An inductance coil has a reactance of 100 Ω. When an AC signal of frequency 1000 Hz is applied to the coil, the applied voltage leads the current by 45°. The self-inductance of the coil is:

1.1 × 10⁻² H
1.1 × 10⁻¹ H
5.5 × 10⁻⁵ H
6.7 × 10⁻⁷ H

Correct answer: 1.1 × 10⁻² H

Solution

A 45 deg phase lead means R = X_L, so X_L = Z*sin45 = 100*0.707 = 70.7 ohm. Then L = X_L/(2*pi*f) = 70.7/(2*pi*1000) = 1.1 x 10^-2 H.

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