In the given circuit , the charge on 4μF capacitor will be

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In the given circuit , the charge on 4 $μF$ capacitor will be

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$13.4 μ C$

$9.6 μ C$

$5.4 μ C$

$24 μ C$

answer is B.

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Detailed Solution

Equivalent simplified circuit will be $C_{e q} = (3 + \frac{4 \times 6}{4 + 6}) = 5.4 μ F$

Total charge $q = C_{e q} . V = 5.4 \times 10 = 54 μ C$

Charge distribution in the circuit will be as

$\frac{q_{1}}{q_{2}} = \frac{2.4}{3} = \frac{4}{5} = k, q_{1} = 4 K, q_{2} = 5 k$

Total charge , $9 k = 54 μ c \Rightarrow k = 6 μ c$

Charge on $4 μ F & 6 μ F$ will be same

$= 4 \times 6 μ c = 24 μ c$

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