A $50\,V$ battery is connected across a $10\, ohm$ resistor. The current is $4.5\, amperes$. The internal resistance of the battery is ............. $ohm$
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(c) $i = \frac{{50}}{{R + r}}$ $ \Rightarrow $ $r = \frac{{50}}{{4.5}} - 10 = \frac{5}{{4.5}} = 1.1\,\Omega $
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