A new flashlight cell of $e.m.f.$ $1.5\, volts$ gives a current of $15\, amps$, when connected directly to an ammeter of resistance $0.04\,\Omega $. The internal resistance of cell is ........... $\Omega$
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(b) Let the internal resistance of cell be $r$,
then $i = \frac{E}{{R + r}}$ $ \Rightarrow $ $15 = \frac{{1.5}}{{0.04 + r}}$ $ \Rightarrow $ $r = 0.06\,\Omega $
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