1 \begin{problem*}{28.71}
2 In Figure~P28.71, suppose the switch has been closed for a time
3 interval sufficiently long for the capacitor to become fully charged.
4 Find \Part{a} the steady state current in each resistor and \Part{b}
5 the charge $Q$ on the capacitor. \Part{c} The switch is now opened at
6 $t=0$. Write an equation for the current in $R_2$ as a function of
7 time and \Part{d} find the time interval required for the charge on
8 the capacitor to fall to one-fifth its initial value.
11 % +---S----12kO--+-----+
17 % +--------------+-----+
24 MultiTerminal S = switchSPST(label="$S$");
25 MultiTerminal R1 = resistor(
26 S.terminal[1], label="$R_1$", value="$12.0\U{k\Ohm}$");
27 pair Pu = R1.terminal[1] + (dx,0); // top junction
28 pair Pul = S.terminal[0] - (dx,0); // upper-left corner
29 pair Pur = Pu + (2u,0); // upper-right corner
30 MultiTerminal C = capacitor(
31 Pur-(0,dy), dir=-90, label="$C$", value="$10.0\U{$\mu$F}$");
32 MultiTerminal R3 = resistor(
33 C.terminal[1]+(0,-dy), -90, label="$R_3$", value="$3.00\U{k\Ohm}$");
34 MultiTerminal R2 = resistor(
35 label="$R_2$", value="$15.0\U{k\Ohm}$", draw=false);
36 R2.centerto(R3.terminal[1], C.terminal[0], offset=Pur.x - Pu.x);
38 pair Pb = (R2.terminal[0].x, R3.terminal[1].y - dy);
39 MultiTerminal V = source(type=DC, label="$V$", value="$9.00\U{V}$", draw=false);
40 V.centerto(R3.terminal[1], C.terminal[0], offset=Pur.x - Pul.x);
43 wire(R1.terminal[1], R2.terminal[1], rlsq);
44 wire(Pu, C.terminal[0], rlsq);
45 wire(C.terminal[1], R3.terminal[0]);
46 wire(R3.terminal[1], Pb, udsq);
47 wire(R2.terminal[0], Pb);
48 wire(Pb, V.terminal[0], rlsq);
49 wire(V.terminal[1], S.terminal[0], udsq);
58 Label the currents $I_1$, $I_2$, and $I_3$ from left to right with
59 each current moving up in its vertical wire. In the steady state,
60 $I_3=\ans{0}$ (otherwise $Q$ would be changing, which would not be
61 ``steady state''). Applying the Kirchhoff's junction rule to the top
69 MultiTerminal I1 = current((0, 0), label="$I_1$");
70 real ilen = I1.terminal[1].x - I1.terminal[0].x;
71 pair P = I1.terminal[1] + (dx, 0);
72 MultiTerminal I2 = current(P - (0, dx), dir=90, label="$I_2$", draw=false);
73 I2.shift((0, -ilen-dx));
75 MultiTerminal I3 = current(P + (dx, 0), dir=180, label="$I_3$", draw=false);
79 wire(I1.terminal[1], I3.terminal[1]);
80 wire(P, I2.terminal[1]);
85 0 &= I_1 + I_2 + I_3 = I_1 + I_2 \\
89 Applying Kirchhoff's loop rule to the left-hand loop moving clockwise
90 from the lower-left corner, we have
97 wire((0, 0), (2u, u), rlsq);
98 wire((0, 0), (2u, u), udsq);
103 MultiTerminal I1 = current(label="$I_1$", draw=false);
104 I1.centerto((0, 0), (0, u));
106 MultiTerminal I2 = current(label="$I_2$", draw=false);
107 I2.centerto((u, 0), (u, u));
109 MultiTerminal I3 = current(label="$I_3$", draw=false);
110 I3.centerto((2u, 0), (2u, u));
113 pair[] points = {(0, 0), (0, u), (u, u), (u, 0)};
114 kirchhoff_loop(points);
118 0 &= V - I_1 R_1 + I_2 R_2 \\
119 V &= I_1 R_1 - I_2 R_2 = I_1 R_1 + I_1 R_2 = I_1 (R_1+R_2) \\
120 I_1 &= \frac{V}{R_1+R_2} = \ans{333\U{$\mu$A}} \\
121 I_2 &= -I_1 = \ans{-333\U{$\mu$A}} \;.
123 So we have $333\U{$\mu$A}$ of current flowing clockwise through the
124 left loop, and no current in the right loop.
