Q. The current passing through a choke coil of 5 henry is decreasing at the rate of $2\ ampere/sec$. The e.m.f. developing across the coil is
A) $10\ V$
B) $- 10\ V$
C) $2.5\ V$
D) $- 2.5\ V$
Explanation
(a) Given $\frac{di}{dt} = 2A/sec.,$ $L = 5\ H\ \therefore e = L\frac{di}{dt} = 5 \times 2 = 10\ V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.