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(B) The circuit consists of two parallel branches connected across a $12\,V$ source.Branch $1$ has resistors of $3\,\Omega$ and $9\,\Omega$ in series.

Vedclass · Accepted Answer

$75$

Vedclass · Answer

(B) The circuit consists of two parallel branches connected across a $12\,V$ source.Branch $1$ has resistors of $3\,\Omega$ and $9\,\Omega$ in series. The current in this branch is $I_1 = \frac{12}{3+9} = 1\,A$.The potential at point $C$ relative to point $A$ is $V_A - V_C = I_1 	imes 3 = 1 	imes 3 = 3\,V$.Branch $2$ has resistors of $4\,\Omega$ and $2\,\Omega$ in series. The current in this branch is $I_2 = \frac{12}{4+2} = 2\,A$.The potential at point $D$ relative to point $A$ is $V_A - V_D = I_2 	imes 4 = 2 	imes 4 = 8\,V$.The potential difference across the capacitor is $V_{CD} = |(V_A - V_D) - (V_A - V_C)| = |8 - 3| = 5\,V$.The energy stored in the capacitor is $U = \frac{1}{2} CV^2 = \frac{1}{2} 	imes 6\,\mu F 	imes (5\,V)^2 = 3 	imes 25 = 75\,\mu J$.Thus,$n = 75$.

In the circuit shown,the energy stored in the capacitor is $n\,\mu J$. The value of $n$ is ..............

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