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.
Correct Answer: 250
Approach Solution - 1
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.
Approach Solution -2
=\(\frac{\frac{V_0}{Z}}{\frac{V_0}{R}}\)
⇒\(\frac{R}{Z}=\frac{1}{\sqrt2}\) and \(\frac{1}{212C}\)−212L=232L−\(\frac{1}{232C}\)
⇒212L=\(\frac{1}{232C}\)
so 400L2=R2
⇒L=\(\frac{5}{20}\)
⇒L=250 mH
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Concepts Used:
LCR Circuit
An LCR circuit, also known as a resonant circuit, or an RLC circuit, is an electrical circuit consist of an inductor (L), capacitor (C) and resistor (R) connected in series or parallel.
Series LCR circuit
When a constant voltage source is connected across a resistor a current is induced in it. This current has a unique direction and flows from the negative to positive terminal. Magnitude of current remains constant.
Alternating current is the current if the direction of current through this resistor changes periodically. An AC generator or AC dynamo can be used as AC voltage source.
