabular}{l l l l} + Symbol & Name & SI units & other unit names \\ + \hline + \Tstrut \vect{F} & Force & kg m/s$^2$ & N (newtons) \\ + q & Charge & C (coulombs) & \\ + \vect{E} & Electric field & kg m/s$^2$ C & N/C = V/m \\ + $\Phi_E$ & Electric flux & kg m$^3$/s$^2$ C & N m$^2$/C \\ + \vect{A} & Area vector & m$^2$ & \\ + $\rho$ (rho) & Charge per unit volume & C/m$^3$ & \\ + $\sigma$ (sigma) & Charge per unit area & C/m$^2$ & \\ + $\lambda$ (lambda) & Charge per unit length & C/m & \\ +\end{tabular} diff --git a/latex/notes/notation/Giancoli_6/chapter_20.tex b/latex/notes/notation/Giancoli_6/chapter_20.tex new file mode 100644 index 0000000..e294bc5 --- /dev/null +++ b/latex/notes/notation/Giancoli_6/chapter_20.tex @@ -0,0 +1,10 @@ +\section{Chapter 20: Electric Potential and Capacitance} + +\begin{tabular}{l l l l} + Symbol & Name & SI units & other unit names \\ + \hline + \Tstrut U & Potential energy & kg m$^2$/s$^2$ & J (joules) \\ + V & Electric potential & kg m$^2$/s$^2$ C & J/C = V (volts) \\ + C & Capacitance & C$^2$ s$^2$ / kg m$^2$ & C/V = F (farads) \\ + $\kappa$ (kappa) & Dielectric constant & Unitless & \\ +\end{tabular} diff --git a/latex/notes/notation/Giancoli_6/instl b/latex/notes/notation/Giancoli_6/instl new file mode 100755 index 0000000..59c3920 --- /dev/null +++ b/latex/notes/notation/Giancoli_6/instl @@ -0,0 +1 @@ +scp main.pdf wking@einstein.physics.drexel.edu:./public_html/course/notes/symbols.pdf diff --git a/latex/notes/notation/Giancoli_6/main.tex b/latex/notes/notation/Giancoli_6/main.tex new file mode 100644 index 0000000..18b4eab --- /dev/null +++ b/latex/notes/notation/Giancoli_6/main.tex @@ -0,0 +1,23 @@ +\documentclass[letterpaper, 10pt]{article} + +\usepackage[author={Trevor King}, + coursetitle={Physics 102}, + classtitle={Notation}, + subheading={Common symbols}]{../../tex/problempack} + +\begin{document} + +\maketitle + +\section{DISCLAIMER} +This is not an exhaustive list, and people aren't always consistent in their notation + (eg. \emph{$\Phi_E$ is electric flux} may not apply to the situation you're dealing with.). +The idea was to give you something to jog your memory. +The SI units I list \emph{are} accurate for the named quantity + (eg. electric field in N/C), but again, people don't always use standard SI units. +Use these at your own risk. + +\input{chapter_19} +\input{chapter_20} + +\end{document} diff --git a/latex/notes/style/Makefile b/latex/notes/style/Makefile new file mode 100644 index 0000000..a4b5451 --- /dev/null +++ b/latex/notes/style/Makefile @@ -0,0 +1,19 @@ +# each MetaPost graphic has it's own file, +# so the basic definitions should get bundled out into an external file +# (encourages reuse anyway, so it's good for you :p) + +all : style.pdf + +view : all + xpdf style.pdf & + +%.pdf : %.asy + asy style.asy + + +semi-clean : + rm -f *.1 *.log *.mp *.mpx *.aux *.out *.cut style_* + rm -f mp*.tex # remove metapost-generated tex + +clean : semi-clean + rm -f *.pdf diff --git a/latex/notes/style/drawing.asy b/latex/notes/style/drawing.asy new file mode 100644 index 0000000..47ef2b6 --- /dev/null +++ b/latex/notes/style/drawing.asy @@ -0,0 +1,29 @@ +import three; + +real u = 1cm; +currentprojection=perspective(50u,-20u,0); +path3 cir = unitcircle3; +path3 cyl[] = cir ^^ (shift((0,0,1))*cir) ^^ (0,1,0)--(0,1,1) ^^ (0,-1,0)--(0,-1,1); +currentpen = blue; +for (int i; i < cyl.length; i+=1) + draw(yscale3(5)*rotate(-90,(1,0,0))*scale3(u)*cyl[i]); +currentpen = red; +for (int i; i < cyl.length; i+=1) + draw(yscale3(5)*rotate(90,(1,0,0))*scale3(u)*cyl[i]); +currentpen = black; + +label("$\rho_a$", (0,2.5u,0)); +label("$\rho_c$", (0,-2.5u,0)); +draw((0,-1u,0)--(0,1u,0), Arrow); +label("$I$", (0, 1u,0), E); + +label("$A$", (0, -5.5u, 0)); +draw((0,-5u,-u)--(0,-5u,u)); +label("$d$", (0, -4.9u /*hack*/, 0), W); + +draw((0,-5u,-1.1u)--(0,0,-1.1u), Arrows); +label("$l$, $R_c$, $V_c$", (0, -2.5u,-1.1u), S); +draw((0,5u,-1.1u)--(0,0,-1.1u), Arrows); +label("$l$, $R_a$, $V_a$", (0, 2.5u,-1.1u), S); +draw((0,-5u,1.1u)--(0,5u,1.1u), Arrows); +label("$R_T$, $V_T$", (0, 0,1.1u), N); diff --git a/latex/notes/style/instl b/latex/notes/style/instl new file mode 100755 index 0000000..642df9d --- /dev/null +++ b/latex/notes/style/instl @@ -0,0 +1,3 @@ +PDIR=`pwd` +DIR=`basename "$PDIR"` +scp $DIR.pdf wking@einstein.physics.drexel.edu:./public_html/course/notes/${DIR}.pdf diff --git a/latex/notes/style/style.asy b/latex/notes/style/style.asy new file mode 100644 index 0000000..eef9b88 --- /dev/null +++ b/latex/notes/style/style.asy @@ -0,0 +1,270 @@ +// Example of a clearly worked problem and reasoning behind it. +// Command-line options to enable stepping and/or reverse video: +// asy [-u stepping=true] [-u reverse=true] slidedemo + +orientation=Landscape; + +import slide; +settings.tex = "pdflatex"; +usersetting(); + +// Commands to generate optional bibtex citations: +// asy slidedemo +// bibtex slidedemo_ +// asy slidedemo +// +bibliographystyle("alpha"); +//texpreamble("\usepackage{problempack}"); // optionclash +texpreamble("\usepackage{amsmath}"); +texpreamble("\newcommand{\U}[1]{\text{ #1}} % units shortcut"); +texpreamble("\newcommand{\E}[1]{\ensuremath{\cdot 10 ^{#1}}} % exponent shortcut"); +texpreamble(bulletcolor("black")); + +string eqnstring(string s, real width=100bp) +{ + return minipage("\begin{align*}"+s+"\end{align*}",width); +} +// Override slide.equations to use the prettier align* env from amsmath +void equations(string s, pen p=itempen) +{ + vbox("\begin{align*}"+s+"\end{align*}",p); +} +// Wrap textblock from the textpos package + +// Generated needed files if they don't already exist. +asy(nativeformat(),"drawing"); + +// Optional background color or header: +// import x11colors; +// fill(background,box((-1,-1),(1,1)),Azure); +// label(background,"Header",(0,startposition.y)); + +pair titlealign = W; +titlepagepen = colorless(titlepagepen); +authorpen = colorless(authorpen); +institutionpen = colorless(authorpen); +pen questionPen = fontsize(18pt); +pen genEqnPen = 0.5*red + fontsize(18pt); +pen partEqnPen = 0.5*blue + fontsize(18pt); +pen solnEqnPen = 0.5*green + fontsize(18pt); +pair solutionposition = startposition-(0,0.5); // paper (-1,-1) to (1,1) + +titlepage(title="Solving physics problems:\\ a detailed solution style", + author="W. Trevor King", + institution="Drexel University", + date="\today", + url="http://www.physics.drexel.edu/$\sim$wking"); + +title("Presenting solutions in an organized manner"); +remark( +"Physics isn't just about getting the ``right answer,'' but also about +demonstrating to others why your answer is right and how you came to +that conclusion. This solution framework makes your argument clearer, +which will help you as you develop the argument, and others when you +try and teach/convince them."); +remark( +"This layout is along the lines of the homework solution format from my +high-school physics class, and I like it a lot. However, if you have +another format that works for you, go ahead and use that. Remember, +the purpose is to make the solution look obvious, with very clear +reasoning, both for you and any potential readers."); + +outline("Procedure"); +item("Copy down the assigned problem"); +item("Draw a picture of the problem, labeling useful quantities"); +item("Make a list of relevant formulas"); +item("Apply the formulas to the particular cases in the problem"); +item("Solve for the unknown quantities"); +item("Are your answers reasonable?"); + +title("Copy down the assigned problem"); +remark("This step ensures you read the problem carefully"); +currentposition=solutionposition; +remark( +" +{ +17.24) \\ +A $10.0\U{m}$ length of wire consists of $5.0\U{m}$ of copper followed +by $5.0\U{m}$ of aluminum, both of diameter $1.0\U{mm}$. A voltage +difference of $85\U{mV}$ is placed across the composite wire. +(a) What is the total resistance (sum) of the two wires? +(b) What is the currnent through the wire? +(c) What are the voltages across the aluminum part and across the copper part? +} +", questionPen); + +title("Draw