In the circuit shown in the figure,four capac | vedclass

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Question

(A) Let the potential at point $A$ be $0 \ V$ and at point $C$ be $10 \ V$. Let the potential at point $B$ be $V_B$.The capacitors $10 \ \mu F$ and $5

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$6$

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(A) Let the potential at point $A$ be $0 \ V$ and at point $C$ be $10 \ V$. Let the potential at point $B$ be $V_B$.The capacitors $10 \ \mu F$ and $5 \ \mu F$ are connected in parallel between $A$ and $B$,so their equivalent capacitance is $C_1 = 10 + 5 = 15 \ \mu F$.The capacitors $4 \ \mu F$ and $6 \ \mu F$ are connected in parallel between $B$ and $C$,so their equivalent capacitance is $C_2 = 4 + 6 = 10 \ \mu F$.Now,the circuit consists of two equivalent capacitors $C_1$ and $C_2$ in series across the $10 \ V$ battery.The potential at point $B$ can be found using the voltage divider rule for capacitors in series: $V_B = V_{total} 	imes \frac{C_2}{C_1 + C_2} = 10 	imes \frac{10}{15 + 10} = 10 	imes \frac{10}{25} = 4 \ V$.The potential difference across the $6 \ \mu F$ capacitor is the potential difference between points $B$ and $C$,which is $V_C - V_B = 10 \ V - 4 \ V = 6 \ V$.

In the circuit shown in the figure,four capacitors are connected to a battery. The potential difference across the $6 \ \mu F$ capacitor is......$V$.

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