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..