127 Applying Kirchhoff's loop rule to the right-hand loop moving clockwise
128 from the lower-right corner, we have
135 wire((0, 0), (2u, u), rlsq);
136 wire((0, 0), (2u, u), udsq);
137 wire((u, 0), (u, u));
141 pair[] points = {(2u, 0), (u, 0), (u, u), (2u, u)};
142 kirchhoff_loop(points);
146 0 &= - I_2 R_2 - \frac{Q}{C} + I_3 R_3 \\
147 I_2 R_2 &= -\frac{Q}{C} \\
148 Q &= -C I_2 R_2 = -(10.0\U{$\mu$F})\cdot(-333\U{$\mu$A})\cdot(15.0\U{k\Ohm})
149 = \ans{50\U{$\mu$C}} \;.
151 You could jump to the final equation for $Q$ by using the capacitor
152 equation $Q=CV$ and noting that the voltage drop over the capacitor
153 must match the voltage drop $V=|I_2R_2|$ over $R_2$ (effectively doing
154 Kirchhoff's loop rule in your head).
156 The top capacitor plate will hold the positive charge, because when
157 the switch was first closed, the capacitor was completely discharged.
158 The current $I_1$ split and flowed down through both the middle and
159 right wire. As time went on, the charge $Q$ built up on the
160 capacitor, and current through the right wire slowed, stopping at
161 equilibrium with the capacitor fully charged.
164 With the switch opened, $I_1=0$. Resolving our earlier junction equation
166 0 &= I_1 + I_2 + I_3 = I_2 + I_3 \\
169 so current will flow up through the right-hand wire ($I_3>0$) and down
170 through the center wire ($I_2<0$) as the capacitor discharges.
171 Repeating our Kirchhoff loop from \Part{b},
173 0 &= - I_2 R_2 - \frac{q}{C} + I_3 R_3 \\
174 \frac{q}{C} &= (R_2+R_3)I_3 = R'I_3 \;,
176 where $R'\equiv R_2+R_3$ is shorthand to allow cleaner formulas for
177 the rest of the problem, and $q$ is the charge on the capacitor
180 As the capacitor discharges, $\deriv{t}{q}<0$, so $I_3=-\deriv{t}{q}$.
181 We can then solve for $q(t)$ by integrating with respect to time.
183 \frac{q}{C} &= -R'\deriv{t}{q} \\
184 \frac{-\dd t}{R'C} &= \frac{\dd q}{q} \\
185 \int_{t=0}^t \frac{-\dd t}{R'C} &= \int_{t=0}^t \frac{\dd q}{q} \\
186 \frac{-t}{R'C} &= \ln(q) - \ln(Q) = \ln\p({\frac{q}{Q}}) \;,
188 where $-\ln(Q)$ is a constant of integration which is determined by
189 the conditions at $t=0$. Raising $e$ to either side this equation, we
192 e^{\frac{-t}{R'C}} &= e^{\ln\p({\frac{q}{Q}})} = \frac{q}{Q} \\
193 q &= Q e^{\frac{-t}{R'C}} \\
195 = -\p({\frac{-1}{R'C}}) Q e^{\frac{-t}{R'C}}
196 = \frac{V_0}{R'} e^{\frac{-t}{R'C}} \;,
198 where $V_0=Q/C$ is the initial voltage across the capacitor. You can
199 see that both the charge on the capacitor and the current through the
200 loop will drop off exponentially with a time constant $R'C$ as the
203 Now we can put together our knowledge of the switch-closed and
204 switch-open cases to write an equation for $I_2$.
208 \frac{-V}{R_1+R_2} = \ans{-333\U{$\mu$A}}
210 -\frac{Q}{R'C} e^{\frac{-t}{R'C}}
211 = \frac{C I_2(t<0) R_2}{R'C} e^{\frac{-t}{R'C}}
212 = \frac{-V R_2}{(R_1+R_2)\cdot(R_2+R_3)} e^{\frac{-t}{R'C}}
213 = \ans{-(278\U{$\mu$A})\cdot e^{\frac{-t}{180\U{ms}}}}
219 Plugging into our formula for $q(t)$
221 \frac{Q}{5} &= q(t) = Q e^{\frac{-t}{R'C}} \\
222 \frac{1}{5} &= e^{\frac{-t}{R'C}} \\
223 \ln\p({\frac{1}{5}}) &= \frac{-t}{R'C} \\
224 t &= -R'C\ln\p({\frac{1}{5}}) = R'C\ln(5) = \ans{290\U{ms}}