Two batteries of emf $E_{1}$ and $E_{2}$ a capacitor of capacitance $C$ and resistor $R$ are connected in a circuit as shown. When the switch S is shifted from position (1) to (2), then the charge flow and work done by battery $E_{2}$ are

Δ q = C (E_{2} - E_{1})

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Δ W = C (E_{2} - E_{1}) E_{2}

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Δ q = C (E_{2} + E_{1})

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Δ W = \frac{q_{2}^{2}}{2 c} - \frac{q_{1}^{2}}{2 c}

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Solution

The correct options are
A $Δ q = C (E_{2} - E_{1})$
B $Δ W = C (E_{2} - E_{1}) E_{2}$

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