* 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.
\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}
\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 |
% | 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}
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
% 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}
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}
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}
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