Question

An ideal inductor of \( \frac{1}{2} \, \text{H} \) is connected in series with a \( 300 \, \Omega \) resistor. If a 20 V, 200 Hz alternating source is connected across the combination, the phase difference between the voltage and current is

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

Hint:

The phase difference in an AC circuit with both inductive and resistive components depends on the ratio of the inductive reactance to the resistance.

Answer 1

The Correct Option is B

Step 1: Understanding the circuit components.
We are given an AC circuit consisting of an ideal inductor and a resistor connected in series. The voltage across the combination is 20 V, and the frequency of the AC source is 200 Hz. The inductance \( L = \frac{1}{2} \, \text{H} \) and the resistance \( R = 300 \, \Omega \).

Step 2: Impedance of the series combination.
The total impedance \( Z \) of a series \( R-L \) circuit is given by:
\[ Z = \sqrt{R^2 + (X_L)^2}, \]
where \( R \) is the resistance and \( X_L \) is the inductive reactance, given by:
\[ X_L = 2 \pi f L. \]
Substituting the given values, \( f = 200 \, \text{Hz} \) and \( L = \frac{1}{2} \, \text{H} \):
\[ X_L = 2 \pi \times 200 \times \frac{1}{2} = 200 \, \Omega. \]
So, the total impedance is:
\[ Z = \sqrt{300^2 + 200^2} = \sqrt{90000 + 40000} = \sqrt{130000} = 360.6 \, \Omega. \]

Step 3: Finding the phase difference.
The phase difference \( \phi \) between the voltage and the current is given by:
\[ \tan \phi = \frac{X_L}{R}. \]
Substituting the values:
\[ \tan \phi = \frac{200}{300} = \frac{2}{3}. \]
Thus, the phase difference \( \phi \) is:
\[ \phi = \tan^{-1} \left( \frac{2}{3} \right). \]
Final Answer:
Thus, the phase difference is:
\[ \boxed{\tan^{-1} \left( \frac{4}{3} \right)}. \]