Question

A 50 Hz AC circuit has a 10 mH inductor and a 2 $\omega$ resistor in series. The value of capacitance to be placed in series in the circuit to make the circuit power factor unity is:

• A. \( 1.014 \times 10^{-6} F \)
• B. \( 1.014 \times 10^{-3} F \)
• C. \( 2.6 \times 10^{-3} F \)
• D. \( 4.125 \times 10^{-3} F \)

Hint:

For resonance in an AC circuit, the condition is: \[ X_L = X_C. \] The capacitance is given by: \[ C = \frac{1}{\omega^2 L}. \]

Answer 1

The Correct Option is B

For power factor = 1, inductive reactance \( X_L \) and capacitive reactance \( X_C \) must be equal: \[ X_L = X_C. \] \[ \omega L = \frac{1}{\omega C}. \]
Step 2: Solving for \( C \) Given: \[ L = 10 mH = 10 \times 10^{-3} H, \quad f = 50 Hz. \] \[ \omega = 2\pi f = 2\pi \times 50. \] \[ C = \frac{1}{\omega^2 L}. \] \[ C = \frac{1}{(2\pi \times 50)^2 \times (10 \times 10^{-3})}. \] \[ C = 1.014 \times 10^{-3} F. \] Thus, the correct answer is: \[ \boxed{1.014 \times 10^{-3} F}. \]