In a series LCR circuit, the inductance L is 10 mH, capacitance C is 100 μ F and resistance R is 20 . The frequency at which impedance is 20 , is

Correct answer: C. (10^3)/(2π) Given L = 10 mH = 10^ -2 H , C = 100 = 10^ -4 F , and R = 20 . The impedance of a series LCR circuit is given by Z = R^2 + (X_L - X_C)^2 . Since Z = 20 = R , the circuit is in resonance, which implies X_L = X_C . ω L = (1)/(ω C) ω = 1 LC Substituting the values of L and C : ω = 1 10^ -2 × 10^ -4 = 1 10^ -6 = 10^3 rad/s The frequency f is given by f = (ω)/(2π) . f = (10^3)/(2π) Hz Answer: (10^3)/(2π)

MHT CET 202622 AprilMorning Shift

Exam: MHT CET
Year: 2026
Date: 22 April
Shift: Morning Shift
Subject: Physics
Chapter: Alternating Current

MHT CET 2026 · 22 April · Morning Shift · Physics · Alternating Current

In a series LCR circuit, the inductance L is 10 mH, capacitance C...

In a series LCR circuit, the inductance L is 10 mH, capacitance C is 100

\mu

F and resistance R is 20

\Omega

. The frequency at which impedance is 20

\Omega

, is

Options

A $\dfrac{10^4}{2\pi}$
B $\dfrac{10^4}{\pi}$
C $\dfrac{10^3}{2\pi}$
D $\dfrac{10^3}{\pi}$

Correct answer

C. $\dfrac{10^3}{2\pi}$

Step-by-step solution

Given

L = 10 \text{ mH} = 10^{-2} \text{ H}

C = 100 \mu\text{F} = 10^{-4} \text{ F}

, and

R = 20 \Omega

.

The impedance of a series LCR circuit is given by

Z = \sqrt{R^2 + \left(X_L - X_C\right)^2}

.

Since

Z = 20 \Omega = R

, the circuit is in resonance, which implies

X_L = X_C

\omega L = \dfrac{1}{\omega C} \Rightarrow \omega = \dfrac{1}{\sqrt{LC}}

Substituting the values of

L

and

C

\omega = \dfrac{1}{\sqrt{10^{-2} \times 10^{-4}}} = \dfrac{1}{\sqrt{10^{-6}}} = 10^3 \text{ rad/s}

The frequency

f

is given by

f = \dfrac{\omega}{2\pi}

f = \dfrac{10^3}{2\pi} \text{ Hz}

Answer:

\dfrac{10^3}{2\pi}

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