From: W. Trevor King Date: Mon, 7 May 2012 20:55:27 +0000 (-0400) Subject: Add solutions and graphics to Serway and Jewett v8's chapter 28 problems. X-Git-Url: http://git.tremily.us/?a=commitdiff_plain;h=92cdb6762e6fe2f7d3558abf8555cedbc1ee623d;p=course.git Add solutions and graphics to Serway and Jewett v8's chapter 28 problems. * Use \EMF instead of \varepsilon. * Graphics and solutions added to 28.2, 28.11, 28.21, 28.26, 28.59. * Assorted typos fixed in the problem text. --- diff --git a/latex/problems/Serway_and_Jewett_8/problem28.02.tex b/latex/problems/Serway_and_Jewett_8/problem28.02.tex index e5c10a1..44f22df 100644 --- a/latex/problems/Serway_and_Jewett_8/problem28.02.tex +++ b/latex/problems/Serway_and_Jewett_8/problem28.02.tex @@ -1,12 +1,65 @@ \begin{problem*}{28.2} Two $1.50\U{V}$ batteries---with their positive terminal in the same direction---are inserted in series into a flashlight. One battery has -an internal resistance of $0.255\U{\Ohm}$, annd the other has an +an internal resistance of $0.255\U{\Ohm}$, and the other has an internal resistance of $0.153\U{\Ohm}$. When the switch is closed, the bulb carries a current of $600\U{mA}$. \Part{a} What is the -bulb's reistance? \Part{b} What fraction of the chemical energy +bulb's resistance? \Part{b} What fraction of the chemical energy tranformed appears as internal energy in the batteries? \end{problem*} \begin{solution} +\begin{center} +\begin{asy} +import Circ; + +real u = 1.5cm; +pair dx = (.1u, 0); + +MultiTerminal B1 = battery(label="$V_1$", value="$1.50\U{V}$"); +MultiTerminal R1 = resistor( + B1.terminal[1] + dx, label="$r_1$", value="$0.255\U{\Ohm}$"); +MultiTerminal B2 = battery( + R1.terminal[1] + dx, label="$V_2$", value="$1.50\U{V}$"); +MultiTerminal R2 = resistor( + B2.terminal[1] + dx, label="$r_2$", value="$0.153\U{\Ohm}$"); +wire(B1.terminal[1], R1.terminal[0]); +wire(R1.terminal[1], B2.terminal[0]); +wire(B2.terminal[1], R2.terminal[0]); +MultiTerminal S = switchSPST(type=open, draw=false); +S.centerto(B2.terminal[0], R2.terminal[1], offset=-u); S.draw(); +MultiTerminal L = lamp(type=illuminating, draw=false); +L.centerto(B1.terminal[0], R1.terminal[1], offset=-u); L.draw(); +wire(R2.terminal[1], S.terminal[1], rlsq, 0.1u); +wire(S.terminal[0], L.terminal[1]); +wire(B1.terminal[0], L.terminal[0], rlsq, -0.1u); +\end{asy} +\end{center} +\Part{a} +Summing the voltages around the loop and using Ohm's law, we have +\begin{align} + 0 &= V_1 - Ir_1 + V_2 - Ir_2 - IR \\ + I (r_1 + r_2 + R) &= V_1 + V_2 \\ + R &= \frac{V_1 + V_2}{I} - r_1 - r_2 + = \frac{1.50\U{V} + 1.50\U{V}}{600\U{mA}} - 0.255\U{\Ohm} - 0.153\U{\Ohm} + = \ans{4.59\U{\Ohm}} \;. +\end{align} + +\Part{b} +The total chemical energy tranformed is +\begin{equation} + P_\text{chem} = I(V_1+V_2) \;. +\end{equation} +The energy lost to the batteries' internal