Question 166 of 180

A capacitor of $50\mu F$ is connected in a circuit as shown in figure. The charge on the upper plate of the capacitor is _________ $\mu C.$

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Answer: 100

At steady state current through branch of capacitor is zero. Potential difference across capacitor is $2V.$ So, $q=50\times 2=100\mu C$

$i=\dfrac{6}{(2+2+2)\times {10}^{3}}=1\times {10}^{-3}Amp$
${V}_{A,B}=$ potential across $50\mu F$
${V}_{A,B}=i\times 2\times {10}^{3}=1\times {10}^{-3}\times 2\times {10}^{3}=2$ volt
Charge on $50\mu F$
$q=CV=50\times {10}^{-6}\times 2\Rightarrow 100\times {10}^{-6}C=100\mu C$

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