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 $C_3$.For capacitors in series,the total charge $

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 $C_3$.For capacitors in series,the total charge $Q_1$ flowing through $C_1$ must equal the sum of the charges on the parallel branches,so $Q_1 = Q_2 + Q_3$.For capacitors in parallel,the potential difference across them is the same,so $V_2 = V_3$.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 ($V_2$ or $V_3$).Therefore,$V = V_1 + V_2$ (or $V = V_1 + V_3$).

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

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