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(C) For a series $R-C$ circuit, the potential across the capacitor at time $t$ is given by $V(t) = E(1 - e^{-t/RC})$.When the capacitor is charged to

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$4: 3$

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(C) For a series $R-C$ circuit, the potential across the capacitor at time $t$ is given by $V(t) = E(1 - e^{-t/RC})$.When the capacitor is charged to $E/2$, we have $E/2 = E(1 - e^{-t/RC})$, which implies $e^{-t/RC} = 1/2$.The charge on the capacitor at this instant is $Q = C(E/2) = CE/2$.The work done by the battery is $W = Q \cdot E = (CE/2) \cdot E = CE^2/2$.The energy stored in the capacitor is $U = Q^2 / (2C) = (CE/2)^2 / (2C) = CE^2/8$.According to the energy conservation principle, the work done by the battery is equal to the sum of the energy stored in the capacitor and the heat dissipated by the resistor: $W = U + H$.Therefore, the heat dissipated $H = W - U = CE^2/2 - CE^2/8 = 3CE^2/8$.The ratio of the work done by the battery to the heat dissipated is $W/H = (CE^2/2) / (3CE^2/8) = (1/2) / (3/8) = 4/3$.

An initially uncharged capacitor $C$ is being charged by a battery of emf $E$ through a resistance $R$. Up to the instant when the capacitor is charged to a potential $E/2$, the ratio of the work done by the battery to the heat dissipated by the resistor is given by

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