Capacitors

JEE Main level, numerical answer, medium

129 of 180

Question 129 of 180

In the circuit shown, the energy stored in the capacitor is nμJn\mu J. The value of nn is _____.

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Answer: 75

The capacitor works as an open circuit in the given configuration.
The potential difference between the two plates of the capacitor will be the difference across the resistors having resistances 3Ω3\Omega and 4Ω4\Omega.
Hence, the potential difference (∆V)\left(∆V\right) across the capacitor can be calculated as follows:
ΔV=(46×12−312×12)V=5V\begin{matrix}\Delta V & = & \left(\dfrac{4}{6}\times 12-\dfrac{3}{12}\times 12\right)V \\ & = & 5V\end{matrix}
The formula to calculate the energy stored in the capacitor is given by
U=12C(∆V)2...(1)U=\dfrac{1}{2}C(∆V{)}^{2}...\left(1\right)
Substitute the values of the known parameters into equation (1) to calculate the required energy stored.
U=12×6μF×(5V)2=75μJ\begin{matrix}U & = & \dfrac{1}{2}\times 6\mu F\times {\left(5V\right)}^{2} \\ & = & 75\mu J\end{matrix}

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