From 0381572db0787a2f85a845f1c5bd94ed1b51f589 Mon Sep 17 00:00:00 2001 From: "W. Trevor King" Date: Fri, 31 Jul 2009 12:21:37 -0400 Subject: [PATCH] Added week 6 recitation solutions --- .../Young_and_Freedman_12/problem25.57.tex | 63 ++++++++++++++++ .../Young_and_Freedman_12/problem25.72.tex | 31 ++++++++ .../Young_and_Freedman_12/problem26.06.tex | 72 +++++++++++++++++++ .../Young_and_Freedman_12/problem26.21.tex | 69 ++++++++++++++++++ .../Young_and_Freedman_12/problem26.22.tex | 54 ++++++++++++++ .../Young_and_Freedman_12/problem26.48.tex | 58 +++++++++++++++ 6 files changed, 347 insertions(+) create mode 100644 latex/problems/Young_and_Freedman_12/problem25.57.tex create mode 100644 latex/problems/Young_and_Freedman_12/problem25.72.tex create mode 100644 latex/problems/Young_and_Freedman_12/problem26.06.tex create mode 100644 latex/problems/Young_and_Freedman_12/problem26.21.tex create mode 100644 latex/problems/Young_and_Freedman_12/problem26.22.tex create mode 100644 latex/problems/Young_and_Freedman_12/problem26.48.tex diff --git a/latex/problems/Young_and_Freedman_12/problem25.57.tex b/latex/problems/Young_and_Freedman_12/problem25.57.tex new file mode 100644 index 0000000..6c47019 --- /dev/null +++ b/latex/problems/Young_and_Freedman_12/problem25.57.tex @@ -0,0 +1,63 @@ +\begin{problem*}{25.57} +An electrical conductor designed to carry large currents has a +circular cross section $2.50\U{mm}$ in diameter and is $14.0\U{m}$ +long. The resistance between the ends is $0.104\U{\Ohm}$. \Part{a} +What is the resistivity of the material? \Part{b} If the +electric-field magnitude in the conductor is $1.28\U{V/m}$, what is +the total current? \Part{c} If the material has $8.5\E{28}$ free +electrons per cubic meter, find the average drift speed under the +conditions of \Part{b}. +\end{problem*} + +\begin{solution} +\Part{a} +Using the resistivity formula (Eqn.~25.10) +\begin{align} + R &= \frac{\rho}{L}{A} & + \rho &= \frac{RA}{L} = \frac{R\pi(d/2)^2}{L} + = \frac{0.104\U{\Ohm}\cdot\pi\cdot(1.25\E{-3}\U{m})^2}{14.0\U{m}} + = \ans{3.65\E{-8}\U{\Ohm$\cdot$}} \;. +\end{align} + +\Part{b} +Because the electric field is constant inside the conductor, the +voltage difference across the conductor is +\begin{equation} + V = EL = 17.9\U{V} \;. +\end{equation} +Using Ohm's law for the current through the conductor +\begin{align} + V &= IR & I &= \frac{V}{R} = \ans{172\U{A}} \;. +\end{align} + +\Part{c} +In a time $\Delta t$, all the electrons within $v\Delta t$ of a given +cross section will drift through that cross section. Since we know +the cross sectional area of the wire is $A$, that corresponds to a +volume of +\begin{equation} + \mathcal{V} = Av\Delta t \;. +\end{equation} +We know the density of electrons $n = 8.5\E{28}\U{e$^-$/m$^3$}$, +so the number of electrons crossing the section is +\begin{equation} + N = n\mathcal{V} = nAv\Delta t +\end{equation} +and the charge crossing is +\begin{equation} + Q = eN = enAv\Delta t \;, +\end{equation} +where $e=1.60\E{-18}\U{C}$ is the charge on one electron. + +The current in the wire is thus +\begin{equation} + I = \frac{Q}{\Delta t} = \frac{enAv\Delta t}{\Delta t} = enAv \;, +\end{equation} +and the drift velocity is +\begin{equation} + v = \frac{I}{enA} + = \frac{172\U{A}} + {1.8\E{-18}\U{C}\cdot8.5\E{28}\U{e$^-$/m$^3$}\cdot\pi(1.25\E{-3}\U{m})^2} + = \ans{2.5\U{mm/s}} \;. +\end{equation} +\end{solution} diff --git a/latex/problems/Young_and_Freedman_12/problem25.72.tex b/latex/problems/Young_and_Freedman_12/problem25.72.tex new file mode 100644 index 0000000..4d3ba34 --- /dev/null +++ b/latex/problems/Young_and_Freedman_12/problem25.72.tex @@ -0,0 +1,31 @@ +\begin{problem*}{25.72} +A typical cost for electric power is $12.0\text{\textcent}$ per +kilowatt-hour. \Part{a} Some people leave their porch light on all +the time. What is the yearly cost to keep a $75\U{W}$ bulb burning +day and night? \Part{b} Suppose your refrigerator uses $400\U{W}$ of +power when it's running, and it runs $8$ hours a day. What is the +yearly cost of operating your refrigerator? +\end{problem*} + +\begin{solution} +This is just a unit conversion problem, but it's good to get a feel +for how much a lightbulb costs. Note that PECO power in Philly is +currently around $15.4\text{\textcent}/kW\cdot\text{hour}$ +(\url{http://www.jea.com/services/electric/rates_quarterly.asp}). + +\Part{a} +\begin{equation} + 75\U{W}\cdot\frac{24\U{hours}}{1\U{day}} + \cdot\frac{365\U{days}}{1\U{year}} + \cdot\frac{\$0.120}{10^3\U{W$\cdot$hour}} + = \ans{\$78.84} +\end{equation} + +\Part{b} +\begin{equation} + 400\U{W}\cdot\frac{8\U{hours}}{1\U{day}} + \cdot\frac{365\U{days}}{1\U{year}} + \cdot\frac{\$0.120}{10^3\U{W$\cdot$hour}} + = \ans{\$140.16} +\end{equation} +\end{solution} diff --git a/latex/problems/Young_and_Freedman_12/problem26.06.tex b/latex/problems/Young_and_Freedman_12/problem26.06.tex new file mode 100644 index 0000000..396766b --- /dev/null +++ b/latex/problems/Young_and_Freedman_12/problem26.06.tex @@ -0,0 +1,72 @@ +\begin{problem*}{26.6} +For the circuit shown in Fig.~26.40 both meters are idealized, the +battery has no appeciable internal resistance, and the ammeter reads +$1.25\U{A}$. \Part{a} What does the voltmeter read? \Part{b} What is +the emf \EMF of the battery? +\begin{center} +\begin{verbatim} + 25.0 ++---V---+ +-/\/\/-A-+ +| | | 15.0 | ++-/\/\/-+--+-/\/\/---+------+ +| 45.0 | 15.0 Z 10.0 | +| +-/\/\/---+ | +| 35.0 E | ++-------/\/\/----i|---------+ +\end{verbatim} +\end{center} +\end{problem*} + +\begin{solution} +It's always a good idea to start off by labeling things. +\begin{center} +\begin{verbatim} + R1=25.0 I1=1.25A ++---V---+ +-/\/\/--<-+ +| | |R2=15.0 I2| ++-/\/\/-+--+-/\/\/--<-a----------+ +| 45.0=Rv |R3a=15.0 Z 10.0=R3b | +| +-/\/\/--<-+ | +| It 35.0=Re E I3 | ++--->---/\/\/----i|--------------+ +\end{verbatim} +\end{center} +\Part{a} +First, we'll get a handle on the $I2$/$I2$/$I3$ situation, since +that's our only handle on current at the moment. Looping +counter-clockwise from $a$ through wires 1 and 2, +\begin{align} + 0 &= -I_1R_1 + I_2R_2 \\ + I_2 &= \frac{R_1}{R_2} I_1 = \frac{25.0\U{\Ohm}}{15.0\U{\Ohm}}1.25\U{A} + = 2.08\U{A} \;. +\end{align} +Looping counter-clockwise from $a$ through wires 1 and 3, +\begin{align} + 0 &= -I_1R_1 + I_3R_{3a} + I_3R_{3b} = -I_1R_1 + I_3(R_{3a}+R_{3b}) \\ + I_2 &= \frac{R_1}{R_{3a}+R_{3b}} I_1 + = \frac{25.0\U{\Ohm}}{(15.0+10.0)\U{\Ohm}}1.25\U{A} + = 1.25\U{A} \;. +\end{align} + +Now that we know all about that little cluster, we can move out to the +larger loop. Using the junction rule at junction $a$ +\begin{align} + 0 &= I_t - I_1 - I_2 - I_3 \\ + I_t &= I_1 + I_2 + I_3 = 4.58\U{A} \;. +\end{align} +Knowing the current in the larger loop, we just use Ohm's law on $R_V$ +\begin{equation} + V = I_t R_V = \ans{206\U{V}} +\end{equation} + +\Part{b} +For this part, we do the full loop, heading counter clockwise from $a$ +and using wire 1 (although you could use any of wires 1, 2, or 3, +since they must all have the same voltage across them) +\begin{align} + 0 &= -I_1R_1 - I_tR_V - I_tR_\EMF + \EMF \\ + \EMF &= I_1R_1 + I_tR_V + I_tR_\EMF + = 1.25\U{A}\cdot25.0\U{\Ohm} + 206\U{V} + 4.58\U{A}\cdot35.0\U{\Ohm} + = \ans{398\U{V}} +\end{align} +\end{solution} diff --git a/latex/problems/Young_and_Freedman_12/problem26.21.tex b/latex/problems/Young_and_Freedman_12/problem26.21.tex new file mode 100644 index 0000000..29013b2 --- /dev/null +++ b/latex/problems/Young_and_Freedman_12/problem26.21.tex @@ -0,0 +1,69 @@ +\begin{problem*}{26.21} +In the circuit shown in Fig.~26.49 find \Part{a} the current in +resistor $R$; \Part{b} the resistance $R$; \Part{c} the unknown emf +\EMF. \Part{d} If the circuit is broken at point $x$, what is the +current in resistor $R$? +\end{problem*} + +\begin{nosolution} +\begin{center} +\begin{verbatim} + 28.0V R ++----|i-----/\/\/-------+ +| E 6.00 4.00A | ++----|i--x--/\/\/---<---+ +| 3.00 6.00A | ++---------/\/\/---->----+ +\end{verbatim} +\end{center} +\end{nosolution} + +\begin{solution} +\begin{center} +\begin{verbatim} + 28.0V R I_1 ++----|i-----/\/\/---<-------+ +| E 6.00 4.00A=I_2 | ++----|i--x--/\/\/---<-------a +| 3.00 6.00A=I_3 | ++---------/\/\/---->--------+ +\end{verbatim} +\end{center} +\Part{a} +Labeling the currents as above, we use Kirchoff's junction rule +summing the currents entering node $a$. +\begin{align} + 0 &= -I_1 - I_2 + I_3 & + I_1 &= I_3 - I_2 = \ans{2.00\U{A}} +\end{align} + +\Part{b} +We'll use a loop rule for this part. We still don't know \EMF in wire +2, so lets loop around 1 and 3 counter-clockwise starting from $a$. +\begin{align} + 0 &= -I_1 R + 28.0\U{V} - I_3 R_3 \\ + R &= \frac{28.0\U{V} - I_3 R_3}{I_1} + = \frac{28.0\U{V} - 6.00\U{A}\cdot3.00\U{\Ohm}}{2.00\U{A}} + = \ans{5.00\U{\Ohm}} +\end{align} + +\Part{c} +Now we'll use a loop involving wire 2, say 2 and 3 counter-clockwise +starting from $a$. +\begin{align} + 0 &= - I_2 R_2 + \EMF - I_3 R_3 \\ + \EMF &= I_2 R_2 + I_3 R_3 + = 4.00\U{A}\cdot6.00\U{\Ohm} + 6.00\U{A}\cdot3.00\U{\Ohm} + = \ans{42.0\U{V}} +\end{align} + +\Part{d} +Breaking the circuit at $x$ means that $I_2\rightarrow0$ so $I_1=I_2=I$. +Applying our same loop rule from \Part{b} +\begin{align} + 0 &= -I R + 28.0\U{V} - I R_3 \\ + I &= \frac{28.0\U{V}}{R+R_3} + = \frac{28.0\U{V}}{8.00\U{\Ohm}} + = \ans{3.50\U{A}} +\end{align} +\end{solution} diff --git a/latex/problems/Young_and_Freedman_12/problem26.22.tex b/latex/problems/Young_and_Freedman_12/problem26.22.tex new file mode 100644 index 0000000..c2ec360 --- /dev/null +++ b/latex/problems/Young_and_Freedman_12/problem26.22.tex @@ -0,0 +1,54 @@ +\begin{problem*}{26.22} +Find the emfs $\EMF_1$ and $\EMF_2$ in the circuit of Fig.