Q. A 5V battery with internal resistance 2$\ \Omega$ and a 2V battery with internal resistance 1$\ \Omega$ are connected to a 10$\ \Omega$ resistor as shown in the figure The current in the 10$\ \Omega$ resistor is
A) 0.27A, $P_{2}to\ P_{1}$
B) 0.03A, $P_{1}\ to\ P_{2}$
C) 0.03A, $P_{2}\ tp\ P_{1}$
D) 0.27A, $P_{1}to\ P_{2}$
Explanation
(c) Let potential of $P_{1}$ is 0 V and potential of $P_{2}$ is $V_{0}$. Now apply KCL at $P_{2}$.
$\frac{V_{0} - 5}{2} + \frac{V_{0} - 0}{10} + \frac{V_{0} - ( - 2)}{1} = 0$
$\Longrightarrow \ \ \ \ V_{0} = \frac{5}{16}$
So, current through 10$\Omega$ resistor is
$\frac{V_{0}}{10}$ from $P_{2}to\ P_{1}$.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.