JEE Main 2014 · 12 Apr · Shift 1 · Physics Q21

PhysicsSingle correct+4 / -1Question 21 of 90

A sinusoidal voltage $V(t)=100\sin(500t)$ is applied across a pure inductance of $L=0.02$ H. The current through the coil is :

10\cos(500t)

-10\cos(500t)

10\sin(500t)

-10\sin(500t)

Answer and worked solution

Official answer: B

$X_L=\omega L=500\times0.02=10$ Ω, so the peak current is 10 A. The current lags the voltage by 90°:
$I=10\sin(500t-\tfrac\pi2)=-10\cos(500t).$

JEE Main 2014 · 12 Apr · Shift 1, Physics question 21