A torch battery consisting of two cells of $1.45\, volts$ and an internal resistance $0.15\,\Omega $, each cell sending currents through the filament of the lamps having resistance $1.5\,ohms$. The value of current will be ....... $A$
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Here two cells are in series.
Therefore total emf $=$ $2E$.
Total resistance $=$ $R + 2r$
$i = \frac{{2E}}{{R + 2r}} = \frac{{2 \times 1.45}}{{1.5 + 2 \times 0.15}} = \frac{{2.9}}{{1.8}} = \frac{{29}}{{18}} = 1.611\,amp$
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