In the adjoining figure,three capacitors C_1, | vedclass

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(C) From the circuit diagram,capacitor $C_1$ is in series with the parallel combination of $C_2$ and $C_3$.$1$. Since $C_2$ and $C_3$ are connected in

Vedclass · Accepted Answer

$Q_1 = Q_2 + Q_3$ and $V = V_1 + V_2$

Vedclass · Answer

(C) From the circuit diagram,capacitor $C_1$ is in series with the parallel combination of $C_2$ and $C_3$.$1$. Since $C_2$ and $C_3$ are connected in parallel,the potential difference across them is the same,so $V_2 = V_3$.$2$. The total charge $Q_1$ flowing through $C_1$ splits into $Q_2$ and $Q_3$ at the junction,so $Q_1 = Q_2 + Q_3$.$3$. The total potential $V$ of the battery is the sum of the potential drop across $C_1$ and the potential drop across the parallel combination of $C_2$ and $C_3$. Since $V_2 = V_3$,let this common potential be $V_p$. Then $V = V_1 + V_p$. Thus,$V = V_1 + V_2$ (or $V = V_1 + V_3$).Comparing this with the given options,option $C$ is correct.

In the adjoining figure,three capacitors $C_1, C_2,$ and $C_3$ are connected to a battery. The correct condition will be (Symbols have their usual meanings):

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