Find the charge on the capacitor $C$ in the following circuit ............. $\mu C$
Medium
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At steady state current will not flow in branch that contains capacitator, so $6 \Omega$ and $2 \Omega$ will become in series so net current $\left(\frac{12}{6+2}\right)$ Resistor of $4 \Omega$ will become short circuited $I_{n e t}=\frac{3}{2} A$
Voltage across capacitor = Voltage across $6 \Omega$ $\frac{q}{C}=i R$
$\frac{q}{2 \times 10^{-6}}=\frac{3}{2} \times 6$
$q=18 \times 10^{-6}$
$q=18 \mu C$
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