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 determine the current and phase angle in a series AC circuit containing resistive, capacitive, and inductive components, first identify the given parameters:

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

Calculate the net reactance (\( X \)) by subtracting the capacitive reactance from the inductive reactance:

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

Determine the circuit's impedance (\( Z \)) using the Pythagorean theorem:

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

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

Apply Ohm's Law to find the current (\( I \)):

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

Calculate the phase angle (\( \phi \)) using the arctangent function:

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

The circuit current is approximately 7.8 A, with a phase angle of \( 45^\circ \) between the current and voltage. The results are:

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