Four capacitors with capacitances C_1 = 1 , m | vedclass

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(D) Let the potential at the left junction be $30 \, V$ and at the right junction be $0 \, V$. For the upper branch,the capacitors $C_1$ and $C_2$ are

Vedclass · Accepted Answer

$13$

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(D) Let the potential at the left junction be $30 \, V$ and at the right junction be $0 \, V$. For the upper branch,the capacitors $C_1$ and $C_2$ are in series. The potential at point $A$ $(V_A)$ is given by the voltage divider rule for capacitors: $V_A = 30 	imes \frac{C_1}{C_1 + C_2} = 30 	imes \frac{1}{1 + 1.5} = 30 	imes \frac{1}{2.5} = 30 	imes 0.4 = 12 \, V$. For the lower branch,the capacitors $C_3$ and $C_4$ are in series. The potential at point $B$ $(V_B)$ is given by: $V_B = 30 	imes \frac{C_3}{C_3 + C_4} = 30 	imes \frac{2.5}{2.5 + 0.5} = 30 	imes \frac{2.5}{3} = 10 	imes 2.5 = 25 \, V$. The potential difference between points $B$ and $A$ is $V_B - V_A = 25 \, V - 12 \, V = 13 \, V$.

Four capacitors with capacitances $C_1 = 1 \, \mu F$,$C_2 = 1.5 \, \mu F$,$C_3 = 2.5 \, \mu F$,and $C_4 = 0.5 \, \mu F$ are connected as shown and are connected to a $30 \, V$ source. The potential difference between points $B$ and $A$ is....$V$

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