Q. In the circuit shown below, the cell has an e.m.f. of $10\ V$ and internal resistance of $1\ ohm.$ The other resistances are shown in the figure. The potential difference $V_{A} - V_{B}$ is
A) $6\ V$
B) $4\ V$
C) $2\ V$
D) $- 2\ V$
Explanation
(d) Total external equivalent resistance $R_{eq} = 4\Omega$
Current supply by cell $i = \frac{E}{R_{eq} + r} = \frac{10}{(4 + 1)} = 2A$
$\therefore\left( V_{A} - V_{B} \right) = \frac{i}{2}\left( R_{2} - R_{1} \right) = \frac{2}{2}(2 - 4) = - 2V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.