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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