A circuit consists of an inductor of inductance 50 mH connected to a voltage source \( V=220\cos(100t)\,\mathrm{V} \). Find the peak current through the circuit.
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 44 A
In a purely inductive ac circuit, the inductive reactance \( X_L=\omega L \) limits the current. The peak current is the peak voltage divided by \( X_L \). Comparing the source voltage with \( v=v_m\cos\omega t \):
\( v_m=220\,\mathrm{V} \)
\( \omega=100\,\mathrm{rad\,s^{-1}} \)
\( L=50\,\mathrm{mH}=0.05\,\mathrm{H} \)
The inductive reactance is:
\( \begin{aligned} X_L &= \omega L \\ &= 100\,\mathrm{rad\,s^{-1}}\times0.05\,\mathrm{H} \\ &= 5\,\Omega \end{aligned} \)
The peak current is:
\( \begin{aligned} i_m &= \frac{v_m}{X_L} \\ &= \frac{220\,\mathrm{V}}{5\,\Omega} \\ &= 44\,\mathrm{A} \end{aligned} \)