A series LCR circuit has $\mathrm{L}=0.01 \mathrm{H}, \mathrm{R}=10 \Omega$ and $\mathrm{C}=1 \mu \mathrm{~F}$ and it is connected to ac voltage of amplitude ($\mathrm{V}_{\mathrm{m}}$) 50 V. At frequency $60\%$ lower than resonant frequency, the amplitude of current will be approximately:

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Alternating Current

A series LCR circuit has

L = 0.01 H, R = 10Ω

and

C = 1 μ F

and it is connected to ac voltage of amplitude (

V_{m}

) 50 V. At frequency

60%

lower than resonant frequency, the amplitude of current will be approximately:

Answer: (C)

The resonant frequency is

ω_{0} = \frac{1}{L C} = \frac{1}{0.01 \times 1 0 ^{- 6}} = 1 0^{4}

rad/s. At 60% lower frequency,

ω^{'} = 0.4 \times 1 0^{4} = 4000

rad/s. The current amplitude is

i_{0} = \frac{V _{0}}{R ^{2} + ( X _{C}^{'} - X _{L}^{'} ) ^{2}}

, where

X_{C}^{'} = \frac{1}{ω ^{'} C} = 250Ω

and

X_{L}^{'} = ω^{'} L = 40Ω

. Thus

i_{0} = \frac{50}{100 + ( 250 - 40 ) ^{2}} = \frac{50}{100 + 44100} \approx \frac{50}{210} \approx 238

mA.

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Single correct option