Q. The charge deposited on 4 $\mu$F capacitor in the circuit is

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Solution:

$6 \mu $ F and 6 $\mu $ F are in series $\therefore$ Voltage ascross 4 $\mu $F = 6 V
$\therefore$ Q $= 6\times 4 \times 10^{-6}$
$= 24 \times 10^{-6}$ C

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