Question

A coil of inductive reactance $31\, \Omega$ has a resistance of 8 $\Omega$. It is placed in series with a condenser of capacitative reactance $25 \,\Omega$. The combination is connected to an a.c. source of 110 V. The power factor of the circuit is :-

• A. 0.33
• B. 0.56
• C. 0.64
• D. 0.8

Answer 1

The Correct Option is D

To find the power factor of the circuit, we need to analyze the components connected in series, which include a coil with an inductive reactance and resistance, and a condenser with capacitive reactance.

Given data:

• Inductive reactance of the coil, \( X_L = 31 \, \Omega \)
• Resistance of the coil, \( R = 8 \, \Omega \)
• Capacitive reactance of the condenser, \( X_C = 25 \, \Omega \)
• Voltage of the AC source, \( V = 110 \, \text{V} \)

In an RLC series circuit, the following formulae apply:

• The net reactance \( X = X_L - X_C \) because the inductive reactance and capacitive reactance oppose each other.
• Substituting the values, \( X = 31 \, \Omega - 25 \, \Omega = 6 \, \Omega \).
• The impedance of the circuit \( Z \) is given by: Z = \sqrt{R^2 + X^2}

Substituting the known values:

Z = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \, \Omega

The power factor (pf) is the cosine of the phase angle \( \theta \) between the voltage and the current, and is given by:

\text{Power Factor} = \frac{R}{Z}

Substituting the values:

\text{Power Factor} = \frac{8}{10} = 0.8

Therefore, the power factor of the circuit is \( 0.8 \). Hence, the correct answer is 0.8.