In a series resonant LCR circuit, the voltage across R is 100 volts and $R=1 \mathrm{k} \Omega$ with $C=2 \mu \mathrm{~F}$. The resonant frequency $\omega$ is $200 \mathrm{rad} / \mathrm{s}$. At resonance the voltage across L is

(A) $4 \times 10^{-3} \mathrm{~V}$

(B) $2.5 \times 10^{-2} \mathrm{~V}$

JEE Main · 2006 · 30 Apr 2006 · Offline · Alternating Current

In a series resonant LCR circuit, the voltage across R is 100 volts and

R = 1 k Ω

with

C = 2 μ F

. The resonant frequency

ω

200 rad / s

. At resonance the voltage across L is

Answer: (D)

At resonance the inductive and capacitive reactances cancel, so the whole supply voltage appears across

R

and the current is

I = \frac{V _{R}}{R} = \frac{100}{1000} = 0.1 A

. The voltage across the inductor equals that across the capacitor at resonance, and the capacitor value makes the latter easiest to use:

V_{L} = V_{C} = \frac{I}{ω C} = \frac{0.1}{200 \times 2 \times 1 0 ^{- 6}} = 250 V

JEE Main · 2006 · 30 Apr 2006 · Offline · Single correct option