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

To find the self-inductance (\(L\)) of the circuit, we can use the formula for the resonance condition in an LCR circuit: \[ \omega = \frac{1}{\sqrt{LC}} \] where \(\omega\) is the angular frequency. Given the problem states two angular frequencies, 212 rad/s and 232 rad/s, at which the current is half its maximum value, these frequencies are related to the bandwidth of the system. The bandwidth (\(\Delta \omega\)) is defined as: \[ \Delta \omega = \frac{R}{L} \] where \(R\) is the resistance. Hence, \(\Delta \omega = \omega_2 - \omega_1 = 232 - 212 = 20 \text{ rad/s}\) Given \(R = 5 \, \Omega\), we rearrange the bandwidth equation to solve for \(L\): \[ L = \frac{R}{\Delta \omega} = \frac{5}{20} = 0.25 \, \text{H} \] Consequently, the self-inductance \(L\) in millihenries is: \[ L = 0.25 \, \text{H} = 250 \, \text{mH} \] Verifying, \(250 \, \text{mH}\) fits within the given range of (250,250). Therefore, the self-inductance of the circuit is 250 mH.