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 45o. The self-inductance of the coil is
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- \(6.7\times10^{-7}\) H
- \(1.1\times10^{-1}\) H
- \(5.5\times10^{-5}\) H
- \(1.1\times10^{-2}\) H
The Correct Option is D
Solution and Explanation
\(X_L = Z\sin\phi = 100\sin 45^\circ = 70.7\ \Omega\).
\(X_L = \omega L = 2\pi f L\), so \(L = \frac{70.7}{2\pi\times1000} = \frac{70.7}{6283} \approx 1.125\times10^{-2}\) H.
\(L \approx 1.1\times10^{-2}\) H.
Answer: (4) \(1.1\times10^{-2}\) H
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