Q. In a L – R circuit of 3 mH inductance and 4$\ \Omega$ resistance, emf $E = 4\cos{1000t\ V}$ is applied. The amplitude of emf is
A) 0.8 A
B) $\frac{4}{7}\ A$
C) 1.0 A
D) $\frac{4}{\sqrt{7}}A$
Explanation
(a) The current is $I = \frac{E_{0}}{\sqrt{R^{2} + \omega^{2}L^{2}}}$
$= \frac{4}{\sqrt{4^{2} + \left( 1000 \times 3 \times 10^{- 3} \right)^{2}}} = 0.8\ $A
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.