~26.50, and +find the potential difference of point $b$ relative to point $a$. +\begin{center} +\begin{verbatim} + 1.00 20.0V 6.00 + +---/\/\/--|i----/\/\/---+ +1.00A v 4.00 1.00 E1 | + a---/\/\/----/\/\/--|i---b +2.00A v 1.00 E2 2.00 | + +---/\/\/--|i----/\/\/---+ +\end{verbatim} +\end{center} +\end{problem*} + +\begin{solution} +First we'll simplify resistors in series, leading to the equivalent +circuit +\begin{center} +\begin{verbatim} +I_3=1.00A 20.0V 7.00=R_3 ++---<-------|i----/\/\/---+ +| I_1 R_1=5.00 E1 | +a---<--------/\/\/--|i----b +| I_2=2.00A E2 3.00=R_2| ++--->-------|i----/\/\/---+ +\end{verbatim} +\end{center} +We know everything about wire 3, so $V_{ba}$ is given by +\begin{equation} + V_{ba} = -20.0\U{V} + I_3R_3 + = -20.0\U{V} + 1.00\U{A}\cdot7.00\U{\Ohm} + = \ans{-13.0\U{V}} +\end{equation} + +We also know everything about wire 2, so we can find $\EMF_2$ by +looping clockwise from $a$ through 3 and 2. +\begin{align} + 0 &= V_{ba} + I_2R_2 + \EMF_2 \\ + \EMF_2 &= -(V_{ba}+I_2R_2) = \ans{7.00\U{V}} +\end{align} + +To find $\EMF_1$ we'll need $I_1$. Applying Kirchoff's junction rule +to $a$ +\begin{align} + 0 &= I_3 + I_1 - I_2 \\ + I_1 &= I_2 - I_3 = 2.00\U{A} - 1.00\U{A} = 1.00\U{A} \;. +\end{align} +Looping clockwise from $a$ through 3 and 1 +\begin{align} + 0 &= V_{ba} + \EMF_1 - I_1R_1 \\ + \EMF_1 &= I_1R_1 - V_{ba} = \ans{18.0\U{V}} +\end{align} +\end{solution} diff --git a/latex/problems/Young_and_Freedman_12/problem26.48.tex b/latex/problems/Young_and_Freedman_12/problem26.48.tex new file mode 100644 index 0000000..a75d8d4 --- /dev/null +++ b/latex/problems/Young_and_Freedman_12/problem26.48.tex @@ -0,0 +1,58 @@ +\begin{problem*}{26.48} +In the circuit shown in Fig.~26.61, $C=5.90\U{$\mu$F}$, $\EMF = +28.0\U{V}$, and the emf has negligible resistance. Initially the +capacitor is uncharged and the switch $S$ is in position 1, The switch +is then moved to position 2, so that the capacitor begins to +charge. \Part{a} What will be the charge on the capacitor a long time +after the switch is moved to position 2? \Part{b} After the switch +has been in position 2 for $3.00\U{ms}$, the charge on the capacitor +is measured to be $110\U{$\mu$C}$. What is the value of the +resistance $R$? \Part{c} How long after the switch is moved to +position 2 will the charge on the capacitor be equal to $99.0\%$ of +the final value found in \Part{a}. +\begin{center} +\begin{verbatim} + +---------. + | / + | S1 S2 + | | | +=== C | --- E + | | - + | | | + +-/\/\-+----+ + R +\end{verbatim} +\end{center} +\end{problem*} + +\begin{solution} +\Part{a} +After a long time in position 2, the capacitor will have become fully +charged, no more current will flow through the circuit, and the +voltage across the capacitor will balance that across the battery. +\begin{equation} + Q = CV = CE = 5.90\U{$\mu$F}\cdot28.0\U{V} = \ans{165\U{$\mu$C}} +\end{equation} + +\Part{b} +Here we use the charging capacitor formula (Eqn.~26.12) +\begin{align} + q &= CE\p({1-e^{-t/RC}}) \\ + \frac{q}{CE} &= 1-e^{-t/RC} \\ + e^{-t/RC} &= 1-\frac{q}{CE} \\ + \frac{-t}{RC} &= \ln\p({1-\frac{q}{CE}}) \\ + R &= \frac{-t}{C\ln\p({1-\frac{q}{CE}})} + = \frac{-3.00\U{ms}} + {5.90\U{$\mu$F}\cdot\ln\p({1-\frac{110\U{$\mu$C}}{165\U{$\mu$C}}})} + = \ans{463\U{\Ohm}} +\end{align} + +\Part{c} +Again, we use the charging-capacitor formula +\begin{align} + \frac{-t}{RC} &= \ln\p({1-\frac{q}{CE}}) \\ + t &= -RC\ln\p({1-\frac{q}{CE}}) + = -463\U{\Ohm}\cdot5.90\U{$\mu$F}\cdot\ln(1-0.99) + = \ans{12.6\U{ms}} +\end{align} +\end{solution} -- 2.26.2