Question

The frequencies at which the current amplitude in an LCR series circuit becomes \(\frac{1}{2}\) times its maximum value, are 212 rad s^–1 and 232 rad s^–1. The resistance value in the Question: the circuit is R = 5 Ω. The self-inductance in the circuit is _____ mH.

Answer 1

Correct Answer: 250

The problem involves finding the self-inductance \(L\) of an LCR series circuit given that the angular frequencies at which the current amplitude becomes half of its maximum value are 212 rad/s and 232 rad/s. The resistance \(R\) is 5 Ω.

In an LCR circuit, the amplitude of current becomes \(\frac{1}{\sqrt{2}}\) times its maximum value at resonant frequency due to the concept of bandwidth. This is related to the full width at half maximum (FWHM) of the resonance curve. The FWHM is given by the difference between the angular frequencies at half maximum: 

FWHM = \(ω_2 - ω_1 = 232 - 212 = 20 \text{ rad/s}\)

The relationship for bandwidth in terms of resistance \(R\) and inductance \(L\) is:

\[\text{FWHM} = \frac{R}{L}\]

Rearranging gives:

\[L = \frac{R}{\text{FWHM}}\]

Substitute the given values:

\[L = \frac{5}{20} = 0.25 \, \text{H}\]

Convert H to mH: \(L = 250 \text{ mH}\)

Verification: The computed self-inductance \(L = 250\) mH fits the range 250,250 provided in the question.

Therefore, the self-inductance of the circuit is 250 mH.

Answer 2

\(\frac{i}{i_{max}}\)=\(\frac{1}{\sqrt2}\)
=\(\frac{\frac{V_0}{Z}}{\frac{V_0}{R}}\)
⇒\(\frac{R}{Z}=\frac{1}{\sqrt2}\) and \(\frac{1}{212C}\)−212L=232L−\(\frac{1}{232C}\)
⇒212L=\(\frac{1}{232C}\)
so 400L²=R²
⇒L=\(\frac{5}{20}\)
⇒L=250 mH