An inductive coil having inductance of 0.04 H and resistance 25 Ω has been connected in series with another inductive coil of inductance 0.2 H and resistance 15 Ω. The whole circuit is powered with 230 V, 50 Hz mains. Calculate the power dissipation…

Coil 1: R₁ = 25 Ω, \( X_1 = 2\pi(50)(0.04) = 12.566\ \Omega \). Coil 2: R₂ = 15 Ω, \( X_2 = 2\pi(50)(0.2) = 62.832\ \Omega \).

Total: R = 25 + 15 = 40 Ω, X = 12.566 + 62.832 = 75.398 Ω.

\( Z = \sqrt{40^2 + 75.398^2} = \sqrt{1600 + 5684.9} = 85.35\ \Omega \).

Current \( I = \dfrac{230}{85.35} = 2.695 \) A.

Power in coil 1: \( P_1 = I^2R_1 = 2.695^2\times25 = \) 181.5 W.

Power in coil 2: \( P_2 = I^2R_2 = 2.695^2\times15 = \) 108.9 W.

Total power = 290.5 W.

Total power factor: \( \cos\phi = \dfrac{R}{Z} = \dfrac{40}{85.35} = \) 0.469 lagging (φ = 62.05°).

Check: VI cos φ = 230 × 2.695 × 0.4686 = 290.5 W ✓. (Coil voltages: V₁ = 2.695 × 27.98 = 75.4 V, V₂ = 2.695 × 64.60 = 174.1 V.)