Capacitors

JEE Main level, numerical answer, medium

160 of 180

Question 160 of 180

As show in the figure, in steady state, the charge stored in the capacitor is _____×10−6C\times {10}^{-6}C.

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Answer: 10

At the steady state, there will be no current through the capacitor. Therefore, potential difference across capacitor VC=10R+rR=10×100100+10=10011V{V}_{C}=\dfrac{10}{R+r}R=\dfrac{10\times 100}{100+10}=\dfrac{100}{11}V
Charge on capacitor is q=CVCq=C{V}_{C}
=(1.1×10−6)(10011)=\left(1.1\times {10}^{-6}\right)\left(\dfrac{100}{11}\right)
⇒q=10μC=10×10−6C\Rightarrow q=10\mu C=10\times {10}^{-6}C.

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