Question

Two circuits, the impedances of which are given by Z₁ = 15 + j12 ohms and Z₂ = 8 − j5 ohms are connected in parallel. If the potential difference across one of the impedance is 250 + j0 V, calculate. 

(i) total current and branch currents 

(ii) total power and power consumed in each branch and 

(iii) overall power-factor and power-factor of each branch.

Answer 1

Phasor-diagram for these currents, in fig.

(i) I₁ = (250 + j 0)/(15 + j12) = 250 ∠ 0º/19.21 ∠38.6º 

= 13 ∠−38.6º amp = 13 (0.78 −j 0.6247) 

= 10.14 −j 8.12 amp 

I₂ = (250 + j 0)/(8 −j5) = 250 ∠0º/9.434 ∠−32º 

= 26.5 ∠ + 32º = 26.5 (0.848 + j 0.530) 

= 22.47 + j14.05 amp 

I = I₁ + I₂ = 32.61 + j 5.93 = 33.15 ∠ + 10.36º 

(ii) Power in branch 1 = 13² × 15 = 2535 watts 

Power in branch 2 = 26.5² × 8 = 5618 watts 

Total power consumed = 2535 + 5618 = 8153 watts 

(iii) Power factor of branch 1 = cos 38.60º = 0.78 lag 

Power factor of branch 2 = cos 32º = 0.848 lead. 

Overall power factor = cos 10.36º = 0.984 lead