question_answer

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                                        [MP PET 1996]

A)            30 calorie

B)            60 calorie

C)            100 calorie

D)            120 calorie

Correct Answer: D

Solution :

                   Resistance of upper branch \[{{R}_{1}}=2+3=5\,\Omega \] Resistance of lower branch \[{{R}_{2}}=4+6=10\,\Omega \] Hence \[\frac{{{i}_{1}}}{{{i}_{2}}}=\frac{{{R}_{2}}}{{{R}_{1}}}=\frac{10}{5}=2\] \[\frac{\text{Heat generated across 3 }\Omega \text{ (}{{\text{H}}_{\text{1}}})}{\text{Heat generated across 6 }\Omega \text{ (}{{\text{H}}_{\text{2}}})}\]\[=\frac{i_{1}^{2}\times 3}{i_{2}^{2}\times 6}=\frac{4}{2}=2\] \ Heat generated across 3 W = 120 cal/sec