The capacitor C is initially uncharged. Switc | vedclass

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(B) $1$. Initially, $S_1$ is closed for a long time, so the capacitor $C$ charges to voltage $E$. The charge on the capacitor is $q_0 = CE$.$2$. At $t

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At time $t = 0$ (just after $S_2$ is closed), the reading of the ammeter is zero.

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(B) $1$. Initially, $S_1$ is closed for a long time, so the capacitor $C$ charges to voltage $E$. The charge on the capacitor is $q_0 = CE$.$2$. At $t = 0$, $S_1$ is opened and $S_2$ is closed. The capacitor now discharges through the series combination of resistors $2R$ and $3R$. The total resistance in the circuit is $R_{eq} = 2R + 3R = 5R$.$3$. The time constant of the circuit is $	au = R_{eq}C = 5RC$.$4$. The current in the circuit at $t = 0$ is $i_0 = \frac{V}{R_{eq}} = \frac{E}{5R}$. Thus, option $A$ is correct and option $B$ is incorrect.$5$. The charge on the capacitor at any time $t$ is $q(t) = q_0 e^{-t/	au} = CE e^{-t/5RC}$. Thus, option $D$ is correct.$6$. The current in the circuit is $i(t) = \frac{dq}{dt} = \frac{E}{5R} e^{-t/5RC}$.$7$. The heat developed in resistance $3R$ is $H = \int_0^t i^2 (3R) dt = \int_0^{5RC \ln 2} (\frac{E}{5R} e^{-t/5RC})^2 (3R) dt = \frac{3E^2}{25R} \int_0^{5RC \ln 2} e^{-2t/5RC} dt$.$8$. Solving the integral: $H = \frac{3E^2}{25R} [-\frac{5RC}{2} e^{-2t/5RC}]_0^{5RC \ln 2} = \frac{3E^2}{25R} (\frac{5RC}{2}) (1 - e^{-2 \ln 2}) = \frac{3}{10} CE^2 (1 - \frac{1}{4}) = \frac{3}{10} CE^2 (\frac{3}{4}) = \frac{9}{40} CE^2$. Thus, option $C$ is correct.$9$. Since option $B$ is the only incorrect statement, it is the answer.

The capacitor $C$ is initially uncharged. Switch $S_1$ is closed for a long time while $S_2$ remains open. Now at $t = 0$, $S_2$ is closed while $S_1$ is opened. All the batteries are ideal and connecting wires are resistanceless. Find the $\text{INCORRECT}$ statement.

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