Let the mesh currents be as shown in Fig..
The different items of the mesh resistance matrix [Em] are :
Z11 = (2 + j1 + j2 −j1) = (2 + j2)
Z22 = (−j2 + 1 −j1 + j2) = (1 −j1)
Z12 = Z21 = − (j2 −j1) = −j1
Hence, the mesh equations in the matrix form are
= 5.46∠− 135 or ∠225
I1 = Δ1/Δ = 19.1 ∠− 49.1/5 = 3.82 ∠−49.1 = 2.5 – j2.89
I2 = Δ2/Δ = 5.46 ∠− 135/5 = 1.1 ∠−135 = − 0.78 −j0.78
Current through branch BC = I1−I2 = 2.5 −j2.89 + 0.78 + j0.78
= 3.28 −j2.11 = 3.49 ∠− 32.75
Drop over branch AB = (2 + j1)(2.5 −j 2.89) = 7.89 −j 3.28
Drop over branch BD = (1 −j2) (−0.78 −j0.78) = 2.34 + j0.78
Drop over branch BC = j 1 (I1 −I2) = j1 (3.28 −j2.11) = 2.11 + j3.28
Power delivered by the sources would be found by using conjugate method.
Using current conjugate, we get
VA1 = 10(2.5 + j2.89) = 25 + j28.9 ;
∴ W1 = 25 W
VA2 = V2 × I2 — because –I2 is the current coming out of the second voltage source. Again, using current conjugate, we have
VA2 = (4.43 + j2.5) (0.78 −j0.78) or W2 = 4.43 × 0.78 + 2.5 × 0.78 = 5.4 W
∴ total power supplied by the two sources = 25 + 5.4 = 30.4 W
Incidentally, the above fact can be verified by adding up the powers dissipated in the three branches of the circuit. It may be noted that there is no power dissipation in the branch BC.
Power dissipated in branch AB = 3.822 × 2 = 29.2 W
Power dissipated in branch BD = 1.12 × 1 = 1.21 W
Total power dissipated = 29.2 + 1.21 = 30.41 W.
