In the given figure,the potential at point A | vedclass

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(C) Capacitors $C_2$ and $C_3$ are in parallel. Hence,their equivalent capacitance is:$C_p = C_2 + C_3 = 8 \ \mu F + 4 \ \mu F = 12 \ \mu F$Now,$C_p$

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$300$

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(C) Capacitors $C_2$ and $C_3$ are in parallel. Hence,their equivalent capacitance is:$C_p = C_2 + C_3 = 8 \ \mu F + 4 \ \mu F = 12 \ \mu F$Now,$C_p$ and $C_1$ are in series. Their equivalent capacitance $C_{eq}$ is:$C_{eq} = \frac{C_1 	imes C_p}{C_1 + C_p} = \frac{6 	imes 12}{6 + 12} = \frac{72}{18} = 4 \ \mu F$The total charge stored by the combination is:$q = C_{eq} 	imes V_{AB} = 4 \ \mu F 	imes 900 \ V = 3600 \ \mu C$In a series combination,the charge on each capacitor is the same. Therefore,the charge on $C_1$ is $3600 \ \mu C$.The potential difference across $C_1$ is:$V_1 = \frac{q}{C_1} = \frac{3600 \ \mu C}{6 \ \mu F} = 600 \ V$Since $V_A - V_P = V_1$,we have:$900 \ V - V_P = 600 \ V$$V_P = 900 \ V - 600 \ V = 300 \ V$

In the given figure,the potential at point $A$ is $900 \ V$ and point $B$ is earthed. What will be the potential at point $P$ (in $V$)?

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