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NEET2022Physics-Series LCR Circuit & Resonance

NEET 2022 Physics Series LCR Circuit & Resonance MCQ Question

Type: MCQ-numerical-Medium-Class 12

A series LCR circuit with inductance 10 H, capacitance 10 μF, resistance 50 Ω is connected to an ac source of voltage, V = 200 sin (100 t) volt. If the resonant frequency of the LCR circuit is v₀ and the frequency of the ac source is v, then:

A

v₀ = 50/π Hz, v = 50 Hz

B

v = 100 Hz, v₀ = 100/π Hz

C

v₀ = v = 50 Hz

D

v₀ = v = 50/π Hz

Correct Answer

Option D

Detailed Explanation

Chapter: Alternating Current

Class: 12 Physics Topic: Series LCR Circuit & Resonance Difficulty: 🟡 Medium

✅ Ans: **D — $

u_0= u=\frac{50}{\pi},\text{Hz}$**

Given:

L=10 H,C=10 μF=10−5FL=10\,H,\qquad C=10\,\mu F=10^{-5}F

The AC voltage is:

V=200sin⁡(100t)V=200\sin(100t)

Comparing with:

V=V0sin⁡(ωt)V=V_0\sin(\omega t)

we get:

ω=100 rad/s\boxed{\omega=100\ rad/s}

Step 1: Find source frequency

ω=2πu\omega=2\pi u u=1002π u=\frac{100}{2\pi} u=50π Hz\boxed{ u=\frac{50}{\pi}\,Hz}

Step 2: Find resonant frequency

For a series LCR circuit:

ω0=1LC\omega_0=\frac{1}{\sqrt{LC}} =110×10−5=\frac{1}{\sqrt{10\times10^{-5}}} =110−4=100 rad/s=\frac{1}{\sqrt{10^{-4}}} =100\ rad/s

Therefore,

u0=ω02π=1002π u_0=\frac{\omega_0}{2\pi} =\frac{100}{2\pi} u0=50π Hz\boxed{ u_0=\frac{50}{\pi}\,Hz}

Hence,

u0=u=50π Hz\boxed{ u_0= u=\frac{50}{\pi}\,Hz}

❌ Why other options are wrong?

  • A: u0=50π, u=50 u_0=\frac{50}{\pi},\ u=50 ❌ Source frequency is not 50 Hz; 100100 is angular frequency, not frequency.
  • B: ❌ Both values are interchanged/incorrect.
  • C: u0=u=50 u_0= u=50 ❌ Confuses angular frequency 100 rad/s100\,rad/s with frequency.
  • D: ✅ Correct.

📌 NCERT Concept

At resonance in a series LCR circuit:

XL=XC\boxed{X_L=X_C}

and

ω0=1LC\boxed{\omega_0=\frac1{\sqrt{LC}}}

At resonance, the source frequency equals the resonant frequency:

u=u0\boxed{ u= u_0}

🧠 NEET Trick

If voltage is written as:

V=V0sin⁡(ωt)V=V_0\sin(\omega t)

the number inside tt is ω\omega, not ff.

Always use:

f=ω2π\boxed{f=\frac{\omega}{2\pi}}

So here:

u0=u=50π Hz\boxed{ u_0= u=\frac{50}{\pi}\,Hz}

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