--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
projects / course.git / commitdiff