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..
![](https://www.sarthaks.com/?qa=blob&qa_blobid=3156296231994382522)