An inductance coil has a reactance of \( 100\,\Omega \). When an ac signal of frequency 1000 Hz is applied to the coil, the applied voltage leads the current by \( 45^\circ \). The self-inductance of the coil is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 1.6\times10^{-2}\,\mathrm{H} \)
The inductive reactance of a coil is \( X_L=\omega L \), with \( \omega=2\pi\nu \). The \( 45^\circ \) lead only shows that the resistance of the coil equals its reactance. The self-inductance follows from the reactance alone.
The angular frequency of the signal is:
\( \begin{aligned} \omega &= 2\pi\nu \\ &= 2\times3.14\times1000\,\mathrm{Hz} \\ &= 6280\,\mathrm{rad\,s^{-1}} \end{aligned} \)
The self-inductance is then:
\( \begin{aligned} L &= \frac{X_L}{\omega} \\ &= \frac{100\,\Omega}{6280\,\mathrm{rad\,s^{-1}}} \\ &\approx 1.6\times10^{-2}\,\mathrm{H} \end{aligned} \)