resistance is +\begin{equation} + P_i = I^2(r_1+r_2) \;, +\end{equation} +so the fraction lost is +\begin{equation} + \frac{P_i}{P_\text{chem}} = \frac{I^2(r_1+r_2)}{I(V_1+V_2)} + = \frac{r_1+r_2}{V_1 + V_2} \cdot I + = \frac{0.255\U{\Ohm} + 0.153\U{\Ohm}}{1.50\U{V} + 1.50\U{V}} + \cdot 600\U{mA} + = \ans{0.0816} = \ans{8.16\%} \;. +\end{equation} \end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem28.11.tex b/latex/problems/Serway_and_Jewett_8/problem28.11.tex index f0f5114..02e6574 100644 --- a/latex/problems/Serway_and_Jewett_8/problem28.11.tex +++ b/latex/problems/Serway_and_Jewett_8/problem28.11.tex @@ -1,12 +1,12 @@ \begin{problem*}{28.11} -A battery with $\varepsilon=6.00\U{V}$ and no internal resistance -supplies current to the circuit shown in Figure~P28.11. When the -double-throw switch $S$ is open as shown in the figure, the current in -the battery is $1.00\U{mA}$. When the switch is closed in position -$a$, the current in the battery is $1.20\U{mA}$. When the switch is -closed in position $b$, the current in the battery is $2.00\U{mA}$. -Find the resistances \Part{a} $R_1$, \Part{b} $R_2$, and \Part{c} -$R_3$. +A battery with $\EMF=6.00\U{V}$ and no internal resistance supplies +current to the circuit shown in Figure~P28.11. When the double-throw +switch $S$ is open as shown in the figure, the current in the battery +is $1.00\U{mA}$. When the switch is closed in position $a$, the +current in the battery is $1.20\U{mA}$. When the switch is closed in +position $b$, the current in the battery is $2.00\U{mA}$. Find the +resistances \Part{a} $R_1$, \Part{b} $R_2$, and \Part{c} $R_3$. +\begin{center} % +---R1---+----R2----+ % | | | % | R2 | @@ -15,8 +15,104 @@ $R_3$. % | b R3 % | | | % +--------+----------+ +\begin{asy} +import Circ; + +MultiTerminal S = switchSPDT(dir=180, label=Label("$S$", align=dir(-70))); +label("$a$", S.terminal[2], align=W); +label("$b$", S.terminal[1], align=W); +MultiTerminal R2v = resistor(S.terminal[2], dir=90, "$R_2$"); +MultiTerminal R1 = resistor( + R2v.terminal[1], dir=180, label=Label("$R_1$", align=N)); +MultiTerminal R2h = resistor(R2v.terminal[1], "$R_2$"); +MultiTerminal R3 = resistor( + (R2h.terminal[1].x, S.terminal[0].y), dir=-90, "$R_3$"); +MultiTerminal B = battery(label=Label("$\EMF$", align=W), draw=false); +B.centerto(R1.terminal[1], (R1.terminal[1].x, R3.terminal[1].y)); +B.draw(); +wire(R1.terminal[1], B.terminal[0]); +wire(B.terminal[1], R3.terminal[1], udsq); +pair bx = (S.terminal[1].x, R3.terminal[1].y); +wire(S.terminal[1], bx); +dot(bx); +wire(S.terminal[0], R3.terminal[0]); +wire(R3.terminal[0], R2h.terminal[1]); +dot(R3.terminal[0]); +dot(R2v.terminal[1]); +\end{asy} +\end{center} \end{problem*} \begin{solution} +When the switch is in position $b$, the vertical $R_2$ resistor +recieves no current. Applying Kirchhoff's loop