Q. 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
A) $30\ calorie$
B) $60\ calorie$
C) $100\ calorie$
D) $120\ calorie$
Explanation
(d) Resistance of upper branch $R_{1} = 2 + 3 = 5\Omega$
Resistance of lower branch $R_{2} = 4 + 6 = 10\Omega$
Hence$\ \frac{i_{1}}{i_{2}} = \frac{R_{2}}{R_{1}} = \frac{10}{5} = 2$
$\frac{Heat\ generated\ across\ 3\ \Omega\ \left( H_{1} \right)}{Heat\ generated\ across\ 6\ \Omega\ (H_{2})} = \frac{i_{1}^{2} \times 3}{i_{2}^{2} \times 6} = \frac{4}{2} = 2$
$\therefore$ Heat generated across$\ 3\ \Omega = 120\ cal/sec$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.