Question

An inductor of self inductance 1 H connected in series with a resistor of 100 $ \Omega $ and an AC supply of 10 V, 50 Hz. Maximum current flowing in the circuit is:

Hint:

The current in an RL circuit is determined by the impedance, which depends on the resistance and the inductive reactance.

Answer 1

Correct Answer: 1

The total impedance of the circuit is given by: \[ Z = \sqrt{R^2 + (L\omega)^2} \] where: - \( R = 100 \, \Omega \), - \( L = 1 \, \text{H} \), - \( \omega = 2 \pi f = 2 \pi \times 50 = 314 \, \text{rad/s} \). Substitute the values: \[ Z = \sqrt{(100)^2 + (314 \times 1)^2} = \sqrt{10000 + 98696} = \sqrt{108696} \approx 330.0 \, \Omega \] The maximum current is given by: \[ I_{\text{max}} = \frac{V_{\text{max}}}{Z} = \frac{10}{330} = 1 \, \text{A} \]
Thus, the correct answer is (1).