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

In an LCR series circuit, the current amplitude becomes $$\frac{1}{\sqrt{2}}$$ times its maximum value at the half-power frequencies $$\omega_1$$ and $$\omega_2$$. Here $$\omega_1 = 212$$ rad/s, $$\omega_2 = 232$$ rad/s, and the resistance is $$R = 5$$ $$\Omega$$.

The bandwidth of the circuit is given by the difference between these frequencies: $$\Delta\omega = \omega_2 - \omega_1 = 232 - 212 = 20 \text{ rad/s}.$$

In a series LCR circuit, the bandwidth is also related to $$R$$ and $$L$$ by $$\Delta\omega = \frac{R}{L}.$$ Solving for $$L$$ gives $$L = \frac{R}{\Delta\omega} = \frac{5}{20} = 0.25 \text{ H}.$$

Converting to millihenry yields $$L = 0.25 \text{ H} = 250 \text{ mH}.$$

Therefore, the self inductance in the circuit is 250 mH.