As shown in the figure, a potential of +1200, | vedclass

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(C) The given circuit can be simplified as follows:$1$. The two capacitors of $4\, \mu F$ and $2\, \mu F$ are connected in parallel between point $P$

Vedclass · Accepted Answer

$400$

Vedclass · Answer

(C) The given circuit can be simplified as follows:$1$. The two capacitors of $4\, \mu F$ and $2\, \mu F$ are connected in parallel between point $P$ and point $B$. Their equivalent capacitance is $C_{eq} = 4\, \mu F + 2\, \mu F = 6\, \mu F$.$2$. Now, the circuit consists of a $3\, \mu F$ capacitor in series with a $6\, \mu F$ capacitor, connected between $A$ and $B$.$3$. Since the capacitors are in series, the charge $Q$ on both must be the same.$4$. Using $Q = CV$, we have $C_1(V_A - V_P) = C_{eq}(V_P - V_B)$.$5$. Substituting the values: $3\, \mu F 	imes (1200 - V_P) = 6\, \mu F 	imes (V_P - 0)$.$6$. $3(1200 - V_P) = 6V_P \implies 1200 - V_P = 2V_P \implies 3V_P = 1200$.$7$. Therefore, $V_P = 400\, V$.

As shown in the figure, a potential of $+1200\, V$ is applied at point $A$, and point $B$ is kept at zero potential. Then the electric potential at point $P$ is.....$V$

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