An alternating e.m.f. of 0.1 V is applied across an LCR series circuit having `R = 2 Omega, C = 40 mu F and L = 100 mH`. At resonance, the voltage drop across the inductor is

Options

A. 10 V
B. 5 V
C. 2.5 V
D. 20 V

Answer

C

Explanation

At resonance, `i = e/R = 0.1/2 = 5 xx 10^(-2) A`
At resonance, `omega = 1/(sqrt(LC))`
`:. V_(L) = iX_(L) = i omega L = (iL)/(sqrt(LC))`
`=(5 xx 10^(-2)xx10^(-1))/(sqrt(10^(-1)xx 40 xx 10^(-6))) = (5 xx 10^(-3))/(2 xx 10^(-3))` ltbr gt`:. V_(L) = 2.5V`.