Q.
The resistors $P,Q\ and\ R$ in the circuit have equal resistance. The battery, of negligible internal resistance, supplies a total power of 12 W. What is the power dissipated by heating in resistor $R?$
A) 2 W
B) 4 W
C) 3 W
D) 6 W
Explanation
(a) Let the resistance of $P,Q\ and\ R\ be\ r$.
The total resistance across the battery is
${\ r}_{total} = r + \frac{r}{2} = \frac{3}{2}r.$
Current through $P,$
$I_{P} = \sqrt{\frac{Power}{r_{total}} =}\sqrt{\frac{12}{\frac{3}{2}r}} = \sqrt{\frac{8}{r}}.$
Current through $R,$
$\ I = \frac{1}{2}I_{P} = \sqrt{\frac{2}{r}}$.
Power dissipated in $R$ is thus
${\ P}_{R} = I^{2}r = \left( \ \frac{2}{r} \right)r =$2 W
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.