A coil when given supply of 20 V D.C. draws a current of 4 A. When the same coil is given supply of 20 V A.C. with 80 Hz, draw a current of 2 A. The self Inductance of the coil is :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 1.7\times10^{-2}\,\mathrm{H} \)
With steady (dc) current an inductor has no reactance, so the dc reading gives only the resistance of the coil. With ac the coil has an impedance \( Z = \sqrt{R^{2} + X_L^{2}} \), where \( X_L = 2\pi f L \).
\( \displaystyle R = \frac{V}{I_{\mathrm{dc}}} = \frac{20}{4} = 5\,\Omega \)
\( \displaystyle Z = \frac{V}{I_{\mathrm{ac}}} = \frac{20}{2} = 10\,\Omega \)
\( \displaystyle X_L = \sqrt{Z^{2} - R^{2}} = \sqrt{100 - 25} = \sqrt{75} \approx 8.66\,\Omega \)
\( \displaystyle L = \frac{X_L}{2\pi f} = \frac{8.66}{2\pi \times 80} = \frac{8.66}{502.7} \approx 1.7\times10^{-2}\,\mathrm{H} \)