In the circuit as shown in figure, if all the symbols have their usual meanings, then identify the correct option.

q_{2} = q_{3}; V_{2} = V_{3}

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q_{1} = q_{2} + q_{3}; V_{2} = V_{3}

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q_{1} = q_{2} + q_{3}; V = V_{1} + V_{2} + V_{3}

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q_{1} + q_{2} + q_{3} = 0; V_{2} = V_{3} = V - V_{1}

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Solution

The correct option is B $q_{1} = q_{2} + q_{3}; V_{2} = V_{3}$

As $C_{1}$ and combination of $(C_{2}, C_{3})$ are in series, so, charge on $C_{1} =$ charge on combination of $(C_{2}, C_{3})$ .
$\Rightarrow q_{1} = q_{2} + q_{3}$

Now, as $(C_{2}, C_{3})$ are in parallel, so, $V_{2} = V_{3}$

Also, $V = V_{1} + V_{2}$ and $V = V_{1} + V_{3}$

Therefore, option $(B)$ is correct.

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