JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Alternating Current

Find the peak current and resonant frequency of the following circuit (as shown in figure).

Answer: (C)

Solution

The source is

V = 30 sin 100 t

, so

ω = 100 rad s^{- 1}

. With

L = 100

mH and

C = 100 μ F

X_{L} = ω L = 100 \times 0.1 = 10 Ω, X_{C} = \frac{1}{ω C} = \frac{1}{100 \times 100 \times 1 0 ^{- 6}} = 100 Ω

The series impedance with

R = 120 Ω

Z = R^{2} + (X_{C} - X_{L})^{2} = 12 0^{2} + 9 0^{2} = 150 Ω

so the peak current is

i_{0} = \frac{V _{0}}{Z} = \frac{30}{150} = 0.2

A. Resonance needs

X_{L} = X_{C}

, which gives

f_{0} = \frac{1}{2 π L C} = \frac{1}{2 π 0.1 \times 1 0 ^{- 4}} \approx 50 Hz

The peak current is

0.2

A and the resonant frequency is

50

Hz.

JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Single correct option