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Initially charge on \( C_{1} \) is \( q_{0} \) and all other capacitors are uncharged. If switch is closed at \( t=0 \), then the charge on \( C_{4} \) after along time of closing the switch is,

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The correct answer is:
\( \frac{q 0}{3} \)
In parallel, total charge distribution is the direct ratio of capacitance.
Hence, .

Hence, .

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