In the figure, a potential of +1200 , V is ap | vedclass

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(C) The circuit consists of a $3 \, \mu F$ capacitor in series with a parallel combination of $4 \, \mu F$ and $2 \, \mu F$ capacitors.First, calculat

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$400$

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(C) The circuit consists of a $3 \, \mu F$ capacitor in series with a parallel combination of $4 \, \mu F$ and $2 \, \mu F$ capacitors.First, calculate the equivalent capacitance of the parallel combination between point $P$ and point $B$:$C_{PB} = 4 \, \mu F + 2 \, \mu F = 6 \, \mu F$.Now, the circuit simplifies to a $3 \, \mu F$ capacitor in series with a $6 \, \mu F$ capacitor connected between $A$ and $B$.Since the capacitors are in series, the charge $Q$ on each is the same. The potential difference $V$ across a capacitor is given by $V = Q/C$, so $V \propto 1/C$.The potential difference is distributed in the inverse ratio of the capacitances:$V_{AP} / V_{PB} = C_{PB} / C_{AP} = 6 \, \mu F / 3 \, \mu F = 2/1$.Given the total potential difference $V_{AB} = 1200 \, V$, we have $V_{AP} + V_{PB} = 1200 \, V$.Substituting $V_{AP} = 2 V_{PB}$ into the equation:$2 V_{PB} + V_{PB} = 1200 \, V$$3 V_{PB} = 1200 \, V$$V_{PB} = 400 \, V$.Since point $B$ is earthed, its potential is $0 \, V$. Therefore, the potential at point $P$ is $V_P - V_B = 400 \, V$, which means $V_P = 400 \, V$.

In the figure, a potential of $+1200 \, V$ is applied to point $A$ and point $B$ is earthed. What is the potential at point $P$ (in $V$)?

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