Power dissipated in circuit

1. The problem statement, all variables and given/known data

Use the node-voltage method to find the total power dissipated in the circuit in the figure if i1 = 2A , i2 = 3A and v1=16V

2. Relevant equations

P = iv
v = iR
P = v^2/R

G = 1/R

[itex]\sum[/itex]G connected to node 1 * v_1 – [itex]\sum[/itex]G between node 1 and 2 * v_2 = Current source into node 1

[itex]\sum[/itex]G connected to node 2 * v_2 – [itex]\sum[/itex]G between node 1 and 2 * v_1 = Current source into node 2

3. The attempt at a solution

Nodal analysis. Going clockwise from the node above the voltage source, each node is labeled 1 through 5.

1:
v_1 = -16

2:
-2 = v_2(1/12+12/25+1/20) – v_3(1/20) – v_4(1/25) + 16(1/12)

3:
5 = -v_2(1/20) + v_3(1/20+1/40) – 0 – v_5(1/40)

4:
0 = 0 – v_2(1/25) + 0 + v_4(1/25+1/40) – v_5(1/40)

5:
-3 = 0 – v_3(1/40) – v_4(1/40) + v_5(1/40+1/40)

Solving, v_2 = -10.8, v_3 = 43.7, v_4 = -18.1, v_5 = -47.2

Power dissipated:
P = [itex]\sum[/itex](v_a – v_b)^2/R
P = (-16+10.8)^2/12 + (-10.8-43.7)^2/20 + (-10.8+18.1)^2/25 + (43.7+47.2)^2/40 + (-18.1+47.2)^2/40
P = 380.6 W (incorrect obviously)

Feedback: "Not quite. Check through your calculations; you may have made a rounding error or used the wrong number of significant figures."

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