a) Use mesh analysis to find the total power developed in the circuit.

COMMENT: Note that superficially it would appear that all three sources would tend to make the current flow upward through them (positive is at top in all three). However, considering KCL at the bottom center node should immediately lead to the conclusion that either all currents are zero (very unlikely) or at least one of the currents up through the sources is negative, thus one will actually be absorbing power, not generating power.

No dependent sources.

No current sources.

Mesh A:

6 I_A + 3 (I_A - I_C) + 1 (I_A - I_B) = 0
10 I_A - I_B - 3 I_C= 0 (equation 1)

Mesh B:

1 (I_B - I_A) + 2 (I_B - I_C) + 115 + 4 I_B - 230 = 0
- I_A + 7 I_B - 2 I_C = 115 (equation 2)

Mesh C:

3 (I_C - I_A) + 460 + 5 I_C - 115 + 2 (I_C - I_B) = 0
- 3 I_A - 2 I_B+ 10 I_C = - 345 (equation 3)

Solving the three equations gives (details not shown here)

I_A = - 10.6 A
I_B = 4.4 A
I_C = - 36.8 A

230 volt source:

P_dev = 230 I_A = 230 (4.4) = 1012 W

115 volt source:

P_dev = 115 (I_C - I_B) = 115 (- 36.8 - 4.4) = - 4738 W
Note that the 115 volt source is actually absorbing power.

460 volt source:

P_dev = 460 (- I_C) = 460 (36.8) = 16928 W

Thus the total power absorbed is the sum of powers developed by the 230 volt source and the 460 volt source:

P_totaldev = 17940 W

We need the power absorbed by each resistor. Remember that for a resistor P_abs = I₂ R.

1 W:

P₁ = (I_A - I_B)₂ (1) = (- 15₂) (1) = 225 W

2 W:

P₂ = (I_B - I_C)₂ (2) = (41.2₂) (2) = 3394.9 W

3W:

P₃ = (I_A - I_C)₂ (3) = (26.2₂) (3) = 2059.3 W

4 W:

P₄ = I_B2 (4) = (4.4₂) (4) = 77.4 W

5 W:

P₅ = I_C2 (5) = (- 36.8₂) (5) = 6771.2 W

6 W:

P₆ = I_A2 (6) = (- 10.6₂) (6) = 674.2 W

The total power absorbed is thus the sum of powers absorbed by the six resistors plus the power absorbed by the 115 volt source. Adding all thsoe together gives

P_totalabs = 17940 W

This is indeed equal to the total power developed.