Q.
The network shown in figure is part of a complete circuit. If a certain instant, the current $i\ is\ $ 5 A and is decreasing at a rate $10^{3}\ As^{- 1},\ then\ \ (V_{B} - V_{A})$ is
A) 20 V
B) 15 V
C) 10 V
D) 5 V
Explanation
(b) Moving from $A\ to\ B$.
$\ V_{B} - V_{A} = \left\lbrack 5 \times 10^{- 3}\left( - 10^{3} \right) + 15 + 1 \times 5 \right\rbrack$
$= 15$ volt
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.