In the circuit shown in figure $C_1 = C_2 = 2$ $\mu F$. Then charge stored in
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$\mathrm{i}=\frac{120}{\mathrm{R}_{\mathrm{eq}}}=\frac{120}{\left(\frac{1}{2}+\frac{1}{2}\right)^{-1}}=40 \mathrm{A}$
Now this I will be divided equally, so $\mathrm{i}_{1}=20 \mathrm{A}, \mathrm{i}_{2}=20 \mathrm{A}$
so $i_{1}=20 A, i_{2}=20 A$
Now $\left(\Delta \mathrm{V}_{2}\right)=0$
$\mathrm{Q}_{\mathrm{C}_{1}}=20 \times 2 \mu \mathrm{C}=40 \mu \mathrm{C}$
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