Q. When two identical batteries of internal resistance $1\ \Omega$ each are connected in series across a resistor $R,$ the rate of heat produced in $R$ is $J_{1}.$ When the same batteries are connected I parallel across $R,\ $ the rate is $J_{2}.$ If $J_{1} = 2.25\ J_{2}$ then the value of $R$ in $\Omega$ is
A) 4
B) 6
C) 4.8
D) 5.16
Explanation
(d) In series $\ \ \ i = \frac{2E}{2\ + \ R}$
$\therefore\ \ \ \ J_{1} = i^{2}R = \left( \frac{2E}{2 + R} \right)^{2}\ .R$
In parallel
$\ \ \ \ \ i = \frac{E}{0.5 + R}$
$\ \ \ J_{2} = i^{2}R = \left( \frac{E}{0.5 + R} \right)^{2}\ .R$
$\ \ \frac{J_{1}}{J_{2}} = 2.25 = \frac{4(0.5\ + R)^{2}}{(2\ + R)^{2}}$
Solving we get, $R = 4\ \Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.