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(A) The work done by a battery of emf $\varepsilon$ to drive a charge $q$ through the circuit is given by $W = \varepsilon q$.Since the current $I$ fl

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$\varepsilon I t$

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(A) The work done by a battery of emf $\varepsilon$ to drive a charge $q$ through the circuit is given by $W = \varepsilon q$.Since the current $I$ flows for time $t$,the total charge transferred is $q = I t$.Therefore,the work done by the battery is $W = \varepsilon I t$.Note: The options provided in the question are potentially confusing. The total work done by the battery is $\varepsilon I t$. The energy dissipated as heat in the external resistor $R$ is $I^2 R t$,and in the internal resistor $r$ is $I^2 r t$. The total energy is $I^2(R+r)t = \varepsilon I t$. Given the standard form of such questions,if the question asks for the work done by the battery,the correct expression is $\varepsilon I t$.

$A$ wire of resistance $R$ is connected across a cell of emf $\varepsilon$ and internal resistance $r$. The current through the circuit is $I$. In time $t$,the work done by the battery to establish the current $I$ is

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