The Power Factor of an RL Circuit

An RL circuit is a type of electrical circuit that consists of a resistor (R) and an inductor (L) connected in series. The power factor of an RL circuit is a measure of how effectively the circuit converts electrical power into useful work. It is defined as the ratio of the real power (P) to the apparent power (S) in the circuit.

Power Factor (PF) = P / S

The power factor can also be calculated using the cosine of the phase angle (θ) between the voltage (V) and current (I) waveforms in the circuit.

Power Factor (PF) = cos(θ)

The power factor can range from 0 to 1, with 1 being the ideal power factor where the real power and apparent power are equal. A power factor of less than 1 indicates that there is reactive power in the circuit, which is power that oscillates back and forth between the source and the load without being consumed.

Given Information:
- Power factor (PF) = 1/√2
- Frequency of AC is doubled

Determining the New Power Factor

When the frequency of an AC circuit is doubled, the inductive reactance (XL) of the inductor in the RL circuit also doubles. This means that the inductor will have a higher impedance at the higher frequency, which affects the phase relationship between the voltage and current waveforms.

Impedance (Z) = √(R^2 + XL^2)

As the inductive reactance increases, the angle between the voltage and current waveforms also increases. This leads to a decrease in the power factor.

To calculate the new power factor, we need to determine the new phase angle (θ) between the voltage and current waveforms. Since the power factor is equal to the cosine of the phase angle, we can use the inverse cosine (arccos) function to find the phase angle.

cos(θ) = 1/√2
θ = arccos(1/√2)

Using a calculator, we find that the phase angle is approximately 45 degrees.

cos(θ) = 1/√2 = cos(45 degrees)

Therefore, the new power factor is equal to the cosine of the new phase angle.

New Power Factor (PF) = cos(45 degrees) = 1/√2 = 1/√(2^2) = 1/√4 = 1/2

The new power factor is 1/2, which is equivalent to 1/√4. Simplifying further, we get 1/√(2^2) = 1/√4 = 1/2.

Conclusion

When the frequency of an RL circuit is doubled, the power factor decreases. In this case, the original power factor of 1/√2 is reduced to 1/2. Therefore, the correct answer is option 'B' - 1/√5.