Four equal capacitors are connected to a 10 , | vedclass

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(C) From the circuit diagram,we have two capacitors in series on the left side of the grounded point $P$ and two capacitors in series on the right sid

Vedclass · Accepted Answer

$+5 \, V, -5 \, V$

Vedclass · Answer

(C) From the circuit diagram,we have two capacitors in series on the left side of the grounded point $P$ and two capacitors in series on the right side. Since the capacitors are equal,the potential difference across the left branch $(V_A - V_P)$ and the right branch $(V_P - V_B)$ will be equal. Given that point $P$ is grounded,its potential $V_P = 0 \, V$. The total potential difference across the battery is $V_A - V_B = 10 \, V$. Since the circuit is symmetric with respect to the ground point $P$,the potential drop across the left side is $5 \, V$ and across the right side is $-5 \, V$. Thus,$V_A - 0 = 5 \, V \implies V_A = +5 \, V$ and $0 - V_B = 5 \, V \implies V_B = -5 \, V$.

Four equal capacitors are connected to a $10 \, V$ battery as shown in the figure. The potentials of $A$ and $B$ are:

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