JEE MainPhysicsAlternating Current
A series RC circuit connected to an ac source of variable frequency has a power factor of 3 2 at a frequency of 150 Hz . The frequency of the source is then adjusted so that the power factor of the circuit becomes 1 2 . The new frequency of the source is :
Options
- A450 Hz
- B50 3 Hz
- C150 3 Hz
- D50 Hz
Correct answer
D. 50 Hz
Step-by-step solution
For a series RC circuit, the power factor is given by = R R^2 + X_C^2 . Initially, ₁ = 3 2 . R R^2 + X_ C1 ^2 = 3 2 Squaring both sides, R^2 R^2 + X_ C1 ^2 = 3 4 4R^2 = 3R^2 + 3X_ C1 ^2 R^2 = 3X_ C1 ^2 X_ C1 = R 3 Finally, the power factor becomes ₂ = 1 2 . R R^2 + X_ C2 ^2 = 1 2 Squaring both sides, R^2 R^2 + X_ C2 ^2 = 1 4 4R^2 = R^2 + X_ C2 ^2 X_ C2 ^2 = 3R^2 X_ C2 = 3 R The capacitive reactance is inversely proportional to frequency: X_C = 1 2 f C 1 f . Therefore, f₂ f₁ = X_ C1 X_ C2 f₂ 150 = R/ 3 3 R = 1 3 f₂