Four capacitors marked with their capacitance | vedclass

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(C) Let the upper branch have capacitors $C_1 = 5 \ \mu F$ $(V_{b1} = 1 \ kV)$ and $C_2 = 4 \ \mu F$ $(V_{b2} = 2 \ kV)$. In series,charge $Q$ is cons

Vedclass · Accepted Answer

$2.25$

Vedclass · Answer

(C) Let the upper branch have capacitors $C_1 = 5 \ \mu F$ $(V_{b1} = 1 \ kV)$ and $C_2 = 4 \ \mu F$ $(V_{b2} = 2 \ kV)$. In series,charge $Q$ is constant. The maximum charge $C_1$ can hold is $Q_1 = C_1 V_{b1} = 5 \ \mu F 	imes 1 \ kV = 5 \ mC$. The maximum charge $C_2$ can hold is $Q_2 = C_2 V_{b2} = 4 \ \mu F 	imes 2 \ kV = 8 \ mC$. To prevent breakdown,$Q \le \min(Q_1, Q_2) = 5 \ mC$. The potential across the upper branch is $V_U = \frac{Q}{C_1} + \frac{Q}{C_2} = 1 \ kV + \frac{5 \ mC}{4 \ \mu F} = 1 \ kV + 1.25 \ kV = 2.25 \ kV$.Let the lower branch have capacitors $C_3 = 2 \ \mu F$ $(V_{b3} = 2 \ kV)$ and $C_4 = 3 \ \mu F$ $(V_{b4} = 1 \ kV)$. The maximum charge $C_3$ can hold is $Q_3 = C_3 V_{b3} = 2 \ \mu F 	imes 2 \ kV = 4 \ mC$. The maximum charge $C_4$ can hold is $Q_4 = C_4 V_{b4} = 3 \ \mu F 	imes 1 \ kV = 3 \ mC$. To prevent breakdown,$Q' \le \min(Q_3, Q_4) = 3 \ mC$. The potential across the lower branch is $V_L = \frac{Q'}{C_3} + \frac{Q'}{C_4} = \frac{3 \ mC}{2 \ \mu F} + \frac{3 \ mC}{3 \ \mu F} = 1.5 \ kV + 1 \ kV = 2.5 \ kV$.The source is connected in parallel to both branches,so $V = V_U = V_L$. To ensure no capacitor breaks down,we must choose the minimum of the two branch potentials: $V = \min(2.25 \ kV, 2.5 \ kV) = 2.25 \ kV$.

Four capacitors marked with their capacitances and breakdown voltages are connected as shown in the figure. Find the maximum $EMF$ of the source such that no capacitor breaks down. (in $kV$)

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