Physics

Physics

Problem

An inductance coil has a reactance of 100Ω100 \Omega . When an AC signal of frequency 1000 Hz1000 \text{ Hz} is applied to the coil, the applied voltage leads the current by 45∘45^\circ . The self-inductance of the coil is : a. 1.1×10−2 H1.1 \times 10^{-2} \text{ H} b. 1.1×10−1 H1.1 \times 10^{-1} \text{ H} c. 5.5×10−5 H5.5 \times 10^{-5} \text{ H} d. 6.7×10−7 H6.7 \times 10^{-7} \text{ H}
Intuitive explanation

Inductive reactance isn't the whole story—a real coil also has resistance, so the impedance is a combination of both. The phase angle tells you the ratio of reactance to resistance: use tan⁡(45°)=XL/R\tan(45°) = X_L / R to find the resistance, then apply XL=2πfLX_L = 2\pi f L to extract the self-inductance from the given reactance and frequency.

Intuitive explanation

Inductive reactance isn't the whole story—a real coil also has resistance, so the impedance is a combination of both. The phase angle tells you the ratio of reactance to resistance: use tan⁡(45°)=XL/R\tan(45°) = X_L / R to find the resistance, then apply XL=2πfLX_L = 2\pi f L to extract the self-inductance from the given reactance and frequency.

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