Potential difference across the terminals of the battery shown in figure is .................... $V$ ($r =$ internal resistance of battery)
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As battery is short circuited

$\mathrm{I}=\frac{\mathrm{E}}{r}=\frac{10}{1}=10 \mathrm{A}$

$v=E-I r$

$V=10-10 \times 1$

$V=0$

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