A coil of inductance 0.50 H and resistance 100 Ω is connected to 240 V, 50 Hz supply. The maximum current in the coil is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 1.82\,\mathrm{A} \)
A coil with resistance is an LR series circuit. Its impedance combines the resistance and the inductive reactance, and the 240 V of the mains is the rms value, so the maximum current is \( \sqrt{2} \) times the rms current.
\( \displaystyle X_L = 2\pi \nu L = 2\pi \times 50 \times 0.50 = 157.1\,\Omega \)
\( \displaystyle Z = \sqrt{R^{2} + X_L^{2}} = \sqrt{100^{2} + 157.1^{2}} = \sqrt{34680} \approx 186.2\,\Omega \)
\( \displaystyle I_{\mathrm{rms}} = \frac{V_{\mathrm{rms}}}{Z} = \frac{240}{186.2} \approx 1.29\,\mathrm{A} \)
\( \displaystyle I_0 = \sqrt{2}\,I_{\mathrm{rms}} = 1.414 \times 1.29 \approx 1.82\,\mathrm{A} \)