question_answer

In the circuit as shown in the figure, the heat produced by \[6\,\,\Omega \] resistance due to current flowing in it is 60cal/s. The heat generated across\[3\,\,\Omega \] resistance per second will be

A)  30cal                                    

B)  60cal

C)  100 cal                                 

D)  120 cal

Correct Answer: D

Solution :

Heat produced \[H={{I}^{2}}Rt\] In both the parts, current is in the ratio of \[2:1\]                 \[\therefore \]  \[{{I}_{1}}=\frac{2I}{3},\]\[{{I}_{2}}=\frac{I}{3}\]                                                 So,          \[\frac{{{H}_{1}}}{{{H}_{2}}}=\frac{I_{1}^{2}{{R}_{1}}t}{I_{2}^{2}{{R}_{2}}t}\]                 or            \[\frac{{{H}_{1}}}{60}={{\left( \frac{21/3}{1/3} \right)}^{2}}\times \frac{3}{6}\]                 or            \[{{H}_{1}}=120cal.\]