a picture of the problem"); +remark( +"This step doublechecks your understanding of the problem, helps you +translate it into your own terms, and makes it clear what quantities +your various symbols refer to."); + +currentposition=solutionposition; +remark( +" +{\ +17.24) \\ +A $10.0\U{m}$ length of wire consists of $5.0\U{m}$ of copper followed +by $5.0\U{m}$ of aluminum, both of diameter $1.0\U{mm}$. A voltage +difference of $85\U{mV}$ is placed across the composite wire. +(a) What is the total resistance (sum) of the two wires? +(b) What is the currnent through the wire? +(c) What are the voltages across the aluminum part and across the copper part? +} +", questionPen); +//new* +figure("drawing"); + +title("Make a list of relevant formulas"); +remark( +"This is the key step, digging deep into your physics knowledge to +dredge up any appropriate formulas. :p"); + +currentposition=solutionposition; +remark( +" +{\ +17.24) \\ +A $10.0\U{m}$ length of wire consists of $5.0\U{m}$ of copper followed +by $5.0\U{m}$ of aluminum, both of diameter $1.0\U{mm}$. A voltage +difference of $85\U{mV}$ is placed across the composite wire. +(a) What is the total resistance (sum) of the two wires? +(b) What is the currnent through the wire? +(c) What are the voltages across the aluminum part and across the copper part? +} +", questionPen); +figure("drawing"); +//new* +pair genEqnPos = currentposition+(-.1,.2); +label(eqnstring(" + R &= \rho \frac{L}{A} \\ + V &= IR \\ + R_{1,2} &= R_1 + R_2 \\ + A &= \pi r^2 = \pi (d/2)^2 +", minipagewidth/3), genEqnPos, E, genEqnPen); + +title("Apply formulas to patricular cases"); +remark( +"Here we simply apply the general equations to the problem at hand, and +look up any constants we don't know."); + +currentposition=solutionposition; +remark( +" +{\ +17.24) \\ +A $10.0\U{m}$ length of wire consists of $5.0\U{m}$ of copper followed +by $5.0\U{m}$ of aluminum, both of diameter $1.0\U{mm}$. A voltage +difference of $85\U{mV}$ is placed across the composite wire. +(a) What is the total resistance (sum) of the two wires? +(b) What is the currnent through the wire? +(c) What are the voltages across the aluminum part and across the copper part? +} +", questionPen); +figure("drawing"); +label(eqnstring(" + R &= \rho \frac{L}{A} \\ + V &= IR \\ + R_{1,2} &= R_1 + R_2 \\ + A &= \pi r^2 = \pi (d/2)^2 +", minipagewidth/3), genEqnPos, E, genEqnPen); +//new* +pair partEqnPos = genEqnPos+(0,-.4); +label(eqnstring(" + R_T &= R_c + R_a & R_c &= \rho_c \frac{l}{A} & R_a &= \rho_a \frac{l}{A} \\ + V_T &= I R_T & V_c &= I R_c & V_a &= I R_a \\ + \rho_c &= 1.68\E{-8}\U{$\Omega$m} \\ + \rho_a &= 2.65\E{-8}\U{$\Omega$m} +", minipagewidth/1.8), partEqnPos, E, partEqnPen); + +title("Solve for the unknown quantities"); +remark("From this point out it's just math."); +currentposition=solutionposition; +remark( +" +{\ +17.24) \\ +A $10.0\U{m}$ length of wire consists of $5.0\U{m}$ of copper followed +by $5.0\U{m}$ of aluminum, both of diameter $1.0\U{mm}$. A voltage +difference of $85\U{mV}$ is placed across the composite wire. +(a) What is the total resistance (sum) of the two wires? +(b) What is the currnent through the wire? +(c) What are the voltages across the aluminum part and across the copper part? +} +", questionPen); +figure("drawing"); +label(eqnstring(" + R &= \rho \frac{L}{A} \\ + V &= IR \\ + R_{1,2} &= R_1 + R_2 \\ + A &= \pi r^2 = \pi (d/2)^2 +", minipagewidth/3), genEqnPos, E, genEqnPen); +label(eqnstring(" + R_T &= R_c + R_a & R_c &= \rho_c \frac{l}{A} & R_a &= \rho_a \frac{l}{A} \\ + V_T &= I R_T & V_c &= I R_c & V_a &= I R_a \\ + \rho_c &= 1.68\E{-8}\U{$\Omega$m} \\ + \rho_a &= 2.65\E{-8}\U{$\Omega$m} +", minipagewidth/1.8), partEqnPos, E, partEqnPen); +//new* +pair solnEqnPos = genEqnPos+(0.9,-.1); +label(eqnstring(" + R_c &= \rho_c \frac{l}{A} = \rho_c \frac{4l}{\pi d^2} = 0.10654\U{$\Omega$}\\ + R_a &= \rho_a \frac{l}{A} = \rho_a \frac{4l}{\pi d^2} = 0.16870\U{$\Omega$}\\ + R_T &= R_c + R_a = 0.28\U{$\Omega$} \\ + V_T &= IR_T \qquad I = V_T/R_T = 0.31\U{A} \\ + V_c &= IR_c = 33\U{mV} \qquad V_a = IR_a = 52\U{mV} +", minipagewidth/1.8), solnEqnPos, SE, solnEqnPen); + +title("Are your answers reasonable?"); +remark( +"Yes. The wire is made out of copper and aluminum, both good +conductors, and has a reasonable length and diameter, so we expect low +resistance. We also expect the more resistive aluminum to have a +higher resistance and a greater voltage drop."); + +currentposition=solutionposition; +remark( +" +{\ +17.24) \\ +A $10.0\U{m}$ length of wire consists of $5.0\U{m}$ of copper followed +by $5.0\U{m}$ of aluminum, both of diameter $1.0\U{mm}$. A voltage +difference of $85\U{mV}$ is placed across the composite wire. +(a) What is the total resistance (sum) of the two wires? +(b) What is the currnent through the wire? +(c) What are the voltages across the aluminum part and across the copper part? +} +", questionPen); +figure("drawing"); +label(eqnstring(" + R &= \rho \frac{L}{A} \\ + V &= IR \\ + R_{1,2} &= R_1 + R_2 \\ + A &= \pi r^2 = \pi (d/2)^2 +", minipagewidth/3), genEqnPos, E, genEqnPen); +label(eqnstring(" + R_T &= R_c + R_a & R_c &= \rho_c \frac{l}{A} & R_a &= \rho_a \frac{l}{A} \\ + V_T &= I R_T & V_c &= I R_c & V_a &= I R_a \\ + \rho_c &= 1.68\E{-8}\U{$\Omega$m} \\ + \rho_a &= 2.65\E{-8}\U{$\Omega$m} +", minipagewidth/1.8), partEqnPos, E, partEqnPen); +pair solnEqnPos = genEqnPos+(0.9,-.1); +label(eqnstring(" + R_c &= \rho_c \frac{l}{A} = \rho_c \frac{4l}{\pi d^2} = 0.10654\U{$\Omega$}\\ + R_a &= \rho_a \frac{l}{A} = \rho_a \frac{4l}{\pi d^2} = 0.16870\U{$\Omega$}\\ + R_T &= R_c + R_a = 0.28\U{$\Omega$} \\ + V_T &= IR_T \qquad I = V_T/R_T = 0.31\U{A} \\ + V_c &= IR_c = 33\U{mV} \qquad V_a = IR_a = 52\U{mV} +", minipagewidth/1.8), solnEqnPos, SE, solnEqnPen); + +title("Some thoughts"); +remark( +"Obviously this formal approach is not neccessary for simple problems +that you can almost do in your head. However, for more complicated +problems, the extra work of drawing a labeled figure and explicitly +writing out the general equations you use will help you solve the +problem faster by making it very easy to remember what each symbol +means, see where you have information that you haven't used in your +solution yet, and get a good overview of your line of reasoning."); +label("This document was produced using Asymptote. + {\tt http://asymptote.sourceforge.net/}", (-0.8,-0.8), NE, fontsize(12pt)); diff --git a/latex/notes/topics/kirchoff/Makefile b/latex/notes/topics/kirchoff/Makefile new file mode 100644 index 0000000..54249eb --- /dev/null +++ b/latex/notes/topics/kirchoff/Makefile @@ -0,0 +1,16 @@ +all : beamer.pdf + +view : beamer.pdf + xpdf -z 300 beamer.pdf & + +clean : semi-clean + rm -f *[.]pdf + +semi-clean : + rm -f *[.]aux *[.]log *[.]out *[.]toc *[.]nav *[.]snm *[.]vrb \ + *_[.]tex *[.]asy *_[.]pre *~ || echo "clean" + +beamer.pdf : *.tex + pdflatex beamer + asy beamer + pdflatex beamer diff --git a/latex/notes/topics/kirchoff/RCcircuits.tex b/latex/notes/topics/kirchoff/RCcircuits.tex new file mode 100644 index 0000000..ef7222b --- /dev/null +++ b/latex/notes/topics/kirchoff/RCcircuits.tex @@ -0,0 +1,34 @@ +\section{RC circuits} + +\begin{frame}[fragile=singleslide] + \frametitle{RC circuits} + + \begin{center} +\begin{asy} + import Circ; + real u = 1cm; + TwoTerminal B = source((0,0), DC, 0, "$\mathcal{E}$", ""); + TwoTerminal R = resistor(B.end+(u,.5u), normal, 90, "", "$R$"); + TwoTerminal C = capacitor(B.beg+(-u,.5u), normal, 90, "$C$", ""); + TwoTerminal I = current((B.end.x, R.end.y+.5u), 180, "", "$I$"); + wire(B.beg, C.beg, rlsq); + wire(B.end, R.beg, rlsq); + wire(I.end, C.end, rlsq); + wire(I.beg, R.end, rlsq); +\end{asy} + \end{center} + + From Kirchoff's loop rule + $\sum_{loop} V_i = \mathcal{E} - IR - Q/C = 0$. + + From the definition of current + $I = \deriv{Q}{t}$. + + So + \begin{align*} + \mathcal{E} - Q/C &= R\frac{Q}{t} \\ + \text{d}t &= \p({\mathcal{E}-Q/C})^{-1} R\text{d}Q \\ + Q &= C\mathcal{E}\p({1-e^{-t/RC}}) & + V_C &= \mathcal{E}\p({1-e^{-t/RC}}) + \end{align*} +\end{frame} diff --git a/latex/notes/topics/kirchoff/beamer.tex b/latex/notes/topics/kirchoff/beamer.tex new file mode 100644 index 0000000..f26afc7 --- /dev/null +++ b/latex/notes/topics/kirchoff/beamer.tex @@ -0,0 +1,62 @@ +%\documentclass[draft]{beamer} +\documentclass{beamer} +%\includeonlyframes{current} + +%\usetheme[secheader]{Boadilla} +\usetheme{Pittsburgh} +\setcounter{secnumdepth}{0} + +\usepackage[inline]{asymptote} % Asymptote graphics environments +%\usepackage{asymptote} % Asymptote graphics environments + +\title{Chapter 19: DC Circuits} +\author{Trevor King} +\institute{Drexel University} +\date{\today} + +\newcommand{\D}{\ensuremath{\Delta}} % Delta symbol +\newcommand{\dg}{\ensuremath{^{\circ}}} % degree symbol ^o +\newcommand{\E}[1]{\cdot10^{#1}} % power of 10 +\newcommand{\U}[1]{\ensuremath{\text{ #1}}} % units shortcut +\newcommand{\deriv}[2]{\ensuremath{\frac{\text{d} #1}{\text{d} #2}}} +\newcommand{\p}[3]{\left#1 #2 \right#3} % parenthesis, e.g. \p({complicated expr}) + +\newcommand{\highlight}[1]{\textcolor{red}{#1}} + +% struts for table spacing +\newcommand{\Tstrut}{\rule{0pt}{2.6ex}} +\newcommand{\Bstrut}{\rule[-1.2ex]{0pt}{0pt}} + +\newenvironment{packed_enum} { + \begin{enumerate} + \setlength{\itemsep}{1pt} + \setlength{\parskip}{0pt} + \setlength{\parsep}{0pt} + } {\end{enumerate}} + +\newenvironment{packed_item} { + \begin{itemize} + \setlength{\itemsep}{1pt} + \setlength{\parskip}{0pt} + \setlength{\parsep}{0pt} + } {\end{itemize}} + +\setcounter{tocdepth}{2} +%\setlength{\parskip}{0pt} + +%-----------------------end preamble------------------------- + +\begin{document} + +\begin{frame} + \titlepage +\end{frame} + +\include{overview} +\include{kirchoff} +\include{seriesParallel} +\include{resistors} +\include{capacitors} +\include{RCcircuits} + +\end{document} diff --git a/latex/notes/topics/kirchoff/capacitors.tex b/latex/notes/topics/kirchoff/capacitors.tex new file mode 100644 index 0000000..5c10d99 --- /dev/null +++ b/latex/notes/topics/kirchoff/capacitors.tex @@ -0,0 +1,68 @@ +\section{Capacitors} + +\begin{frame}[fragile=singleslide] + \frametitle{Capacitors in series} + + \begin{center} +\begin{asy} + import Circ; + TwoTerminal Ca = capacitor((0,0), normal, 0, "$C_a$, $V_a$", ""); + TwoTerminal Cb = capacitor(Ca.end+1cm, normal, 0, "$C_b$, $V_b$", ""); + wire(Ca.end, Cb.beg, nsq); + dot("a", Ca.beg, W); + dot("b", Cb.end, E); + pair mid = (Ca.end+Cb.beg)/2; + real u = 5mm; + draw((Ca.beg-(0,u))--(Cb.end-(0,u)), Arrows); + label("$V_T$, $C_T$", (mid-(0,u)), S); +\end{asy} + \end{center} + By Kirchoff's junction rule, $Q_a = Q_b = Q_T$ (because there are no intervening junctions). + + \vskip -\baselineskip + \begin{align*} + V_T &= V_a + V_b = Q_a/C_a + Q_b/C_b = Q_T(1/C_a + 1/C_b) \\ + V_T &= Q_T/C_T + \end{align*} + So the effective capacitance is give by + \begin{equation*} + C_T = \p({\frac{1}{C_a} + \frac{1}{C_b}})^{-1} + \end{equation*} +\end{frame} + +\begin{frame}[fragile=singleslide] + \frametitle{Capacitors in parallel} + + \begin{center} +\begin{asy} + import Circ; + real u = 20pt; + TwoTerminal Ca = capacitor((0,+u), normal, 0, "$C_a$, $V_a$", ""); + TwoTerminal Cb = capacitor((0,-u), normal, 0, "", "$C_b$, $V_b$"); + pair aj = (Ca.beg.x-12pt, 0); + pair a = aj-(12pt,0); + pair bj = (Cb.end.x+12pt, 0); + pair b = bj+(12pt,0); + wire(a, aj, nsq); + wire(aj, Ca.beg, udsq); + wire(aj, Cb.beg, udsq); + wire(b, bj, nsq); + wire(bj, Ca.end, udsq); + wire(bj, Cb.end, udsq); + dot("a", a, W); + dot("b", b, E); + draw(aj--bj, Arrows); + label("$V_T$, $C_T$", (aj+bj)/2, N); +\end{asy} + \end{center} + By Kirchoff's loop rule, $V_a - V_b = 0$ so $V_a = V_b = V_T$. + + By Kirchoff's junction rule, $Q_T = Q_a + Q_b$ + + \vskip -\baselineskip + \begin{align*} + Q_T &= Q_a + Q_b = C_a V_T + C_b V_T = (C_a + C_b) V_T \\ + Q_T &= C_T V_T + \end{align*} + So the effective capacitance is give by $C_T = C_a+C_b$. +\end{frame} diff --git a/latex/notes/topics/kirchoff/instl b/latex/notes/topics/kirchoff/instl new file mode 100755 index 0000000..9d6e2f8 --- /dev/null +++ b/latex/notes/topics/kirchoff/instl @@ -0,0 +1,3 @@ +PDIR=`pwd` +DIR=`basename "$PDIR"` +scp beamer.pdf wking@einstein.physics.drexel.edu:./public_html/course/notes/${DIR}_notes.pdf diff --git a/latex/notes/topics/kirchoff/kirchoff.tex b/latex/notes/topics/kirchoff/kirchoff.tex new file mode 100644 index 0000000..2e57161 --- /dev/null +++ b/latex/notes/topics/kirchoff/kirchoff.tex @@ -0,0 +1,77 @@ +\section{Kirchoff's laws} + +\begin{frame}[fragile=singleslide] + \frametitle{Kirchoff's loop rule} + + \begin{center} +\begin{asy} + import Circ; + real u = 2cm, h = 1.5cm; + TwoTerminal B = source((-u,0), DC, 0, "V", "9\U{V}"); + TwoTerminal Ra = resistor((-u,h), normal, 0, "$20\U{$\Omega$}$", ""); + TwoTerminal Rb = resistor((u,h), normal, 0, "$10\U{$\Omega$}$", ""); + TwoTerminal I = current((u,0), 0, "I", ""); + wireU(Ra.beg, B.beg, -12pt, rlsq); + wire(Ra.end, Rb.beg, nsq); + wireU(Rb.end, B.end, 12pt, rlsq); + dot("a", B.end, NE); + dot("b", Rb.end, NE); + dot("c", Rb.beg, NW); + dot("d", Ra.end, NE); + dot("e", Ra.beg, NW); + dot("f", B.beg, NW); +\end{asy} +\vspace{\baselineskip} +\begin{asy} + import graph; + size(5cm,3cm,IgnoreAspect); + string[] lb = {"f", "a", "b", "c", "d", "e", "f"}; + real[] x = { 0, 1, 2, 3, 4, 5, 6}; + real[] y = { 0, 9, 9, 6, 6, 0, 0}; + real[] ys = {0,3,6,9}; + draw(graph(x,y), red, MarkFill[0]); + xaxis(Bottom, LeftTicks(new string(real x) { + return lb[round(x % lb.length)];})); + yaxis("potential $V$",Left,RightTicks(Ticks=ys)); +\end{asy} + \end{center} +\vspace{-\baselineskip} + \begin{align*} + \sum_{\text{loop}} \D V_i = 0 && \text{Conserving energy} + \end{align*} +\end{frame} + +\begin{frame}[fragile=singleslide] + \frametitle{Kirchoff's junction rule} + + \begin{center} +\begin{asy} + import Circ; + real u = 2cm, h = 1.5cm; + TwoTerminal B = source((-u,0), DC, 0, "V", "9\U{V}"); + TwoTerminal Rll = resistor((-u,h), normal, 0, "", ""); + TwoTerminal Rlr = resistor((0,h), normal, 0, "", ""); + TwoTerminal Rul = resistor((-u,2h), normal, 0, "", ""); + TwoTerminal Rur = resistor((0,2h), normal, 0, "", ""); + TwoTerminal Ib = current((u,0), 0, "$I_1$", ""); // battery + TwoTerminal Il = current((u,h), 0, "$I_2$", ""); // lower resistors + TwoTerminal Iu = current((u,2h), 0, "$I_3$", ""); // upper resistors + pair a = Il.end+(12pt, 0); + wireU(Rll.beg, B.beg, -12pt, rlsq); + wireU(Rul.beg, B.beg, -12pt, rlsq); + wire(B.end, Ib.beg, nsq); + wire(Rll.end, Rlr.beg, nsq); + wire(Rlr.end, Il.beg, nsq); + wire(Rul.end, Rur.beg, nsq); + wire(Rur.end, Iu.beg, nsq); + wire(Ib.end, a, rlsq); + wire(Il.end, a, nsq); + wire(Iu.end, a, rlsq); + dot("a", a, E); +\end{asy} + \end{center} +\vspace{-\baselineskip} + \begin{align*} + \sum_{\text{junction}} I_i = 0 && \text{Conserving charge, steady state} + \end{align*} +\end{frame} diff --git a/latex/notes/topics/kirchoff/overview.tex b/latex/notes/topics/kirchoff/overview.tex new file mode 100644 index 0000000..9f92258 --- /dev/null +++ b/latex/notes/topics/kirchoff/overview.tex @@ -0,0 +1,38 @@ +\section{Overview} + +\begin{frame}[fragile=singleslide] + \frametitle{DC circuits} + + \begin{packed_item} + \item \emph{Direct Current} (DC) circuits are important + \begin{packed_item} + \item often used in digital electronic devices and almost + anything that is battery powered. + \end{packed_item} + \item The physics of DC currents is straightforward + \begin{packed_item} + \item Kirchoff's loop rule + \item Kirchoff's junction rule + \item Understanding the individual components + \begin{packed_item} + \item Resistors + \item Capacitors + \item \ldots + \end{packed_item} + \end{packed_item} + \end{packed_item} + + \begin{center} +\begin{asy} + import Circ; + real u = 2cm; + TwoTerminal B = source((-u,0), DC, 0, "V", "5\U{V}"); + TwoTerminal R = resistor((-u,u), normal, 0, "R", "$10\U{$\Omega$}$"); + TwoTerminal C = capacitor((u,u), normal, 0, "C", "$15\U{$\mu$F}$"); + TwoTerminal I = current((u,0), 