Question

Two impedances consist of (resistance of 15 ohms and series-connected inductance of 0.04 H) and (resistance of 10 ohms, inductance of 0.1 H and a capacitance of 100 μF, all in series) are connectd in series and are connected to a 230 V, 50 Hz a.c. source. Find : (i) Current drawn, (ii) Voltage across each impedance, (iii) Individual and total power factor. Draw the phasor diagram.

Answer 1

Let suffix 1 be used for first impedance, and 2 for the second one. At 50 Hz,

Z1 = 15 + j (314 × 0.04) = 15 + j 12.56 ohms

Z1 = √(152 + 12.562) = 19.56 ohm,

Impedance-angle, θ1 = cos−1 (15/19.56) = + 40°,

Z2 = 10 + j (314 × 0.1) − j {1/(314 × 100 × 16−6 )}

= 10 + j 31.4 − j 31.85 = 10 −j 0.45

= 10.01 ohms, Impedance angle, θ2 = − 2.56°,

Total Impedance, Z = Z1 = Z2 = 15 + j 12.56 + 10 − j 0.85

= 25 + j 12.11 = 27.78 = 25.85°

For this, Phase-angle of + 25.85°, the power-factor of the total impedance

= cos 25.85° = 0.90, Lag.

Current drawn = 230/27.78 = 8.28 Amp, at 090 lagging p.f.

V1 = 8.28 × 19.56 = 162 Volts

V2 = 8.28 × 10.01 = 82.9 Volts

Individual Power-factor

cos θ1 = cos 40° = 0.766 Lagging

cos θ2 = cos 2.56° = 0.999 leading

Phasor diagram : In case of a series circuit, it is easier to treat the current as a reference. The phasor diagram is drawn as in Fig..