Question

Two impedance Z₁ = (8 + j6) and Z₂ = (3 − j4) are in parallel. If the total current of the combination is 25 A, find the current taken and power consumed by each impedance.

Answer 1

Z₁ = (8 + j6) = 10 ∠ 36.87º ; Z₂ = (3 −j4) = 5∠ − 53.1º

Z = Z₁Z₂/(Z₁ + Z₂) = ((10 ∠ 36.87º) (5 ∠- 53.1º))/((8 + j6) + (3 - j4)) 

= (50 ∠ - 16.23º)/(11 + j2) 

= (50 ∠- 16.23º)/(11.18 ∠ 10.3º) = 4.47 ∠-26.53º

Let I = 25 ∠0º ; V = I Z = 25 ∠0º × 4.47 ∠−26.53º = 111.75 ∠−26.53º

I₁ = V/Z₁ = 111.75 ∠−26.53º/10∠36.87º = 11.175 ∠−63.4º 

I₂ = 111.75 ∠−26.53/5 ∠−53.1º = 22.35 ∠26.57º 

Now, the phase difference between V and I₁ is 63.4º − 26.53º = 36.87º with current lagging. 

Hence, cos φ₁ = cos 36.87º = 0.8. Power consumed in Z₁ = VI₁ cos φ = 11.175 × 111.75 × 0.8 = 990 W 

Similarly, φ₂ = 26.57 − (− 26.53) = 53.1º ; cos 53.1º = 0.6 

Power consumed in Z₂ = VI₂ cos φ₂ = 111.75 × 22.35 × 0.6 = 1499 W