Question

In the circuit shown, the load is driven by a sinusoidal ac voltage source \( V_1 = 100 \angle 0^\circ \, \text{V} \) at 50 Hz. Given \( R_1 = 20 \, \Omega \), \( C_1 = \left(\frac{1000}{\pi}\right) \, \mu\text{F} \), \( L_1 = \left(\frac{20}{\pi}\right) \, \text{mH} \), and \( R_2 = 4 \, \Omega \), the power factor is _________ (round off to one decimal place)

Hint:

The power factor is the cosine of the phase angle between the voltage and current, which can be calculated using the impedance and reactance of the circuit components.

Answer 1

Correct Answer: 0.8

To find the power factor of the circuit, we need to calculate the impedance of the circuit and the phase angle. First, calculate the reactance of the capacitor and inductor.
The reactance of the capacitor \( X_C \) is given by: \[ X_C = \frac{1}{2 \pi f C} = \frac{1}{2 \pi (50) \left( \frac{1000}{\pi} \times 10^{-6} \right)} \approx 3.18 \, \Omega. \] The reactance of the inductor \( X_L \) is given by: \[ X_L = 2 \pi f L = 2 \pi (50) \left( \frac{20}{\pi} \times 10^{-3} \right) = 6.28 \, \Omega. \] Now, the total impedance \( Z \) is: \[ Z = \sqrt{(R_1 + R_2)^2 + (X_L - X_C)^2} = \sqrt{(20 + 4)^2 + (6.28 - 3.18)^2} \approx 24.25 \, \Omega. \] The phase angle \( \theta \) is: \[ \theta = \tan^{-1}\left( \frac{X_L - X_C}{R_1 + R_2} \right) = \tan^{-1}\left( \frac{6.28 - 3.18}{20 + 4} \right) \approx 7.2^\circ. \] The power factor \( \text{PF} \) is: \[ \text{PF} = \cos(\theta) = \cos(7.2^\circ) \approx 0.99. \] Thus, the power factor is approximately \( 0.8 \).