Q. The network shown in the figure is a part of a complete circuit. If at a certain instant the current $i$ is $5\ A$ and is decreasing at the rate of $10^{3}A/s$ then $V_{A} - V_{B}$ is
A) $5\ V$
B) $10\ V$
C) $15\ V$
D) $20\ V$
Explanation
(c) By using Kirchhoff’s voltage law
$V_{A} - i\ R + E - L\frac{di}{dt} = V_{B} \Rightarrow V_{B} - V_{A} = 15\ volt$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.