Question

7. A coil has a resistance of 50 ohms and a reactance of 70 ohms. A capacitor is connected
in parallel to produce resonance. The source voltage is 120 V. What is the power drawn
by the circuit?

Answer 1

At parallel resonance the capacitive susceptance cancels the inductive susceptance. For the coil Z = R + jX (R = 50 Ω, X = 70 Ω)

Admittance of the coil: Y_L = 1/(R + jX) = R/(R^2+X^2) − j X/(R^2+X^2).

Set capacitor susceptance B_C = X/(R^2+X^2) so
X_C = (R^2 + X^2)/X = (50^2 + 70^2)/70 = 7400/70 ≈ 105.71 Ω.

Total conductance at resonance G = R/(R^2+X^2) = 50/7400 = 0.00675676 S.

Power drawn P = V^2 G = 120^2 × 0.00675676 ≈ 97.3 W.

(So the circuit draws about 97.3 W; the capacitor reactance required is ≈105.7 Ω.)