Q. Two identical batteries each of emf 2 V and internal resistance 1$\ \Omega$ are available to produce heat in an external resistance by passing current through it. The maximum Joulean power that can be developed across the resistance using these batteries it.
A) 2W
B) 3.2 W
C) 1.28 W
D) 8/9 W
Explanation
(a) When two batteries are in series to the external resistance
$R,$ total resistance of the circuit
$= \ R + 2r = R + 2 \times 1 = R + 2\Omega$
Total emf of batteries = 2+2 = 4 V
$\therefore$ Joulean power across $R$
$= \left( \ \frac{4}{R + 2} \right)^{2} \times R = \frac{4^{2}R}{(R - 2)^{2} + 8\ R}$.
Joulean power will be maximum if $R - 2 = 0$
or $R = 2\Omega$
$\therefore$ Maximum joulean power$\ = \frac{4^{2} \times 2}{(2\ 2)^{2} + 8 \times 2} =$ 2 watt
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.