An inductor of self inductance 1 H connected in series with a resistor of 100 π ohm

Question

An inductor of self inductance 1 H connected in series with a resistor of \(100 \ \pi\) ohm and an ac supply of \(100 \ \pi\) volt, 50 Hz. Maximum current flowing in the circuit is ______ A.

Answer 1

Answer is : (1)

Impedance of circuit

\(Z = \sqrt{R^2 + (X_L)^2} = \sqrt{R^2+ (\omega_L)^2}\)

\(= \sqrt{(100 \pi)^2+ (2 \pi \times 50 \times1)^2}\)

\(= \sqrt{(100 \pi)^2+ (100 \pi)^2}\)

\(= \sqrt 2 \times 100 \pi\)

\(I_{rms} = \frac{V}{2} = \frac{100 \pi}{\sqrt 2 \times 100 \pi} = \frac{1}{\sqrt2}\)

\(I_{max} = \sqrt{2} I_{\text{rms}}= \sqrt 2 \times \frac{1}{\sqrt 2} = 1 \text{Ampere}\)