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In the given circuit, the switch S was initially open. If the switch is closed now, then amount of charge flown hrough the switch S is

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$30 μ C$

$15 μ C$

$90 μ C$

$40 μ C$

answer is A.

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Detailed Solution

When the switch is closed

then, $q_{2} = \frac{2}{(2 + 3)} \times [120 \times \frac{5 \times 3}{(5 + 3)}] = 90 μ C$
$q_{1} = \frac{1}{(1 + 2)} \times [120 \times \frac{5 \times 3}{(5 + 3)}] = 75 μ C ∴ Δ q = 90 - 75 = 15 μ C$

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