Question 151 of 180

In an electrical circuit drawn below the amount of charge stored in the capacitor is _______ $\mu C.$

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Answer: 60

In steady state there will be no current in branch of capacitor, so no voltage drop across ${R}_{2}=5\Omega$
${I}_{2}=0$
${I}_{1}={I}_{3}=\dfrac{10}{4+6}=1A$
${V}_{{R}_{3}}={V}_{c}+{V}_{{R}_{2}}\left({V}_{{R}_{2}}=0\right)$
$\Rightarrow {I}_{3}{R}_{3}={V}_{c}$
$\Rightarrow {V}_{c}=1\times 6=6$ volt
Therefore, charge on capacitor ${q}_{c}=C{V}_{c}=10\times 6=60\mu C$

JEE Main Physics practice question on Capacitors (Charging and Discarging of capacitors), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.