Question

An AC power supply of 220 V at 50 Hz, a resistor of 20\( \Omega \), a capacitor of reactance 25\( \Omega \) and an inductor of reactance 45\( \Omega \) are connected in series. The corresponding current in the circuit and the phase angle between the current and the voltage is, respectively :

• A. \( 7.8 \text{ A and } 45^\circ \)
• B. \( 15.6 \text{ A and } 20^\circ \)
• C. \( 15.6 \text{ A and } 45^\circ \)
• D. \( 7.8 \text{ A and } 30^\circ \)

Hint:

Calculate the impedance of the series RLC circuit using \(Z = \sqrt{R^2 + (X_L - X_C)^2}\). Then find the current using \(I = V/Z\). The phase angle \( \phi \) is given by \( \tan \phi = (X_L - X_C)/R \).

Answer 1

The Correct Option is A

To compute the current and phase angle in a series AC circuit with resistive, capacitive, and inductive components, let's start by identifying the parameters:

• Voltage (\( V \)): 220 V
• Resistor (\( R \)): 20 \( \Omega \)
• Capacitive reactance (\( X_C \)): 25 \( \Omega \)
• Inductive reactance (\( X_L \)): 45 \( \Omega \)

Firstly, calculate the net reactance \( X \):

\[ X = X_L - X_C = 45 \, \Omega - 25 \, \Omega = 20 \, \Omega \]

Now, find the impedance \( Z \) of the circuit using the formula:

\[ Z = \sqrt{R^2 + X^2} = \sqrt{20^2 + 20^2} \]

\[ Z = \sqrt{400 + 400} = \sqrt{800} \approx 28.28 \, \Omega \]

The current \( I \) in the circuit is given by Ohm’s Law:

\[ I = \frac{V}{Z} = \frac{220}{28.28} \approx 7.78 \, \text{A} \]

Finally, calculate the phase angle \( \phi \) using the tangent function:

\[ \phi = \tan^{-1}\left(\frac{X}{R}\right) = \tan^{-1}\left(\frac{20}{20}\right) = \tan^{-1}(1) = 45^\circ \]

Thus, the current in the circuit is approximately 7.8 A, and the phase angle between the current and voltage is \( 45^\circ \). Therefore, the answer is:

\( \mathbf{7.8 \text{ A and } 45^\circ} \)