A series LCR circuit is connected to an AC source of 220 V, 50 Hz. The circuit contains a resistance R = 80Ω, an inductor of inductive reactance X_L=70Ω, and a capacitor of capacitive reactance X_C=130Ω. The power factor of circuit is x/10. The value of x is :

Solution

Physics sir · 3 steps

Find the net reactance first, then the power factor is just R divided by the impedance.

1. Net reactance. In a series LCR circuit the two reactances oppose each other, so subtract them: X_C - X_L = 130 - 70 = 60 Ω. The circuit is capacitive, but for the power factor only the size of the difference matters.
2. Impedance. Z = √(R² + (X_C - X_L)²) = √(80² + 60²) = √(6400 + 3600) = 100 Ω.

   cos φ = R/Z = 80/100 = 8/10

3. So. The power factor is 8/10, so x = 8.

The common slip: 6. Taking (X_C - X_L)/Z = 60/100 gives 6. That ratio is sin φ, not the power factor.

The idea to keep. Power factor is cos φ = R/Z. Only the resistor uses power, so the resistance always goes on top. The 220 V and 50 Hz do not matter here.

Answer from the NTA final answer key. Solution by PrepWiser.