In a series LR circuit, power of 400W is dissipated from a source of 250 V, 50 Hz. The power factor of the circuit is 0.8. In order to bring the power factor to unity, a capacitor of value C is added in series to the L and R. Taking the value of C as ( n/3π ) μF, then value of n is __________.

Solution

Physics sir · 4 steps

Use the power to find the current, then Z, R and X_L; the capacitor must cancel X_L exactly.

1. Find current and Z. Average power is P = Vᵣₘₛ Iᵣₘₛ cos φ, so I = 400/(250 × 0.8) = 2 A. Then Z = 250/2 = 125 Ω.
2. Split Z. cos φ = R/Z gives R = 125 × 0.8 = 100 Ω, and sin φ = 0.6 gives X_L = 125 × 0.6 = 75 Ω. Check: I² R = 4 × 100 = 400 W.
3. Unity power factor. Power factor becomes 1 when the circuit is at resonance, that is X_C = X_L = 75 Ω, with ω = 2π × 50 = 100π rad s⁻¹.

   C = 1/(ω X_C) = 1/(100π × 75) F = 10⁶/7500π μF = 400/3π μF

4. So. Comparing with n/3π μF gives n = 400.

The common slip: 300. Swapping cosine and sine, so that R = 75 Ω and X_L = 100 Ω, gives n = 300. Remember cos φ goes with R.

The idea to keep. Power factor cos φ = R/Z splits the impedance: R = Z cos φ, X = Z sin φ. Unity power factor in series always means X_C = X_L.

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