The four identical capacitors in the circuit | vedclass

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(D) $1$. When the switch is at position $A$, capacitor $C_1$ is connected to the battery of voltage $V_0$. Thus, $C_1$ gets charged to $Q_1 = C V_0$.$

Vedclass · Accepted Answer

$Q_1 = 3Q_3$

Vedclass · Answer

(D) $1$. When the switch is at position $A$, capacitor $C_1$ is connected to the battery of voltage $V_0$. Thus, $C_1$ gets charged to $Q_1 = C V_0$.$2$. When the switch is moved to position $B$, the charged capacitor $C_1$ is connected in parallel with the series combination of $C_2, C_3,$ and $C_4$. Let the equivalent capacitance of the series combination be $C_{eq}$. Since $C_2 = C_3 = C_4 = C$, we have $1/C_{eq} = 1/C + 1/C + 1/C = 3/C$, so $C_{eq} = C/3$.$3$. The total charge $Q_1$ is now shared between $C_1$ and the series combination $C_{eq}$. Let the final common potential be $V$. Then $Q_1 = (C + C/3)V = (4/3)CV$. Since the initial charge was $CV_0$, we have $CV_0 = (4/3)CV$, which gives $V = (3/4)V_0$.$4$. The charge on $C_1$ is $Q_1' = CV = (3/4)CV_0$. The charge on the series combination is $Q_{series} = (C/3)V = (1/4)CV_0$. Since $C_2, C_3, C_4$ are in series, they all have the same charge $Q_2 = Q_3 = Q_4 = (1/4)CV_0$.$5$. Comparing $Q_1'$ and $Q_3$, we see $Q_1' = 3 	imes (1/4)CV_0 = 3 Q_3$. Thus, $Q_1 = 3Q_3$ is correct.

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