The potential difference between the points P | vedclass

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Question

(A) In the upper branch,$C_1$ and $C_2$ are in series across the potential $E$. The potential at point $P$ is given by the voltage divider rule for ca

Vedclass · Accepted Answer

$\frac{(C_1C_4 - C_2C_3)E}{(C_1 + C_3)(C_2 + C_4)}$

Vedclass · Answer

(A) In the upper branch,$C_1$ and $C_2$ are in series across the potential $E$. The potential at point $P$ is given by the voltage divider rule for capacitors:$V_P = \frac{C_1}{C_1 + C_2} E$In the lower branch,$C_3$ and $C_4$ are in series across the potential $E$. The potential at point $Q$ is given by:$V_Q = \frac{C_3}{C_3 + C_4} E$Therefore,the potential difference between points $P$ and $Q$ is:$V_P - V_Q = \left( \frac{C_1}{C_1 + C_2} - \frac{C_3}{C_3 + C_4} \right) E$$V_P - V_Q = \left( \frac{C_1(C_3 + C_4) - C_3(C_1 + C_2)}{(C_1 + C_2)(C_3 + C_4)} \right) E$$V_P - V_Q = \left( \frac{C_1C_3 + C_1C_4 - C_3C_1 - C_3C_2}{(C_1 + C_2)(C_3 + C_4)} \right) E$$V_P - V_Q = \frac{(C_1C_4 - C_2C_3)E}{(C_1 + C_2)(C_3 + C_4)}$

The potential difference between the points $P$ and $Q$ in the adjoining circuit will be :-

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