rule gives +\begin{align} + 0 &= \EMF - I R_3 - I R_2 - I R_1 \\ + \frac{\EMF}{I} &= R_1 + R_2 + R_3 \;. +\end{align} + +When the switch is in position $a$, the two $R_2$ resistors are in +parallel, so they can be replaced by an equivalent resistance +\begin{equation} + R_2' = (1/R_2 + 1/R_2)^{-1} = R_2/2 \;. +\end{equation} +After you've made this replacement, there is only a single loop in the +circuit. Applying Kirchhoff's loop rule gives +\begin{align} + 0 &= \EMF - I_a R_3 - I_a R_2' - I_a R_1 \\ + \frac{\EMF}{I_a} &= R_1 + \frac{R_2}{2} + R_3 \;. +\end{align} + +When the switch is in position $b$, the vertical $R_2$ resistor +recieves no current and $R_3$ is shorted. Kirchoff's loop rule gives +\begin{align} + 0 &= \EMF - I_b R_2 - I_b R_1 \\ + \frac{\EMF}{I_b} &= R_1 + R_2 \;. +\end{align} +This gives three equations with three unknowns. Solve however you +like. +\begin{align} + \begin{pmatrix} + \frac{6.00\U{V}}{1.00\E{-3}\U{A}} \\ + \frac{6.00\U{V}}{1.20\E{-3}\U{A}} \\ + \frac{6.00\U{V}}{2.00\E{-3}\U{A}} + \end{pmatrix} + &= + \begin{pmatrix} + 1 & 1 & 1 \\ + 1 & \frac{1}{2} & 1 \\ + 1 & 1 & 0 + \end{pmatrix} + \begin{pmatrix} + R_1 \\ + R_2 \\ + R_3 + \end{pmatrix} \\ + \begin{pmatrix} + R_1 \\ + R_2 \\ + R_3 + \end{pmatrix} + =& + \begin{pmatrix} + 1 & 1 & 1 \\ + 1 & \frac{1}{2} & 1 \\ + 1 & 1 & 0 + \end{pmatrix}^{-1} + \begin{pmatrix} + 6.00\U{k\Ohm} \\ + 5.00\U{k\Ohm} \\ + 3.00\U{k\Ohm} + \end{pmatrix} + = + \ans{ + \begin{pmatrix} + 1.00 \\ + 2.00 \\ + 3.00 + \end{pmatrix} + \U{k\Ohm} + } \;. +\end{align} \end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem28.21.tex b/latex/problems/Serway_and_Jewett_8/problem28.21.tex index d3cdf8d..3fec279 100644 --- a/latex/problems/Serway_and_Jewett_8/problem28.21.tex +++ b/latex/problems/Serway_and_Jewett_8/problem28.21.tex @@ -6,6 +6,7 @@ battery. \Part{c} Find the energy delivered to each resistor. \Part{d} Identify the type of energy storage transformation that occurs in the operation of the circuit. \Part{e} Find the total amount of energy transformed into internal energy in the resistors. +\begin{center} % +-----+----3O--+ % | 5O | % | | 1O @@ -14,7 +15,204 @@ amount of energy transformed into internal energy in the resistors. % 8O ___ ___ % | - 4V - 12V % +-----+--------+ +\begin{asy} +import Circ; + +real u = 2.3cm; + +MultiTerminal B1 = battery(dir=90, label="$V_1$", value="$4.00\U{V}$"); +MultiTerminal R3 = resistor( + B1.terminal[1], dir=90, label="$R_3$", value="$1.00\U{\Ohm}$"); +MultiTerminal R2 = resistor( + R3.terminal[1], dir=90, label="$R_2$", value="$5.00\U{\Ohm}$"); +real rlen = R2.terminal[1].y - R2.terminal[0].y; +MultiTerminal R1 = resistor(label="$R_1=8.00\U{\Ohm}$", draw=false); +R1.centerto(B1.terminal[0], R2.terminal[1], offset=rlen); R1.draw(); +MultiTerminal R4 = resistor(R2.terminal[1], label="$R_4=3.00\U{\Ohm}$"); +MultiTerminal