Q. The current through a $4.6\ H$ inductor is shown in the following graph. The induced emf during the time interval $t = 5\ milli - sec$ to $6\ milli - sec$ will be
A) $10^{3}V$
B) $- 23 \times 10^{3}\ V$
C) $23 \times 10^{3}\ V$
D) Zero
Explanation
(c) Rate of decay of current between $t = 5\ ms$ to $6\ ms = \frac{di}{dt} = -$ (Slope of the line $BC$)
$= - \left( \frac{5}{1 \times 10^{- 3}} \right) = - 5 \times 10^{3}A/s$. Hence induced emf
$e = - L\frac{di}{dt} = - 4.6 \times \left( - 5 \times 10^{3} \right) = 23 \times 10^{3}V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.