In the steady state, the energy stored in the capacitor is :

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$\frac{1}{2} C {(\frac{E_{1} R_{1} + E_{1} R_{2}}{r_{1} + r_{2} + R_{1} + R_{2}})}^{2}$

$\frac{1}{2} C {(E_{1} + E_{2})}^{2}$

$\frac{1}{2} C {(E_{1} - E_{2})}^{2}$

$\frac{1}{2} C {(E_{2} + \frac{E_{1} R_{1}}{r_{1} + R_{1} + R_{2}})}^{2}$

answer is B.

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Detailed Solution

When the capacitor plate acquire full cahrge 0 q , there will be no current in the capacitor arm. Applying Kirchhoff’s second law to the current carrying circuit

$i (R_{1} + R_{2}) = E_{1} - {ir}_{1} or i = \frac{E_{1}}{r_{1} + R_{1} + R_{2}}$
Now $V_{a} - V_{b} = - {iR}_{1} = - \frac{E_{1} R_{1}}{r_{1} + R_{1} + R_{2}}$
and $V_{C} = \frac{q_{0}}{C} = E_{2} + {iR}_{1} = E_{2} + \frac{E_{1} R_{1}}{r_{1} + R_{1} + R_{2}}$
Now energy stored in the capacitor
$U = \frac{1}{2} {CV}_{c}^{2}; = \frac{1}{2} C {[E_{2} + \frac{E_{1} R_{1}}{r_{1} + R_{1} + R_{2}}]}^{2}$