Two impedances z1 6 j8 ohm and Z2 16 j12 ohm are connect

Two impedances z_1 = (6 - j8) ohm and Z_2 = (16 + j12) ohm are connected in parallel. If the total current of the combination is (20 + j10) amperes, find the complexor for power taken by each impedance.

Solution

Ans) As two impedances in parallel Z₁=6-j8 ohms and Z₂=16+j12

Total Current I=20+j10 A

Using Current divison rule

Current through Z₁=I₁=I*(Z₂/(Z₁+Z₂))=(20+j10)*(16+j12)/(6-j8+16+j12)

I₁=(200+j400)/(22+j4)=12+j16 A

Current through Z₂=I₂=I*(Z1/(Z1+Z2))=(20+j10)*(6-j8)/(6-j8+16+j12)

I₂=(200-j100)/(22+j4)=8-j6 A

Voltage across Z₁ And Z₂=V=V₁=V₂=I₁*Z₁=I₂*Z₂

Let\'s take V=I₁ * Z₁=(12+j16)*(6-j8)=200+j0=200 V

Now Complex power Of

Z₁=S₁=V I₁*=200*(12-j16)=2400-j3200 VA

Z₂=S₂=V I₂*=200*(8+j6)=1600+j1200 VA