0, "I", ""); + wireU(R.beg, B.beg, -12pt, rlsq); + wire(R.end, C.beg, nsq); + wireU(C.end, B.end, 12pt, rlsq); +\end{asy} + \end{center} +\end{frame} diff --git a/latex/notes/topics/kirchoff/resistors.tex b/latex/notes/topics/kirchoff/resistors.tex new file mode 100644 index 0000000..00e97d0 --- /dev/null +++ b/latex/notes/topics/kirchoff/resistors.tex @@ -0,0 +1,69 @@ +\section{Resistors} + +\begin{frame}[fragile=singleslide] + \frametitle{Resistors in series} + + \begin{center} +\begin{asy} + import Circ; + TwoTerminal Ra = resistor((0,0), normal, 0, "$R_a$, $V_a$", ""); + TwoTerminal I = current(Ra.end, 0, "$I$", ""); + TwoTerminal Rb = resistor(I.end, normal, 0, "$R_b$, $V_b$", ""); + dot("a", Ra.beg, W); + dot("b", Rb.end, E); + real u = 5mm; + draw((Ra.beg-(0,u))--(Rb.end-(0,u)), Arrows); + label("$V_T$, $R_T$", (I.mid-(0,u)), S); +\end{asy} + \end{center} + By Kirchoff's junction rule, $I_a = I_b$ (because there are no intervening junctions). + + \vskip -\baselineskip + \begin{align*} + V_T &= V_a + V_b = IR_a + IR_b = I(R_a + R_b) \\ + V_T &= IR_T + \end{align*} + So the effective resistance is give by + \begin{equation*} + R_T = R_a + R_b + \end{equation*} +\end{frame} + +\begin{frame}[fragile=singleslide] + \frametitle{Resistors in parallel} + + \begin{center} +\begin{asy} + import Circ; + real u = 20pt; + TwoTerminal Ra = resistor((0,+u), normal, 0, "$R_a$, $V_a$", ""); + TwoTerminal Rb = resistor((0,-u), normal, 0, "", "$R_b$, $V_b$"); + pair aj = (Ra.beg.x, 0); + pair a = aj-(12pt,0); + pair bj = (Rb.end.x, 0); + pair b = bj+(12pt,0); + wire(a, aj, nsq); + wire(aj, Ra.beg, nsq); + wire(aj, Rb.beg, nsq); + wire(b, bj, nsq); + wire(bj, Ra.end, nsq); + wire(bj, Rb.end, nsq); + dot("a", a, W); + dot("b", b, E); + draw(aj--bj, Arrows); + label("$V_T$, $R_T$", (aj+bj)/2, N); +\end{asy} + \end{center} + By Kirchoff's loop rule, $V_a - V_b = 0$ so $V_a = V_b = V_T$. + + By Kirchoff's junction rule, $I_T = I_a + I_b$ + + \vskip -\baselineskip + \begin{align*} + V_T &= I_a R_a = I_b R_b \qquad \qquad \qquad + I_b = I_a \frac{R_a}{R_b} \\ + I_T &= I_a + I_b = I_a \p({1+R_a/R_b}) \\ + V_T &= I_a R_a = \frac{I_T}{1 + R_a/R_b} R_a = I_T \p({1/R_a+1/R_b})^{-1} = I_T R_T + \end{align*} + So the effective resistance is give by $R_T = \p({1/R_a+1/R_b})^{-1}$ +\end{frame} diff --git a/latex/notes/topics/kirchoff/seriesParallel.tex b/latex/notes/topics/kirchoff/seriesParallel.tex new file mode 100644 index 0000000..dd825f1 --- /dev/null +++ b/latex/notes/topics/kirchoff/seriesParallel.tex @@ -0,0 +1,59 @@ +\section{Elements in Series and in Parallel} + +\begin{frame}[fragile=singleslide] + \frametitle{Elements in series} + + \begin{center} +\begin{asy} + import Circ; + real u = 2cm, h = 1.5cm; + TwoTerminal B = source((-u,0), DC, 0, "V", "9\U{V}"); + TwoTerminal Ra = resistor((-u,1.5h), normal, 0, "R$_\text{a}$", ""); + TwoTerminal Rb = resistor(Ra.end, normal, 0, "R$_\text{b}$", ""); + TwoTerminal Rc = resistor((u,2h), normal, 0, "R$_\text{c}$", ""); + TwoTerminal Rd = resistor((u,h), normal, 0, "R$_\text{d}$", ""); + pair m = (Rc.beg.x - 12pt, 1.5h); // middle junction point + pair r = (Rd.end.x + 12pt, 1.5h); // right junction point + + wireU(Ra.beg, B.beg, -12pt, rlsq); + wireU(Rc.end, r, 12pt, rlsq); + wireU(Rd.end, r, 12pt, rlsq); + wireU(r, B.end, 12pt, rlsq); + wire(Rb.end, m, nsq); + wire(Rc.beg, m, rlsq); + wire(Rd.beg, m, rlsq); +\end{asy} + \end{center} + + Two elements are in series if you cannot make a loop crossing one of + them without also crossing the other. +\end{frame} + +\begin{frame}[fragile=singleslide] + \frametitle{Elements in parallel} + + \begin{center} +\begin{asy} + import Circ; + real u = 2cm, h = 1.5cm; + TwoTerminal B = source((-u,0), DC, 0, "V", "9\U{V}"); + TwoTerminal Ra = resistor((-u,1.5h), normal, 0, "R$_\text{a}$", ""); + TwoTerminal Rb = resistor(Ra.end, normal, 0, "R$_\text{b}$", ""); + TwoTerminal Rc = resistor((u,2h), normal, 0, "R$_\text{c}$", ""); + TwoTerminal Rd = resistor((u,h), normal, 0, "R$_\text{d}$", ""); + pair m = (Rc.beg.x - 12pt, 1.5h); // middle junction point + pair r = (Rd.end.x + 12pt, 1.5h); // right junction point + + wireU(Ra.beg, B.beg, -12pt, rlsq); + wireU(Rc.end, r, 12pt, rlsq); + wireU(Rd.end, r, 12pt, rlsq); + wireU(r, B.end, 12pt, rlsq); + wire(Rb.end, m, nsq); + wire(Rc.beg, m, rlsq); + wire(Rd.beg, m, rlsq); +\end{asy} + \end{center} + + Two elements are in parallel if you can make a loop crossing both of + them which crosses no other elements. +\end{frame} diff --git a/latex/notes/topics/linear_algebra/Makefile b/latex/notes/topics/linear_algebra/Makefile new file mode 100644 index 0000000..6be3b9b --- /dev/null +++ b/latex/notes/topics/linear_algebra/Makefile @@ -0,0 +1,20 @@ +# each MetaPost graphic has it's own file, +# so the basic definitions should get bundled out into an external file +# (encourages reuse anyway, so it's good for you :p) + +all : main.pdf + +view : all + xpdf main.pdf & + +%.pdf : %.tex *.tex + pdflatex $< + ../../tex/make_mp.sh + pdflatex $< + +semi-clean : + rm -f *.1 *.log *.mp *.mpx *.aux *.out *.cut + rm -f mp*.tex # remove metapost-generated tex + +clean : semi-clean + rm -f *.pdf diff --git a/latex/notes/topics/linear_algebra/instl b/latex/notes/topics/linear_algebra/instl new file mode 100755 index 0000000..b6cb458 --- /dev/null +++ b/latex/notes/topics/linear_algebra/instl @@ -0,0 +1,3 @@ +PDIR=`pwd` +DIR=`basename "$PDIR"` +scp main.pdf wking@einstein.physics.drexel.edu:./public_html/course/notes/${DIR}.pdf diff --git a/latex/notes/topics/linear_algebra/main.tex b/latex/notes/topics/linear_algebra/main.tex new file mode 100644 index 0000000..5f7b9f0 --- /dev/null +++ b/latex/notes/topics/linear_algebra/main.tex @@ -0,0 +1,353 @@ +\documentclass[letterpaper, 10pt]{article} + +\usepackage[author={Trevor King}, + coursetitle={Physics 102}, + classtitle={Linear algebra introduction}, + subheading={Solving systems of linear equations automatically}, + loose]{../../tex/problempack} + +%-----------------------end preamble------------------------- + +\begin{document} + +\maketitle + +\section{Writing matrix equations} + +All battery-resistor-current problems produce systems of \emph{linear equations}. +Linear equations are just a equations where the unknowns (usually the currents $I_i$) are never multiplied together. +For example from Recitation 6, Problem 35, we have +\begin{align} + 0 &= I_1 - I_2 - I_3 \\ + 0 &= \epsilon_2 - I_2 R_2 - I_1 R_1 \\ + 0 &= \epsilon_3 - I_3 R_3 - I_1 R_1 \;, +\end{align} +from Kirchhoff's laws. +We're given everything except the currents, and no terms have two currents multiplied by each-other in them. +We can rewrite these equations and adjust the spacing a bit to get +\begin{align} + 0 &= -I_1 \;\;+\;\; I_2 \;\;+\;\; I_3 \label{eq.K1} \\ + \epsilon_2 &= R_1 I_1 + R_2 I_2 + 0\cdot I_3 \\ + \epsilon_3 &= R_1 I_1 + 0\cdot R_2 + R_3 I_3 \label{eq.K3}\;. +\end{align} +We can see that each line has the same format +\begin{equation} + b = a_1 I_1 + a_2 I_2 + a_3 I_3 = \sum_{j=1}^3 a_j I_j \;, +\end{equation} +with constant $a_i$s and $b$s. + +We can rewrite the system of equations as a \emph{matrix} equation +\begin{equation} + \begin{pmatrix} + 0 \\ + \epsilon_2 \\ + \epsilon_3 + \end{pmatrix} + = + \begin{pmatrix} + -1 & 1 & 1 \\ + R_1 & R_2 & 0 \\ + R_1 & 0 & R_3 + \end{pmatrix} + \begin{pmatrix} + I_1 \\ + I_2 \\ + I_3 + \end{pmatrix} \label{eq.KM} +\end{equation} +Comparing this to Eqns.~\ref{eq.K1}--\ref{eq.K3}, we can see that all we've done is erased a few equals signs in the middle, drawn some big parenthesis, and dragged the unknown currents off by themselves, giving them their own parenthesis. +We've also rotated order we wrote the currents in +\begin{equation} + \clubsuit I_1 + \diamondsuit I_2 + \heartsuit I_3 + \longrightarrow + \begin{pmatrix} + I_1 \\ + I_2 \\ + I_3 + \end{pmatrix} \;. +\end{equation} +Things look fairly different at first, so go back and compare Eqn.~\ref{eq.KM} with Eqns.