JEE MAIN - Physics (2024 - 1st February Evening Shift - No. 22)
In an electrical circuit drawn below the amount of charge stored in the capacitor is _______ $\mu$ C.
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Answer
60
Explanation
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In steady state there will be no current in branch of capacitor, so no voltage drop across $\mathrm{R}_2=5 \Omega$
$$ \begin{aligned} & \mathrm{I}_2=0 \\\\ & I_1=I_3=\frac{10}{4+6}=1 \mathrm{~A} \\\\ & \mathrm{~V}_{\mathrm{R}_3}=\mathrm{V}_{\mathrm{c}}+\mathrm{V}_{\mathrm{R}_2}, \quad \mathrm{~V}_{\mathrm{R}_2}=0 \end{aligned} $$
$\begin{aligned} & \mathrm{I}_3 \mathrm{R}_3=\mathrm{V}_{\mathrm{c}} \\\\ & \mathrm{V}_{\mathrm{c}}=1 \times 6=6 \text { volt } \\\\ & \mathrm{q}_{\mathrm{c}}=\mathrm{CV}_{\mathrm{c}}=10 \times 6=60 \mu \mathrm{C}\end{aligned}$
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