In the circuit as shown in the figure, the he | vedclass

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Question

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

Vedclass · Accepted Answer

$120$

Vedclass · Answer

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{{{m{Heat \,generated \,across \,3 }}\Omega {m{ (}}{{m{H}}_{m{1}}})}}{{{m{Heat \,generated \,across\, 6 }}\Omega {m{ (}}{{m{H}}_{m{2}}})}}$$ = \frac{{i_1^2 	imes 3}}{{i_2^2 	imes 6}} = \frac{4}{2} = 2$

 Heat generated across $3$ $\Omega$ = $120\, cal/sec$

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 ................. $calorie$

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