~\ref{eq.K1}--\ref{eq.K3} until you get comfortable with the changes. + +\section{Solving matrix equations} + +All the things we could do when we had regular equations, we can still do with the equations written in matrix form. We'll go through and solve these side by side so you can see a solution in action and become more familiar with the matrix notation. +The only difference in solving strategy is that we keep all the symbols on the same side they started on, and add or subtract equations instead of plugging in. +\begin{align} + \begin{matrix} + 0 &= &-I_1 &+ &I_2 &+ &I_3 \\ + \epsilon_2 &= &R_1 I_1 &+ &R_2 I_2 &+ &0 \\ + \epsilon_3 &= &R_1 I_1 &+ &0 &+ &R_3 I_3 + \end{matrix} + && + \begin{pmatrix} + 0 \\ + \epsilon_2 \\ + \epsilon_3 + \end{pmatrix} + = + \begin{pmatrix} + -1 & 1 & 1 \\ + R_1 & R_2 & 0 \\ + R_1 & 0 & R_3 + \end{pmatrix} + \begin{pmatrix} + I_1 \\ + I_2 \\ + I_3 + \end{pmatrix} +\end{align} +Lets solve the middle equation for $I_2$ by dividing by $R_2$. +\begin{align} + \begin{matrix} + 0 &= &-I_1 &+ &I_2 &+ &I_3 \\ + \frac{\epsilon_2}{R_2} &= &\frac{R_1}{R_2}I_1 &+ &I_2 &+ &0 \\ + \epsilon_3 &= &R_1 I_1 &+ &0 &+ &R_3 I_3 + \end{matrix} + && + \begin{pmatrix} + 0 \\ + \frac{\epsilon_2}{R_2} \\ + \epsilon_3 + \end{pmatrix} + = + \begin{pmatrix} + -1 & 1 & 1 \\ + \frac{R_1}{R_2}& 1 & 0 \\ + R_1 & 0 & R_3 + \end{pmatrix} + \begin{pmatrix} + I_1 \\ + I_2 \\ + I_3 + \end{pmatrix} +\end{align} +Now we can subtract (just like plugging in) the second equation to the first, to get rid of the $I_2$ in the first equation in terms of $I_1$. +\begin{align} + \begin{matrix} + \frac{-\epsilon_2}{R_2} &= &-\p({\frac{R_1}{R_2}+1})I_1 &+ &0 &+ &I_3 \\ + \frac{\epsilon_2}{R_2} &= &\frac{R_1}{R_2}I_1 &+ &I_2 &+ &0 \\ + \epsilon_3 &= &R_1 I_1 &+ &0 &+ &R_3 I_3 + \end{matrix} + && + \begin{pmatrix} + \frac{-\epsilon_2}{R_2} \\ + \frac{\epsilon_2}{R_2} \\ + \epsilon_3 + \end{pmatrix} + = + \begin{pmatrix} + -\frac{R_1}{R_2}-1 & 0 & 1 \\ + \frac{R_1}{R_2} & 1 & 0 \\ + R_1 & 0 & R_3 + \end{pmatrix} + \begin{pmatrix} + I_1 \\ + I_2 \\ + I_3 + \end{pmatrix} +\end{align} +Now lets solve the last equation for $I_3$ by dividing by $R_3$\ldots +\begin{align} + \begin{matrix} + \frac{-\epsilon_2}{R_2} &= &-\p({\frac{R_1}{R_2}+1})I_1 &+ &0 &+ &I_3 \\ + \frac{\epsilon_2}{R_2} &= &\frac{R_1}{R_2}I_1 &+ &I_2 &+ &0 \\ + \frac{\epsilon_3}{R_3} &= &\frac{R_1}{R_3}I_1 &+ &0 &+ &I_3 + \end{matrix} + && + \begin{pmatrix} + \frac{-\epsilon_2}{R_2} \\ + \frac{\epsilon_2}{R_2} \\ + \frac{\epsilon_3}{R_3} + \end{pmatrix} + = + \begin{pmatrix} + -\frac{R_1}{R_2}-1 & 0 & 1 \\ + \frac{R_1}{R_2} & 1 & 0 \\ + \frac{R_1}{R_3} & 0 & 1 + \end{pmatrix} + \begin{pmatrix} + I_1 \\ + I_2 \\ + I_3 + \end{pmatrix} +\end{align} +\ldots and subtracting the last equation to the first, to get rid of the $I_3$ in the first equation in terms of $I_1$. +\begin{align} + \begin{matrix} + \frac{-\epsilon_3}{R_3}-\frac{\epsilon_2}{R_2} &= &-\p({\frac{R_1}{R_3}+\frac{R_1}{R_2}+1})I_1 &+ &0 &+ &0 \\ + \frac{\epsilon_2}{R_2} &= &\frac{R_1}{R_2}I_1 &+ &I_2 &+ &0 \\ + \frac{\epsilon_3}{R_3} &= &\frac{R_1}{R_3}I_1 &+ &0 &+ &I_3 + \end{matrix} + && + \begin{pmatrix} + \frac{-\epsilon_3}{R_3}-\frac{\epsilon_2}{R_2} \\ + \frac{\epsilon_2}{R_2} \\ + \frac{\epsilon_3}{R_3} + \end{pmatrix} + = + \begin{pmatrix} + -\frac{R_1}{R_3}-\frac{R_1}{R_2}-1 & 0 & 0 \\ + \frac{R_1}{R_2} & 1 & 0 \\ + \frac{R_1}{R_3} & 0 & 1 + \end{pmatrix} + \begin{pmatrix} + I_1 \\ + I_2 \\ + I_3 + \end{pmatrix} +\end{align} +Now we have an equation with just $I_1$ and the $R_i$ and $\epsilon_i$ that were given. +Dividing through by the junk in front of $I_1$ we have +\begin{align} + \begin{matrix} + \frac{\frac{\epsilon_3}{R_3}+\frac{\epsilon_2}{R_2}}{\frac{R_1}{R_3}+\frac{R_1}{R_2}+1} &= &I_1 &+ &0 &+ &0 \\ + \frac{\epsilon_2}{R_2} &= &\frac{R_1}{R_2}I_1 &+ &I_2 &+ &0 \\ + \frac{\epsilon_3}{R_3} &= &\frac{R_1}{R_3}I_1 &+ &0 &+ &I_3 + \end{matrix} + && + \begin{pmatrix} + \frac{\frac{\epsilon_3}{R_3}+\frac{\epsilon_2}{R_2}}{\frac{R_1}{R_3}+\frac{R_1}{R_2}+1} \\ + \frac{\epsilon_2}{R_2} \\ + \frac{\epsilon_3}{R_3} + \end{pmatrix} + = + \begin{pmatrix} + 1 & 0 & 0 \\ + \frac{R_1}{R_2} & 1 & 0 \\ + \frac{R_1}{R_3} & 0 & 1 + \end{pmatrix} + \begin{pmatrix} + I_1 \\ + I_2 \\ + I_3 + \end{pmatrix} +\end{align} +A solution! We know have an explicit expression for $I_1$. +Going back and subtracting $R_1/R_2$ time the first equation from the middle lets us solve for $I_2$. +\begin{align} + \begin{matrix} + \frac{\frac{\epsilon_3}{R_3}+\frac{\epsilon_2}{R_2}}{\frac{R_1}{R_3}+\frac{R_1}{R_2}+1} &= &I_1 &+ &0 &+ &0 \\ + \frac{\epsilon_2}{R_2} - \frac{1}{R_2}\frac{\frac{\epsilon_3}{R_3}+\frac{\epsilon_2}{R_2}}{\frac{1}{R_3}+\frac{1}{R_2}+\frac{1}{R_1}} &= &0 &+ &I_2 &+ &0 \\ + \frac{\epsilon_3}{R_3} &= &\frac{R_1}{R_3}I_1 &+ &0 &+ &I_3 + \end{matrix} + && + \begin{pmatrix} + \frac{\frac{\epsilon_3}{R_3}+\frac{\epsilon_2}{R_2}}{\frac{R_1}{R_3}+\frac{R_1}{R_2}+1} \\ + \frac{\epsilon_2}{R_2} - \frac{1}{R_2}\frac{\frac{\epsilon_3}{R_3}+\frac{\epsilon_2}{R_2}}{\frac{1}{R_3}+\frac{1}{R_2}+\frac{1}{R_1}} \\ + \frac{\epsilon_3}{R_3} + \end{pmatrix} + = + \begin{pmatrix} + 1 & 0 & 0 \\ + 0 & 1 & 0 \\ + \frac{R_1}{R_3} & 0 & 1 + \end{pmatrix} + \begin{pmatrix} + I_1 \\ + I_2 \\ + I_3 + \end{pmatrix} +\end{align} +And finally subtracting $R_1/R_3$ time the first equation from the last lets us solve for $I_3$. +\begin{align} + \begin{matrix} + \frac{\frac{\epsilon_3}{R_3}+\frac{\epsilon_2}{R_2}}{\frac{R_1}{R_3}+\frac{R_1}{R_2}+1} &= &I_1 &+ &0 &+ &0 \\ + \frac{\epsilon_2}{R_2} - \frac{1}{R_2}\frac{\frac{\epsilon_3}{R_3}+\frac{\epsilon_2}{R_2}}{\frac{1}{R_3}+\frac{1}{R_2}+\frac{1}{R_1}} &= &0 &+ &I_2 &+ &0 \\ + \frac{\epsilon_3}{R_3} - \frac{1}{R_3}\frac{\frac{\epsilon_3}{R_3}+\frac{\epsilon_2}{R_2}}{\frac{1}{R_3}+\frac{1}{R_2}+\frac{1}{R_1}} &= &0 &+ &0 &+ &I_3 + \end{matrix} + && + \begin{pmatrix} + \frac{\frac{\epsilon_3}{R_3}+\frac{\epsilon_2}{R_2}}{\frac{R_1}{R_3}+\frac{R_1}{R_2}+1} \\ + \frac{\epsilon_2}{R_2} - \frac{1}{R_2}\frac{\frac{\epsilon_3}{R_3}+\frac{\epsilon_2}{R_2}}{\frac{1}{R_3}+\frac{1}{R_2}+\frac{1}{R_1}} \\ + \frac{\epsilon_3}{R_3} - \frac{1}{R_3}\frac{\frac{\epsilon_3}{R_3}+\frac{\epsilon_2}{R_2}}{\frac{1}{R_3}+\frac{1}{R_2}+\frac{1}{R_1}} \\ + \end{pmatrix} + = + \begin{pmatrix} + 1 & 0 & 0 \\ + 0 & 1 & 0 \\ + 0 & 0 & 1 + \end{pmatrix} + \begin{pmatrix} + I_1 \\ + I_2 \\ + I_3 + \end{pmatrix} + = + \begin{pmatrix} + I_1 \\ + I_2 \\ + I_3 + \end{pmatrix} +\end{align} + +Yuck! Why would anyone want to solve equations like that? +You have to do all the same work in the matrix version, but you're rewriting everything at each step! +What gives? +The power of linear algebra is that, while it is long and tedious for us (see above), it is very easy for a computer (see below). +Understanding why means taking a step back and getting an abstract view of the solution we just worked through. + +\section{Solving matrix equations with a computer} + +When we started, we had an equation like this +\begin{equation} + \begin{bmatrix} + \epsilon + \end{bmatrix} + = + \begin{bmatrix} + R + \end{bmatrix} + \begin{bmatrix} + I + \end{bmatrix} +\end{equation} +Wait a second, the $\epsilon$s were voltages, so that looks a lot like plain old $V=IR$! +We know how to get the current then, it's just +\begin{equation} + R^{-1} V = I \;. +\end{equation} +Maybe we can think about our matrix equation like this, we just need to figure out what $[R]^{-1}$ means. +What is $R^{-1}$ anyway? +It's the \emph{inverse} of $R$; the thing that, when multiplied by $R$, gives one. +$[R]^{-1}$ is also just the inverse of $[R]$, so $[R]^{-1}\cdot[R] = 1$. + +Inverting a matrix by hand is basically what we were doing in our solution above, +but if we can get our computer (or calculator) to find $[R]^{-1}$ for us, we don't have to do any of the messy algebra. +We can just get our solution via +\begin{equation} + \begin{bmatrix} + R + \end{bmatrix}^{-1} + \begin{bmatrix} + \epsilon + \end{bmatrix} + = + \begin{bmatrix} + I + \end{bmatrix} +\end{equation} + +On the TI-83+, that's pretty much all there is to it. +You can enter your $[R]$ and $[\epsilon]$ matrices in the {\tt [2nd] MATRX $\rightarrow$ EDIT} menu (calling them $[A]$ and $[B]$. +Then just type out +\begin{equation} + [A]^{-1}*B +\end{equation} +using the {\tt [2nd] MATRX $\rightarrow$ NAMES} menu to generate the $[A]$ and $[B]$ symbols. + +On the TI-89, you can enter the matrices straight from the command line, using comas to separate the columns and semi-colons to separate the rows. Using the numbers from Problem 35 that gives +\begin{align} +&[-1,1,1;8,6,0;8,0,4] \rightarrow A \\ +&[0;4;12] \rightarrow I \\ +&A^{-1}*I \\ +& \qquad\qquad + \begin{pmatrix} + 0.8462 \\ + -0.4615 \\ + 1.3077 + \end{pmatrix} +\end{align} +(I don't have a TI-89, so if this is wrong, let me know\ldots). + + +\end{document} diff --git a/latex/problems/Giancoli_6/problem17.40.tex b/latex/problems/Giancoli_6/problem17.40.tex new file mode 100644 index 0000000..e18f780 --- /dev/null +++ b/latex/problems/Giancoli_6/problem17.40.tex @@ -0,0 +1,113 @@ +\begin{problem*}{40} +A $C_1 = 7.7\U{$\mu$F}$ capacitor is charged by a $V = 125\U{V}$ +battery (Fig. 17-29a) and then is disconnected from the battery. When +this capacitor ($C_1$) is then connected (Fig. 17-29b) to a second +(initially uncharged) capacitor, $C_2$, the final voltage on each +capacitor is $V_2 = 15\U{V}$. What is the value of $C_2$? +[\emph{Hint}: charge is conserved.] +\end{problem*} + +\empaddtoprelude{ + input makecirc; % circuit drawing functions + initlatex(""); + pair A, B, Cl, Cr, Dl, Dr, El, Er; + numeric a; + a := 3cm; + A := (-3a/4,0); + B := (3a/4, 0); + def drawA = + % add elements + centreof.b(A+(1,0), A-(1,0), bat); + battery.B(c.b, phi.b, "V", ""); + centreof.c(B.B.n, B.B.p, cap); + capacitor.C(c.c+(0,a), normal, phi.c, "C_1", ""); + % add wiring along the bottom + wire(B.B.n, A+(a/2,0), nsq); + wire(B.B.p, A-(a/2,0), nsq); + % add wiring along the sides and top + wire(A+(a/2,0), C.C.l, udsq); + wire(A-(a/2,0), C.C.r, udsq); + puttext.bot("($a$)", A-(0,24pt)); + Cl = C.C.r; + Cr = C.C.l; + enddef; + def drawB = + % add elements + centreof.d(B+(1,0), B-(1,0), cap); + capacitor.D(c.d, normal, phi.d, "C_2", ""); + centreof.e(C.D.l, C.D.r, cap); + capacitor.E(c.e+(0,a), normal, phi.e, "C_1", ""); + % add wiring along the bottom + wire(C.D.l, B+(a/2,0), nsq); + wire(C.D.r, B-(a/2,0), nsq); + % add wiring along the sides and top + wire(B+(a/2,0), C.E.l, udsq); + wire(B-(a/2,0), C.E.r, udsq); + puttext.bot("($b$)", B-(0,24pt)); + Dl = C.D.r; + Dr = C.D.l; + El = C.E.r; + Er = C.E.l; + enddef; +} + +\begin{nosolution} +\begin{center} +\begin{empfile}[5p] +\begin{emp}(0cm, 0cm) + drawA; + drawB; +\end{emp} +\end{empfile} +\end{center} +\end{nosolution} + +\begin{solution} +\begin{center} +\begin{empfile}[5] +\begin{emp}(0cm, 0cm) + drawA; + drawB; + puttext.ulft("$Q_{1a}$", Cl); + puttext.urt("$-Q_{1a}$", Cr); + puttext.ulft("$Q_{2b}$", Dl); + puttext.urt("$-Q_{2b}$", Dr); + puttext.ulft("$Q_{1b}$", El); + puttext.urt("$-Q_{1b}$", Er); + labeloffset := 9pt; + puttext.top("$V$", (Cl+Cr)/2); + puttext.top("$V_2$", (Dl+Dr)/2); + puttext.top("$V_2$", (El+Er)/2); +\end{emp} +\end{empfile} +\end{center} + +Because the voltage drop across $C_1$ in situation $a$ is the same as +the voltage drop across the battery ($V$), we have +$$ + Q_{1a} = C_1 V +$$ +When we connect $C_2$ in situation $b$, this charge redistributes +between $C_1$ and $C_2$. Because charge is conserved, we know +$$ + Q_{1a} = Q_{1b} + Q_{2b} +$$ +We also know that the voltage drop across both capacitors in situation +$b$ must be equal ($\text{both} = V_2$), so +\begin{align*} + Q_{1b} &= C_1 V_2 \\ + Q_{2b} &= C_2 V_2 +\end{align*} +Plugging each of these formulas for charge ($Q_{1a}$, $Q_{1b}$, and +$Q_{2b}$) into the charge conservation formula yeilds +\begin{align*} + C_1 V &= C_1 V_2 + C_2 V_2 \\ + C_1 (V - V_2) &= C_2 V_2 \\ + V_2 C_2 &= C_1 (V - V_2) \\ + C_2 &= C_1 \p({\frac{V}{V_2} - \frac{V_2}{V_2}}) \\ + C_2 &= C_1 \p({\frac{V}{V_2} - 1}) \\ + C_2 &= 7.7\U{$\mu$F} \cdot \p({\frac{125\U{V}}{15\U{V}} - 1}) + = \ans{56\U{$\mu$F}} +\end{align*} + +\end{solution} diff --git a/latex/problems/Giancoli_6/problem19.02.tex b/latex/problems/Giancoli_6/problem19.02.tex new file mode 100644 index 0000000..05344df --- /dev/null +++ b/latex/problems/Giancoli_6/problem19.02.tex @@ -0,0 +1,18 @@ +\begin{problem*}{2} +Four $1.5\U{V}$ cells are connected in series to a $12\U{\Ohm}$ +lightbulb. If the resulting current is $0.45\U{A}$, what is the +internal resistance of each cell, assuming they are identical and +neglecting the wires. +\end{problem*} + +\begin{solution} +This is simply an application of the procedure outlined in Question 13. +The external resistance is the lightbulb $R_{ext}=12\U{\Ohm}$. +The total internal resistance is the sum of all the individual cell resistances $r_{int}=6r$. +The total voltage is the sum of all the individual cell voltages $V = 6\cdot 1.5\U{V} = 9\U{V}$. +Putting these together we have +\begin{align*} +r_{int} = 6r &= \frac{V}{I} - R_{ext} = \frac{9\U{V}}{0.45\U{A}} - 12\U{\Ohm} \\ + r &= \ans{1.3\U{\Ohm}} +\end{align*} +\end{solution} diff --git a/latex/problems/Giancoli_6/problem19.07.tex b/latex/problems/Giancoli_6/problem19.07.tex new file mode 100644 index 0000000..f511132 --- /dev/null +++ b/latex/problems/Giancoli_6/problem19.07.tex @@ -0,0 +1,19 @@ +\begin{problem*}{7} +A $650\U{\Ohm}$ and a $2200\U{\Ohm}$ resistor are connected in series +with a $12\U{V}$ battery. What is the voltage across the +$2200\U{\Ohm0}$ resistor? +\end{problem*} + +\begin{solution} +First we find the total current in the circuit. The two resistances, +$R_1 = 650\U{\Ohm}$ and $R_2 = 2200\U{\Ohm}$, in series provide an +effective resistance of $R_e = R_1 + R_2$. By Kirchoff's loop rule +\begin{align*} + V - I R_e &= 0 \\ + I &= \frac{V}{R_e} = \frac{V}{R_1 + R_2} +\end{align*} +And applying Ohm's law to the second resistor +\begin{align*} + V_2 = I R_2 = \frac{V R_2}{R_1 + R_2} = \ans{9.3\U{V}} +\end{align*} +\end{solution} diff --git a/latex/problems/Giancoli_6/problem19.15.tex b/latex/problems/Giancoli_6/problem19.15.tex new file mode 100644 index 0000000..6dd4418 --- /dev/null +++ b/latex/problems/Giancoli_6/problem19.15.tex @@ -0,0 +1,20 @@ +\begin{problem*}{15} +Eight $7.0\U{W}$ Christmas tree lights are connected in series to each +other and to a $110\U{V}$ source. What is the resistance of each +bulb. +\end{problem*} + +\begin{solution} +Let $V = 110\U{V}$ be the source voltage, $P_1 = 7.0\U{W}$ be the +power of one bulb, and $R_1$ be the resistance of one bulb. By +Kirchoff's loop rule +\begin{align*} + V - 8IR_1 &= 0 \\ + I &= \frac{V}{8R_1} +\end{align*} +And we can find $R_1$ by considering the power dissipated by the bulb +\begin{align*} + P &= IV_1 = I^2 R_1 = \p({\frac{V}{8}})^2\frac{1}{R_1} \\ + R_1 &= \p({\frac{V}{8}})^2\frac{1}{P} = \ans{27\U{\Ohm}} +\end{align*} +\end{solution} diff --git a/latex/problems/Giancoli_6/problem19.24.tex b/latex/problems/Giancoli_6/problem19.24.tex new file mode 100644 index 0000000..38fdf69 --- /dev/null +++ b/latex/problems/Giancoli_6/problem19.24.tex @@ -0,0 +1,31 @@ +\begin{problem*}{24} +Determine the terminal voltage of each battery in Fig.~19-44. +\begin{center} +\begin{asy} + import Circ; + real u = 3cm; + TwoTerminal ra = resistor((0,0), normal, 0, "$r_1 = 1.0\U{\Ohm}$", ""); + TwoTerminal Ba = source(ra.end, DC, 0, "", "$\mathcal{E}_1 = 12\U{V}$"); + TwoTerminal rb = resistor(ra.beg+(0,-u), normal, 0, "$r_2 = 2.0\U{\Ohm}$", ""); + TwoTerminal Bb = source(rb.end, DC, 0, "", "$\mathcal{E}_2 = 18\U{V}$"); + TwoTerminal R = resistor(Bb.end+(0.5u,0.25u), normal, 90, "", "$R = 6.6\U{\Ohm}$"); + wireU(rb.beg, ra.beg, -24pt, rlsq); + wire(Bb.end, R.beg, rlsq); + wire(Ba.end, R.end, rlsq); +\end{asy} +\end{center} +\end{problem*} + +\begin{solution} +From Kirchoff's loop rule +\begin{align*} + \mathcal{E}_1 - IR - \mathcal{E}_2 - Ir_2 - Ir_1 &= 0 \\ + I(R+r_1+R_2) &= \mathcal{E}_1-\mathcal{E}_2 \\ + I &= \frac{\mathcal{E}_1-\mathcal{E}_2}{R+r_1+r_2} = 0.625\U{A} +\end{align*} + +So the voltage across the top battery is +$$V_1 = \mathcal{E}_1 - I r_1 = \ans{17\U{V}} \qquad (17.375\U{V})$$ +and the voltage across the bottom battery is +$$V_2 = \mathcal{E}_2 - I r_2 = \ans{11\U{V}} \qquad (10.75\U{V})$$ +\end{solution} diff --git a/latex/problems/Giancoli_6/problem19.31.tex b/latex/problems/Giancoli_6/problem19.31.tex new file mode 100644 index 0000000..771032c --- /dev/null +++ b/latex/problems/Giancoli_6/problem19.31.tex @@ -0,0 +1,92 @@ +\begin{problem*}{31} + Calculate the currents in each resistor of Fig.