In the circuit as shown in the figure, the heat produced by $6\, ohm$ resistance due to current flowing in it is $60$ calorie per second. The heat generated across $3\, ohm$ resistance per second will be ................. $calorie$
Medium
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Resistance of upper branch ${R_1} = 2 + 3 = 5\,\Omega $
Resistance of lower branch ${R_2} = 4 + 6 = 10\,\Omega $
Hence $\frac{{{i_1}}}{{{i_2}}} = \frac{{{R_2}}}{{{R_1}}} = \frac{{10}}{5} = 2$
$\frac{{{\rm{Heat \,generated \,across \,3 }}\Omega {\rm{ (}}{{\rm{H}}_{\rm{1}}})}}{{{\rm{Heat \,generated \,across\, 6 }}\Omega {\rm{ (}}{{\rm{H}}_{\rm{2}}})}}$$ = \frac{{i_1^2 \times 3}}{{i_2^2 \times 6}} = \frac{4}{2} = 2$
Heat generated across $3$ $\Omega$ = $120\, cal/sec$
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