Question

A circuit containing inductance of \(\frac{1}{6\pi}\ \text{H}\) and a resistance of \(15\ \Omega\) in series. If an AC voltage of \(100\ \text{V}\) and \(60\ \text{Hz}\) is applied to the above circuit, then the current in the circuit and phase difference between voltage and current respectively are

• A. \(4\ \text{A}\) and \(\tan^{-1}\left(\frac{4}{5}\right)\)
• B. \(5.3\ \text{A}\) and \(\tan^{-1}\left(\frac{3}{4}\right)\)
• C. \(4\ \text{A}\) and \(\tan^{-1}\left(\frac{4}{3}\right)\)
• D. \(5.3\ \text{A}\) and \(\tan^{-1}\left(\frac{4}{3}\right)\)

Hint:

For a series \(R-L\) circuit, \[ X_L=2\pi fL, \] \[ Z=\sqrt{R^2+X_L^2}, \] and \[ \tan\phi=\frac{X_L}{R}. \] Current lags behind voltage by the phase angle \(\phi\).

Answer 1

The Correct Option is C

Step 1: Find inductive reactance.
\( X_L = 2\pi f L = 2\pi\times60\times\frac{1}{6\pi} = 20\,\Omega \).

Step 2: Current and phase angle.
Impedance: \( Z = \sqrt{R^2+X_L^2} = \sqrt{225+400} = \sqrt{625} = 25\,\Omega \).
\( I = V/Z = 100/25 = 4\text{ A} \). Phase: \( \tan\phi = X_L/R = 20/15 = 4/3 \Rightarrow \phi = \tan^{-1}(4/3) \).

\[ \boxed{I = 4\text{ A},\; \phi = \tan^{-1}\!\left(\tfrac{4}{3}\right)} \]