~19-49. +\end{problem*} + +\begin{nosolution} +\begin{center} +\begin{asy} + import Circ; + real u = 3cm; + TwoTerminal Bc = source((0,0), DC, 90, "", "$3.0\U{V}$"); + TwoTerminal Rcb = resistor(Bc.beg, normal, -90, "$10\U{\Ohm}$", ""); + TwoTerminal Rca = resistor(Bc.end, normal, 180, "", "$2\U{\Ohm}$"); + pair Jtop = Rca.end, Jbot = (Jtop.x,Rcb.end.y); + TwoTerminal Rb = resistor(Jtop, normal, -90, "$6\U{\Ohm}$", ""); + TwoTerminal Ba = source(Jtop, DC, 180, "", "$6.0\U{V}$"); + TwoTerminal Rab = resistor(Jbot, normal, 180, "$8\U{\Ohm}$", ""); + TwoTerminal Raa = resistor(Rab.end, normal, 90, "$12\U{\Ohm}$", ""); + wire(Ba.end, Raa.end, rlsq); + wire(Rab.beg, Jbot, nsq); + wire(Jbot, Rb.end, nsq); + wire(Jbot, Rcb.end, rlsq); +\end{asy} +\end{center} +\end{nosolution} + +\begin{solution} +\begin{center} +\begin{asy} + import Circ; + TwoTerminal Bc = source((0,0), DC, 90, "", "$3.0\U{V}$"); + TwoTerminal Rcb = resistor(Bc.beg, normal, -90, "$10\U{\Ohm}$", ""); + TwoTerminal Rca = resistor(Bc.end, normal, 180, "", "$2\U{\Ohm}$"); + pair Jtop = Rca.end, Jbot = (Jtop.x,Rcb.end.y); + TwoTerminal Ic = current((Jbot+Rcb.end)/2, 0, "", "$I_3$"); + TwoTerminal Rb = resistor(Jtop, normal, -90, "$6\U{\Ohm}$", ""); + TwoTerminal Ib = current(Rb.end, -90, "", "$I_2$"); + TwoTerminal Ba = source(Jtop, DC, 180, "", "$6.0\U{V}$"); + TwoTerminal Ia = current(Ba.end, 180, "$I_1$", ""); + TwoTerminal Rab = resistor(Jbot, normal, 180, "$8\U{\Ohm}$", ""); + TwoTerminal Raa = resistor(Rab.end, normal, 90, "$12\U{\Ohm}$", ""); + wire(Ia.end, Raa.end, rlsq); + wire(Jbot, Ib.end, nsq); + wire(Jbot, Ic.beg, nsq); + wire(Ib.end, Rb.end, nsq); + wire(Ic.end, Rcb.end, rlsq); + dot("a", Jbot, S); +\end{asy} +\end{center} +Label the resistors from left to right: $R_1 = 12\U{\Ohm}$, $R_2 = +8\U{\Ohm}$, $R_3 = 6\U{\Ohm}$, $R_4 = 2\U{\Ohm}$, and $R_5 = +10\U{\Ohm}$. + +Label the batteries from left to right: $V_1 = 6.0\U{V}$ and $V_2 = +3.0\U{V}$. + +Applying Kirchoff's junction rule to junction $a$ we have +$$I_1 + I_2 - I_3 = 0$$ + +Applying Kirchoff's loop rule to the left-hand loop we have +$$V_1 - I_1 (R_1 + R_2) + R_3 I_2 = 0$$ +where we \emph{add} the voltage change over $R_3$ because we cross it +\emph{against} the direction of the current $I_2$. + +Applying Kirchoff's loop rule to the right-hand loop we have +$$V_2 - R_4 I_3 - R_3 I_2 - R_5 I_5 = V_2 - I_3 (R_4 + R_5) - R_3 I_2 = 0$$ + +We now have three equations for three unknowns (the $I_i$). +Solving the loop rools for $I_1$ and $I_3$ we have +\begin{align*} + I_1 &= \frac{V_1 + R_3 I_2}{R_1 + R_2} = \frac{V_1 + R_3 I_2}{R_{12}} \\ + I_3 &= \frac{V_2 - R_3 I_2}{R_4 + R_5} = \frac{V_2 - R_3 I_2}{R_{45}} +\end{align*} +where we have used the equivalent resistances $R_{12} \equiv R_1 + +R_2$ and $R_{45} \equiv R_4 + R_5$ to save writing later. We can then +plug those currents into the junction rule and solve for $I_2$ +\begin{align*} + \frac{V_1 + R_3 I_2}{R_{12}} + I_2 - \frac{V_2 - R_3 I_2}{R_{45}} &= 0 \\ + \frac{V_1}{R_{12}} + \frac{R_3}{R_{12}} I_2 + I_2 - \frac{V_2}{R_{45}} + \frac{R_3}{R_{45}}I_2 &= 0 \\ + \p({\frac{R_3}{R_{12}} + 1 + \frac{R_3}{R_{45}}})\cdot I_2 &= \frac{V_2}{R_{45}} - \frac{V_1}{R_{12}} \\ + I_2 &= \frac{\frac{V_2}{R_{45}} - \frac{V_1}{R_{12}}}{\frac{R_3}{R_{12}} + 1 + \frac{R_3}{R_{45}}} \\ + I_2 &= \ans{-28\U{mA}} +\end{align*} +Where the $-$ sign means the true current is in the opposite direction +to the one we have assigned (so the true current flows upward in the +figure). We can now plug this current in to find $I_1$ and $I_3$. +\begin{align*} + I_1 &= \frac{V_1 + R_3 I_2}{R_{12}} = \ans{292\U{mA}} \\ + I_3 &= \frac{V_2 - R_3 I_2}{R_{45}} = \ans{264\U{mA}} +\end{align*} + +Double-checking our algebra, we see $I_1 + I_2 - I_3 = 292 - 27 - 264 = -1\U{mA} \approx 0$ where difference of $1\U{mA}$ is due to rounding errors from forcing our answers to milli-Volt precision. +\end{solution} diff --git a/latex/problems/Giancoli_6/problem19.58.tex b/latex/problems/Giancoli_6/problem19.58.tex new file mode 100644 index 0000000..be0fa0f --- /dev/null +++ b/latex/problems/Giancoli_6/problem19.58.tex @@ -0,0 +1,111 @@ +\begin{problem*}{58} + A $45\U{V}$ battery of negligable internal resistance is connected + to a $38\U{k\Ohm}$ and a $27\U{k\Ohm}$ resistor in series. What + reading will a voltmeter, of internal resistance $95\U{k\Ohm}$, + give when used to measure the voltage across each resistor? What is + the percent inaccuracy due to meter resistance for each case? +\end{problem*} + +\begin{solution} +Case 1: \\ +The original situation looks like +\begin{center} +\begin{asy} + import Circ; + real u = 0.5cm; + TwoTerminal B = source((0,0), DC, 90, "$45\U{V}$", "$V$"); + pair a = B.end+(0,u); + pair b = B.beg-(0,u); + TwoTerminal Ra = resistor(a, normal, 0, "$38\U{k\Ohm}$", "$R_1$"); + TwoTerminal Rb = resistor(Ra.end, normal, 0, "$27\U{k\Ohm}$", "$R_2$"); + TwoTerminal I = current((Rb.end.x, (a.y+b.y)/2), -90, "", "$I$"); + wire(Rb.end, I.beg, nsq); + wire(I.end, b, udsq); + wire(b, B.beg, nsq); + wire(a, B.end, nsq); +\end{asy} +\end{center} +Using Kirchoff's loop rule +\begin{align*} + V - IR_1 - IR_2 &= 0 \\ + I &= \frac{V}{R_1+R_2} +\end{align*} +so +\begin{align*} + V_1 &= IR_1 = \frac{VR_1}{R_1+R_2} \approx 26.3\U{V} \\ + V_2 &= IR_2 = \frac{VR_2}{R_1+R_2} \approx 18.7\U{V} +\end{align*} + +Case 2: \\ +With the voltmeter across $R_1$ we have +\begin{center} +\begin{asy} + import Circ; + real u = 0.5cm; + TwoTerminal B = source((0,0), DC, 90, "$45\U{V}$", "$V$"); + pair a = B.end+(0,u); + pair b = B.beg-(0,u); + TwoTerminal Ra = resistor(a, normal, 0, "$38\U{k\Ohm}$", "$R_1$"); + TwoTerminal Ia = current(Ra.end, 0, "", "$I_1$"); + TwoTerminal Rv = resistor(a+(0,4u), normal, 0, "$95\U{k\Ohm}$", "$R_v$"); + TwoTerminal Iv = current(Rv.end, 0, "", "$I_v$"); + TwoTerminal Rb = resistor(Ia.end, normal, 0, "$27\U{k\Ohm}$", "$R_2$"); + TwoTerminal I = current((Rb.end.x, (a.y+b.y)/2), -90, "", "$I_T$"); + wire(Rb.end, I.beg, nsq); + wire(I.end, b, udsq); + wire(b, B.beg, nsq); + wire(a, B.end, nsq); + wire(a, Rv.beg, nsq); + wire(Iv.end, Ia.end, rlsq); +\end{asy} +\end{center} +Using our formula for resistors in parallel, we can bundle $R_v$ and $R_1$ into a single resistor $R_1'$, where +$$ R_1' = \p({\frac{1}{R_1} + \frac{1}{R_v}})^{-1} = 27.14285714\ldots\U{k\Ohm} $$ + +Once we've done that, we have the same situation as in Case 1, but with +\begin{align*} + R_1 &\rightarrow R_1' \\ + I &\rightarrow I_T +\end{align*} +so +\begin{align*} + V_1' &= \frac{VR_1'}{R_1'+R_2} = + \frac{ V \p({\frac{1}{R_1}+\frac{1}{R_v}})^{-1} } + { \p({\frac{1}{R_1}+\frac{1}{R_v}})^{-1} + R_2} \approx \ans{22.6\U{V}} \\ + \frac{V_1'}{V_1} &= \frac{R_1'(R_1+R_2)}{R_1(R_1'+R_2)} + = \frac{ \p({\frac{1}{R_1}+\frac{1}{R_v}})^{-1}\cdot(R_1+R_2) } + { R_1\cdot\p[{\p({\frac{1}{R_1}+\frac{1}{R_v}})^{-1} + R_2}] } + = \frac{ \p({\frac{R_v+R_1}{R_1 R_v}})^{-1}\cdot(R_1+R_2) } + { R_1\cdot\p[{\p({\frac{R_v+R_1}{R_1 R_v}})^{-1} + R_2}] } + = \frac{ \frac{R_1 R_v}{R_v+R_1}\cdot(R_1+R_2) } + { R_1\cdot\p({\frac{R_1 R_v}{R_v+R_1} + \frac{R_2(R_v+R_1)}{R_v+R_1}}) } \\ + &=\frac{ R_1 \frac{R_v (R_1+R_2)}{R_v+R_1} } + { R_1\cdot\p({\frac{R_1 R_v + R_2(R_v+R_1)}{R_v+R_1}}) } + = \frac{ R_v R_1 + R_v R_2 } + { R_1 R_v + R_2 R_v + R_1 R_2 } + = \frac{ R_v (R_1 + R_2) } + { R_v (R_1 + R_2) + R_1 R_2 } \\ + &= 0.8575\ldots \\ +\end{align*} +Obviously, we could plug in known numbers and solve for +$\frac{V_1'}{V_1}$ after the first equality above, but crunching +through some simplifying algebra reveals the pretty spectacular final +form, from which you can