Question 155 of 180

The charge accumulated on the capacitor connected in the following circuit is ______$μC$.
(Given $C=150\mu F)$

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Answer: 400

Under the balanced condition, the capacitor will act as an open circuit.
The current $\left({i}_{1}\right)$ through the $1\Omega$ resistor is given by
$\begin{matrix}{i}_{1} & = & \dfrac{10V}{1\Omega +2\Omega } \\ & = & \dfrac{10}{3}A\end{matrix}$
The current $\left({i}_{2}\right)$ through the $6\Omega$ resistor is given by
$\begin{matrix}{i}_{2} & = & \dfrac{10V}{6\Omega +4\Omega } \\ & = & 1A\end{matrix}$
From the above diagram, it can be written that
${V}_{A}+{i}_{1}\times 1-{i}_{2}\times 6={V}_{B} \Rightarrow {V}_{A}-{V}_{B}=1\times 6-\dfrac{10}{3}\times 1 =\dfrac{8}{3}V$
Hence, the charge on the capacitor is given by
$\begin{matrix}Q & = & C\left({V}_{A}-{V}_{B}\right) \\ & = & 150\mu F\times \dfrac{8}{3}V \\ & = & 400\mu C\end{matrix}$

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.