A resistance of 10 Ω and inductance L are connected in series with an alternating source of 100 V, 50 Hz. If the phase angle between the voltage and the current is 45°, the value of inductance L is, approximately:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 31.8 mH
In a series LR circuit the phase angle satisfies \( \tan\phi = X_L/R \). A phase angle of 45° means \( X_L = R \).
\( \displaystyle \tan\phi = \frac{X_L}{R} \)
\( \displaystyle X_L = R\tan 45^\circ = 10\,\Omega \)
\( \displaystyle L = \frac{X_L}{2\pi f} = \frac{10}{2\pi \times 50} \)
\( \displaystyle L \approx 0.0318\,\mathrm{H} = 31.8\,\mathrm{mH} \)
The supply voltage (100 V) is not needed.