trivially see that the fractional error in +Case 3 will be the same as that for Case 2. + +Finally the percent error is given by +$$ \text{Error}_1 = 1-\frac{V_1'}{V_1} = 0.1425\ldots \approx \ans{14\%} $$ + +Case 3: \\ +With the voltmeter across $R_2$ we have the same situation as Case 1, but with +\begin{align*} + X_1 &\leftrightarrow X_2 +\end{align*} +for any symbol $X$ (i.e. $R_1 \leftrightarrow R_2$, \ldots). +However, the equation for error in $V_1'$ is not effected by this exchange, so +$$ \text{Error}_2 = \text{Error}_1 \approx \ans{14\%} $$ +The voltage $V_2'$ measured is given by +$$ V_2' = \frac{VR_2'}{R_2'+R_1} = + \frac{ V \p({\frac{1}{R_2}+\frac{1}{R_v}})^{-1} } + { \p({\frac{1}{R_2}+\frac{1}{R_v}})^{-1} + R_1} \approx \ans{16.0\U{V}} $$ +\end{solution} diff --git a/latex/problems/Giancoli_6/question19.04.tex b/latex/problems/Giancoli_6/question19.04.tex new file mode 100644 index 0000000..2d29fcf --- /dev/null +++ b/latex/problems/Giancoli_6/question19.04.tex @@ -0,0 +1,28 @@ +\begin{problem*}{Q4} +Two lightbulbs of resistance $R_1$ and $R_2$ ($R_2 > R_1$) are +connected in series. Which is brighter? What if they are connected +in parallel? Explain. +\end{problem*} + +\begin{solution} +\Part{a} +In series, the same current $I$ flows through both bulbs, +so the power (proporional to the brightness) can be found via +\begin{align*} + R_1 &< R_2 \\ + I^2 R_1 &< I^2 R_2 \\ + P_1 = I^2 R_1 &< I^2 R_2 = P_2 \\ + P_1 &< P_2 +\end{align*} + +\Part{a} +In series, both bulbs see the same voltage $V$, +so the power (proporional to the brightness) can be found via +\begin{align*} + R_1 &< R_2 \\ + \frac{1}{R_1} &> \frac{1}{R_2} \\ + \frac{V^2}{R_1} &> \frac{V^2}{R_2} \\ + P_1 = \frac{V^2}{R_1} &> \frac{V^2}{R_2} = P_2 \\ + P_1 &> P_2 +\end{align*} +\end{solution} diff --git a/latex/problems/Giancoli_6/question19.07.tex b/latex/problems/Giancoli_6/question19.07.tex new file mode 100644 index 0000000..e395d38 --- /dev/null +++ b/latex/problems/Giancoli_6/question19.07.tex @@ -0,0 +1,25 @@ +\begin{problem*}{Q7} +If two identical resistors are connected in series to a battery, does +the battery have to supply more power or less power than when only one +of the resistors is connected? Explain. +\end{problem*} + +\begin{solution} +We know that the power provided by the battery is given by +$$P = IV$$ +so the power supplied increases if the current $I$ increases (because $V$ remains constant for batteries). + +From Kirchoff's loop rule, we know the voltage drop across the +resistors is the same as the voltage gain across the battery. +$$V_b = V_R$$ +We also know that the voltage across the resistors relates to the current via Ohm's law +$$V_R = IR$$ +Finally, we know that the effective resistance of two identical resistors in parallel is given by +$$R_2 = R_1 + R_1 = 2R_1$$ + +Putting these together in the case of a single resistor, we find a current of +$$I_1 = \frac{V_R}{R_1} = \frac{V_b}{R_1}$$ +and in the case of the two resistors in series +$$I_2 = \frac{V_R}{R_2} = \frac{V_b}{2R_1} = \frac{I_1}{2}$$ +So with two resistors in series, we have less current and need less power. +\end{solution} diff --git a/latex/problems/Giancoli_6/question19.13.tex b/latex/problems/Giancoli_6/question19.13.tex new file mode 100644 index 0000000..0143b6f --- /dev/null +++ b/latex/problems/Giancoli_6/question19.13.tex @@ -0,0 +1,29 @@ +\begin{problem*}{Q13} +Explain in detail how you could measure the internal resistance of a +battery. +\end{problem*} + +\begin{solution} + \begin{center} +\begin{asy} + import Circ; + real u = 1cm; + TwoTerminal B = source((0,0), DC, 0, "", "$V$"); + TwoTerminal r = resistor(B.end, normal, 0, "", "$r$"); + TwoTerminal R = resistor((0,u), normal, 0, "$R$", ""); + TwoTerminal I = current((r.end.x, R.end.y), 180, "", "$I$"); + wire(B.beg, R.beg, udsq); + wire(r.end, I.beg, udsq); + wire(I.end, R.end, nsq); +\end{asy} + \end{center} +Make a circuit using a known resistance $R$ to connect the two +terminals of the battery, and measure the current $I$. + +From Kirchoff's loop rule +\begin{align*} + V - Ir - IR &= 0 \\ + Ir &= V - IR \\ + r &= \ans{\frac{V}{I} - R} +\end{align*} +\end{solution} diff --git a/latex/problems/README b/latex/problems/README index e179ce2..e061863 100644 --- a/latex/problems/README +++ b/latex/problems/README @@ -1,4 +1,2 @@ A collection of LaTeX source (using my problempack.sty and wtk_cmmds.sty) for intro-physics problems that I've covered over the years. - -References are to Serway & Jewett, 4th Edition unless otherwise specified. diff --git a/latex/problems/equation27.07.tex b/latex/problems/Serway_and_Jewett_4/equation27.07.tex similarity index 100% rename from latex/problems/equation27.07.tex rename to latex/problems/Serway_and_Jewett_4/equation27.07.tex diff --git a/latex/problems/problem01.60.tex b/latex/problems/Serway_and_Jewett_4/problem01.60.tex similarity index 100% rename from latex/problems/problem01.60.tex rename to latex/problems/Serway_and_Jewett_4/problem01.60.tex diff --git a/latex/problems/problem01.62.tex b/latex/problems/Serway_and_Jewett_4/problem01.62.tex similarity index 100% rename from latex/problems/problem01.62.tex rename to 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They are tied to the came ceiling hook by light +strings of length $L$. When a horizontal uniform electric field $E$ +is turned on, the balls hang with an angle $\theta$ between the +strings (Fig.~21.46). Assume that the force one ball exerts on the +other is much smaller than the force exerted by the horizontal +electric field. \Part{a} Which ball (the right or the left) is +positive, and which is negative? \Part{b} Find the angle $\theta$ +between the strings in terms of $E$, $q$, $m$, and $g$. \Part{c} As +the electric field is gradually increased in strength, what does ytour +result from \Part{b} give for the largest possible angle $\theta$? +\end{problem*} + +\begin{solution} +\end{solution} diff --git a/latex/problems/Young_and_Freedman_12/problem21.86.tex b/latex/problems/Young_and_Freedman_12/problem21.86.tex new file mode 100644 index 0000000..8a2cbd0 --- /dev/null +++ b/latex/problems/Young_and_Freedman_12/problem21.86.tex @@ -0,0 +1,15 @@ +\begin{problem*}{86} +In an inkjet printer, letters are built up by squirting drops of ink +at the paper from a rapidly moving nozzle. The ink drops, which have +a mass of $1.4\E{-8}\U{g}$ each, leave the nozzle and travel toward +the paper at $20\U{m/s}$, passing through a charging unit that gives +each drop a positive charge $q$ by removing some electrons from it. +The drops then pass between parallel deflecting plates $2.0\U{cm}$ +long where there is a uniform vertical electric field with magnitude +$8.0\E{4}\U{N/C}$. If a drop is to be delected $0.30\U{mm}$ by the +time it reaches the end of the deflection plates, what magnitude of +charge must be given to the drop? +\end{problem*} + +\begin{solution} +\end{solution} diff --git a/latex/problems/problem09.limo.T.limo.py b/latex/problems/wking/problem09.limo.T.limo.py similarity index 100% rename from latex/problems/problem09.limo.T.limo.py rename to latex/problems/wking/problem09.limo.T.limo.py diff --git a/latex/problems/problem09.limo.T.tex b/latex/problems/wking/problem09.limo.T.tex similarity index 100% rename from latex/problems/problem09.limo.T.tex rename to latex/problems/wking/problem09.limo.T.tex diff --git a/latex/problems/problem12.T.bombardier.jpg b/latex/problems/wking/problem12.T.bombardier.jpg similarity index 100% rename from latex/problems/problem12.T.bombardier.jpg rename to latex/problems/wking/problem12.T.bombardier.jpg diff --git a/latex/problems/problem12.T.tex b/latex/problems/wking/problem12.T.tex similarity index 100% rename from latex/problems/problem12.T.tex rename to latex/problems/wking/problem12.T.tex diff --git a/latex/problems/problem28.compton-cat.T.tex b/latex/problems/wking/problem28.compton-cat.T.tex similarity index 100% rename from latex/problems/problem28.compton-cat.T.tex rename to latex/problems/wking/problem28.compton-cat.T.tex -- 2.26.2