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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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The correct answer is:
$I^2 R t$
The given situation is shown in the diagram.
In time $t$, work done by the battery to establish the current $I$ is given as.
$\begin{aligned} W & =V q=V I t & {[\because q=I t] } \\ & =I R \cdot I t & {[\because V=I R] } \\ & =I^2 R t & \end{aligned}$
In time $t$, work done by the battery to establish the current $I$ is given as.
$\begin{aligned} W & =V q=V I t & {[\because q=I t] } \\ & =I R \cdot I t & {[\because V=I R] } \\ & =I^2 R t & \end{aligned}$
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