B2 = battery( + label=Label("$V_2=12.0\U{V}$", align=E), draw=false); +two_terminal_centerto(B1, B2, offset=-u); B2.draw(); +MultiTerminal R5 = resistor( + label=Label("$R_5=1.00\U{\Ohm}$", align=E), draw=false); +R5.centerto(B2.terminal[1], (B2.terminal[1].x, R4.terminal[1].y)); R5.draw(); +wire(R1.terminal[0], B2.terminal[0], udsq); +dot(B1.terminal[0]); +wire(R1.terminal[1], R4.terminal[0], udsq); +dot(R2.terminal[1]); +wire(R4.terminal[1], R5.terminal[1], rlsq); +wire(B2.terminal[1], R5.terminal[0]); +\end{asy} +\end{center} \end{problem*} \begin{solution} +\Part{a} +Label the currents $I_1$, $I_2$, and $I_3$ from left to right with +each current moving up in its vertical wire. Applying the Kirchhoff's +junction rule to the top junction. +\begin{center} +\begin{asy} +import Circ; + +real dx = 6pt; + +MultiTerminal I1 = current((0, 0), label="$I_1$"); +real ilen = I1.terminal[1].x - I1.terminal[0].x; +pair P = I1.terminal[1] + (dx, 0); +MultiTerminal I2 = current(P - (0, dx), dir=90, label="$I_2$", draw=false); +I2.shift((0, -ilen-dx)); +I2.draw(); +MultiTerminal I3 = current(P + (dx, 0), dir=180, label="$I_3$", draw=false); +I3.shift((ilen, 0)); +I3.draw(); + +wire(I1.terminal[1], I3.terminal[1]); +wire(P, I2.terminal[1]); +dot(P); +\end{asy} +\end{center} +\begin{equation} + 0 = I_1 + I_2 + I_3 \;. +\end{equation} + +Applying Kirchhoff's loop rule to the left-hand loop moving clockwise +from the lower-left corner, we have +\begin{center} +\begin{asy} +import Circ; + +real u = 2cm; + +wire((0, 0), (2u, u), rlsq); +wire((0, 0), (2u, u), udsq); +wire((u, 0), (u, u)); +dot((u, 0)); +dot((u, u)); + +MultiTerminal I1 = current(label="$I_1$", draw=false); +I1.centerto((0, 0), (0, u)); +I1.draw(); +MultiTerminal I2 = current(label="$I_2$", draw=false); +I2.centerto((u, 0), (u, u)); +I2.draw(); +MultiTerminal I3 = current(label="$I_3$", draw=false); +I3.centerto((2u, 0), (2u, u)); +I3.draw(); + +pair[] points = {(0, 0), (0, u), (u, u), (u, 0)}; +kirchhoff_loop(points); +\end{asy} +\end{center} +\begin{align} + 0 &= -I_1 R_1 + I_2 R_2 + I_2 R_3 - V_1 \\ + V_1 &= -R_1 I_1 + (R_2 + R_3) I_2 \;. +\end{align} + +Applying Kirchhoff's loop rule to the right-hand loop moving clockwise +from the lower-right corner, we have +\begin{center} +\begin{asy} +import Circ; + +real u = 2cm; + +wire((0, 0), (2u, u), rlsq); +wire((0, 0), (2u, u), udsq); +wire((u, 0), (u, u)); +dot((u, 0)); +dot((u, u)); + +pair[] points = {(2u, 0), (u, 0), (u, u), (2u, u)}; +kirchhoff_loop(points); +\end{asy} +\end{center} +\begin{align} + 0 &= V_1 - I_2 R_3 - I_2 R_2 + I_3 R_4 + I_3 R_5 - V_2 \\ + V_2 - V_1 &= -(R_2 + R_3) I_2 + (R_4 + R_5) I_3 \;. +\end{align} + +This gives three equations with three unknowns. Solve however you +like. +\begin{align} + \begin{pmatrix} + 0 \\ + V_1 \\ + V_2 - V_1 \\ + \end{pmatrix} + &= + \begin{pmatrix} + 1 & 1 & 1 \\ + -R_1 & R_2 + R_3 & 0 \\ + 0 & -(R_2 + R_3) & R_4 + R_5 + \end{pmatrix} + \begin{pmatrix} + I_1 \\ + I_2 \\ + I_3 + \end{pmatrix} \\ + \begin{pmatrix} + I_1 \\ + I_2 \\ + I_3 + \end{pmatrix} + =& + \begin{pmatrix} + 1\U{\Ohm} & 1\U{\Ohm} & 1\U{\Ohm} \\ + -8.00\U{\Ohm} & 6.00\U{\Ohm} & 0 \\ + 0 & -6.00\U{\Ohm} & 4.00\U{\Ohm} + \end{pmatrix}^{-1} + \begin{pmatrix} + 0\U{V} \\ + 4.00\U{V} \\ + 8.00\U{V} + \end{pmatrix} + = + \ans{ + \begin{pmatrix} + -846\U{mA} \\ + -462\U{mA} \\ + 1.31\U{A} + \end{pmatrix} + } \;. +\end{align} +The fact that $I_1$ and $I_2$ are negative means that the current +flows in the opposite direction to the direction we picked for those +branches. Since we picked all the currents to flow up, the current in +the left and center branches actually flows down. + +\Part{b} +The power delivered by a battery (like any other circuit element) is +given by $P=IV$. The energy delivered by each battery is +\begin{align} + U_{V_1} &= P_1 \Delta t = I_2 V_1 \Delta t + = (-462\E{-3}\U{A})\cdot(4.00\U{V})\cdot + \p({2.00\U{min}\cdot\frac{60\U{s}}{1\U{min}}}) + = \ans{-222\U{J}} \\ + U_{V_2} &= P_2 \Delta t = I_3 V_2 \Delta t + = (1.31\U{A})\cdot(12.0\U{V})\cdot + \p({2.00\U{min}\cdot\frac{60\U{s}}{1\U{min}}}) + = \ans{1.88\U{kJ}} \;. +\end{align} +$U_{V_1}$ is negative because $V_1$ is absorbing energy as it is +charged by $I_2$. + +\Part{c} +The power delivered to a resistor is $P=IV=I^2R$. +\begin{align} + U_{R_1} &= I_1^2 R_1 \Delta t = \ans{687\U{J}} \\ + U_{R_2} &= I_2^2 R_2 \Delta t = \ans{127\U{J}} \\ + U_{R_3} &= I_2^2 R_3 \Delta t = \ans{25.6\U{J}} \\ + U_{R_4} &= I_3^2 R_4 \Delta t = \ans{616\U{J}} \\ + U_{R_5} &= I_3^2 R_5 \Delta t = \ans{205\U{J}} \;. +\end{align} + +\Part{d} +Chemical energy in $V_2$ is being converted into internal energy +(heat) in the resistors and increased chemical energy in $V_1$. + +\Part{e} +\begin{equation} + U_R = U_{R_1} + U_{R_2} + U_{R_3} + U_{R_4} + U_{R_5} + = \ans{1.66\U{kJ}} \;. +\end{equation} \end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem28.26.tex b/latex/problems/Serway_and_Jewett_8/problem28.26.tex index 0d33b3c..81162d1 100644 --- a/latex/problems/Serway_and_Jewett_8/problem28.26.tex +++ b/latex/problems/Serway_and_Jewett_8/problem28.26.tex @@ -6,9 +6,116 @@ The following equations describe an electric circuit: I_1 + I_3 - I_2 &= 0 \end{align} \Part{a} Draw a [possible] diagram of the circuit. \Part{b} -Calculate the unknowns sand identify the physical meaning of each +Calculate the unknowns and identify the physical meaning of each unknown. \end{problem*} \begin{solution} +\Part{a} +The last equation looks like a junction rule, for a junction like +this: +\begin{center} +\begin{asy} +import Circ; + +real dx = 6pt; + +MultiTerminal I1 = current((0, 0), label="$I_1$"); +real ilen = I1.terminal[1].x - I1.terminal[0].x; +pair P = I1.terminal[1] + (dx, 0); +MultiTerminal I2 = current(P, dir=-90, label=Label("$I_2$", align=dir(-20))); +MultiTerminal I3 = current( + P + (dx, 0), dir=180, label=Label("$I_3$", align=N), draw=false); +I3.shift((ilen, 0)); +I3.draw(); + +wire(I1.terminal[1], I3.terminal[1]); +wire(P, I2.terminal[1]); +dot(P); +\end{asy} +\end{center} + +The first equation looks like a loop rule traveling over the $I_1$ and +$I_2$ branches, and the second equation looks like a loop rule +traveling over the $I_2$ and $I_3$ branches. Luckily, the $I_2$ +portions match between these equations, and we can fill in the circuit +elements. +\begin{center} +\begin{asy} +import Circ; + +real u = 2cm; + +MultiTerminal R1 = resistor(dir=90, value="$220\U{\Ohm}$"); +MultiTerminal V1 = battery(R1.terminal[1], dir=90, value="$5.80\U{V}$"); +MultiTerminal R2 = resistor(value="$370\U{\Ohm}$", draw=false); +R2.centerto(R1.terminal[0], V1.terminal[1], offset=-u); R2.draw(); +pair a = (R2.center.x, V1.terminal[1].y); +pair b = (a.x, R1.terminal[0].y); +MultiTerminal R3 = resistor( + a+(u,0), dir=-90, value=Label("$150\U{\Ohm}$", align=E)); +MultiTerminal V3 = battery(b+(u,0), dir=90, value="$3.10\U{V}$"); +MultiTerminal I1 = current(label="$I_1$", draw=false); +I1.centerto(V1.terminal[1], a); I1.draw(); +MultiTerminal I2 = current(label="$I_2$", draw=false); +I2.centerto(a, R2.terminal[1]); I2.draw(); +MultiTerminal I3 = current(label=Label("$I_3$", align=N), draw=false); +I3.centerto(R3.terminal[0], a); I3.draw(); +wire(V1.terminal[1], I1.terminal[0]); +wire(I1.terminal[1], I2.terminal[0], rlsq); +wire(I2.terminal[1], R2.terminal[1]); +wire(a, I3.terminal[1]); +wire(I3.terminal[0], R3.terminal[0]); +dot(a); +wire(R1.terminal[0], R2.terminal[0], rlsq); +wire(b, V3.terminal[0]); +dot(b); +\end{asy} +\end{center} + +\Part{b} +The unknowns are the currents through each branch of the circuit. You +can solve for them however you like. +\begin{align} + \begin{pmatrix} + 5.80\U{V} \\ + 3.10\U{V} \\ + 0 + \end{pmatrix} + &= + \begin{pmatrix} + 220\U{\Ohm} & 370\U{\Ohm} & 0 \\ + 0 & 370\U{\Ohm} & 150\U{\Ohm} \\ + 1\U{\Ohm} & -1\U{\Ohm} & 1\U{\Ohm} + \end{pmatrix} + \begin{pmatrix} + I_1 \\ + I_2 \\ + I_3 + \end{pmatrix} \\ + \begin{pmatrix} + I_1 \\ + I_2 \\ + I_3 + \end{pmatrix} + =& + \begin{pmatrix} + 220\U{\Ohm} & 370\U{\Ohm} & 0 \\ + 0 & 370\U{\Ohm} & 150\U{\Ohm} \\ + 1\U{\Ohm} & -1\U{\Ohm} & 1\U{\Ohm} + \end{pmatrix}^{-1} + \begin{pmatrix} + 5.80\U{V} \\ + 3.10\U{V} \\ + 0 + \end{pmatrix} + = + \ans{ + \begin{pmatrix} + 11.0 \\ + 9.10 \\ + -1.87 + \end{pmatrix} \U{mA} + } \;. +\end{align} \end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem28.59.tex b/latex/problems/Serway_and_Jewett_8/problem28.59.tex index af6fe20..3c61fc8 100644 --- a/latex/problems/Serway_and_Jewett_8/problem28.59.tex +++ b/latex/problems/Serway_and_Jewett_8/problem28.59.tex @@ -5,13 +5,89 @@ then closed. After $1.00\U{ms}$ have elapsed, \Part{a} how much charge remains on the $3.00\U{$\mu$F}$ capacitor? \Part{b} How much charge remains on the $2.00\U{$\mu$F}$ capacitor? \Part{c} What is the current in the resistor at this time? +\begin{center} % +---||---+ % | 3 | % +---||---+ % S 2 | % | | % +--5O----+ +\begin{asy} +import Circ; + +real u = 1.5cm; + +MultiTerminal C1 = capacitor(label="$C_1$", value="$3.00\U{$\mu$F}$"); +MultiTerminal C2 = capacitor( + label="$C_2$", value="$2.00\U{$\mu$F}$", draw=false); +two_terminal_centerto(C1, C2, offset=-u); C2.draw(); +MultiTerminal R = resistor(label="$R$", value="$500\U{\Ohm}$", draw=false); +two_terminal_centerto(C2, R, offset=-u); R.draw(); +pair a = C2.center - (u,0); +pair b = C2.center + (u,0); +MultiTerminal S = switchSPST(type=open, draw=false); +S.centerto(a, (a.x, R.terminal[0].y)); S.draw(); +wire(S.terminal[0], C1.terminal[0], udsq); +wire(a, C2.terminal[0]); +dot(a); +wire(S.terminal[1], R.terminal[0], udsq); +wire(R.terminal[1], b, rlsq); +wire(b, C1.terminal[1], udsq); +wire(b, C2.terminal[1]); +dot(b); +\end{asy} +\end{center} \end{problem*} \begin{solution} +The capacitors $C_1$ and $C_2$ are in parallel, so we can replace them +with an equivalent capacitance $C=C_1+C_2$. +\begin{center} +\begin{asy} +import Circ; + +real u = 1.5cm; + +MultiTerminal C = capacitor(label="$C$"); +MultiTerminal R = resistor(label="$R$", draw=false); +two_terminal_centerto(C, R, offset=-u); R.draw(); +wire(R.terminal[0], C.terminal[0], udsq); +wire(R.terminal[1], C.terminal[1], udsq); +\end{asy} +\end{center} +Now this looks like a standard discharging capacitor, so +\begin{align} + q &= q_0 e^{\frac{-t}{RC}} = CV_0 e^{\frac{-t}{RC}} \\ + V &= V_0 e^{\frac{-t}{RC}} \\ + I &= I_0 e^{\frac{-t}{RC}} = \frac{V_0}{R} e^{\frac{-t}{RC}} +\end{align} + +\Part{a} +To find $q_1(t=1.00\U{ms})$, we'll use the capacitor equation. +\begin{equation} + q_1(t=1.00\U{ms}) = C_1 V_1(t=1.00\U{ms}) + = C_1 V(t=1.00\U{ms}) + = C_1 V_0 e^{\frac{-1.00\U{ms}}{RC}} + = 3.00\U{$\mu$F} \cdot 12.0\U{V} + \cdot e^{\frac{-1.00\E{-3}\U{s}}{500\U{\Ohm}\cdot(3.00+2.00)\U{$\mu$F}}} + = \ans{24.1\U{$\mu$C}} \;. +\end{equation} + +\Part{b} +This is the same as \Part{a}, with $q_1 \rightarrow q_2$ and $C_1 +\rightarrow C_2$. +\begin{equation} + q_2(t=1.00\U{ms}) = C_2 V_0 e^{\frac{-1.00\U{ms}}{RC}} + = 2.00\U{$\mu$F} \cdot 12.0\U{V} + \cdot e^{\frac{-1.00\E{-3}\U{s}}{500\U{\Ohm}\cdot(3.00+2.00)\U{$\mu$F}}} + = \ans{16.1\U{$\mu$C}} \;. +\end{equation} + +\Part{c} +\begin{equation} + I(t=1.00\U{ms}) = \frac{V_0}{R} e^{\frac{-t}{RC}} + = \frac{12.0\U{V}}{500\U{\Ohm}} + \cdot e^{\frac{-1.00\E{-3}\U{s}}{500\U{\Ohm}\cdot(3.00+2.00)\U{$\mu$F}}} + = \ans{16.1\U{mA}} \;. +\end{equation} \end{solution}