Question

An inductive coil of resistance 15 ohms and inductive reactance 42 ohms is connected in parallel with a capacitor of capacitive reactance 47.6 ohms. The combination is energized from a 200 V, 33.5 Hz a.c. supply. Find the total current drawn by the circuit and its power factor. Draw to the scale the phasor diagram of the circuit.

Answer 1

Z₁ = 15 + j42, Z₁ = 44.6 ohms, cos φ₁ = 15.44.6 = 0.3363 

φ₁ = 70.40 Lagging, I₁ = 200/44.6 = 4.484 amp 

I_c = 200/47.6 = 4.2 amp 

I = 4.484 (0.3355 −j0.942) + j4.2 = 1.50 −j0.025 = 1.5002 − ∠ 1º 

For the circuit in Fig. , the phasor diagram is drawn in Fig..

Power Calculation 

Power etc. can be calculated by the method of conjugates as explained in 

Branch A 

The current conjugate of (10 + j 0) is (10 −j 0) 

∴ VIA = (120 + j 160) (10 −j 0) = 1200 + j 1600 

∴ kW = 1200/1000 = 1.2 

∴ kVAR = 1600/1000 = 1.6. 

The fact that it is positive merely shows the reactive

volt-amperes are due to a lagging current* kVA = √(1.2² + 1.6²) = 2

Branch B 

The current conjugate of (− 4.0 + j8) is (− 4.0 −j8) 

∴ VI_B = (120 + j160) (− 4 −j8) = 800 −j1600 

∴ kW = 800/1000 = 0.8 

∴ kVAR = − 1600/1000 = − 1.6 

The negative sign merely indicates that reactive volt-amperes are due to the leading current

∴  KVA = √[0.8² + (- 1.6)²] = 1.788

Y = Y_A + Y_B = (0.03 −j0.04) + (0.02 + j0.04) = 0.05 + j0 

I = VY = (120 + j160) (0.05 + j0) = 6 + j8 = 10 ∠53º8′ 

or I = I_A + I_B = (10 + j0) + (− 4 + j8) = 6 + j8 (same as above)

Circuit p.f. = cos 0º = 1 (ä current is in phase with voltage)