An inductor of inductance 200 mH is connected to an a.c. source of peak emf 210 V and frequency 50 Hz. The amplitude of alternating current would be-
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 3.34 A
In a purely inductive circuit the current amplitude is the peak emf divided by the inductive reactance \( X_L = \omega L = 2\pi f L \).
\( L = 200\,\mathrm{mH} = 0.2\,\mathrm{H} \)
\( f = 50\,\mathrm{Hz},\ E_0 = 210\,\mathrm{V} \)
\( \displaystyle X_L = 2\pi f L = 2 \times 3.14 \times 50\,\mathrm{Hz} \times 0.2\,\mathrm{H} \)
\( \displaystyle X_L \approx 62.8\,\Omega \)
\( \displaystyle I_0 = \frac{E_0}{X_L} = \frac{210\,\mathrm{V}}{62.8\,\Omega} \)
\( \displaystyle I_0 \approx 3.34\,\mathrm{A} \)