A battery having $e.m.f.$ $5\,\,V$ and internal resistance $0.5$ $\Omega$ is connected with a resistance of $4.5 \,\Omega$ then the voltage at the terminals of battery is ............. $V$
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(a) $i = \frac{E}{{R + r}} = \frac{5}{{4.5 + 0.5}} = 1\,A$
$V = E - ir = 5 - 1 \times 0.5 = 4.5\,{\rm{Volt}}$
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