# Homework Help: Power dissipated in circuit

1. Feb 17, 2014

### zealeth

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

$\sum$G connected to node 1 * v_1 - $\sum$G between node 1 and 2 * v_2 = Current source into node 1

$\sum$G connected to node 2 * v_2 - $\sum$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 = $\sum$(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."

2. Feb 18, 2014

### Staff: Mentor

Hi zealeth! http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

Check your equation for node 2.

You started by saying let V1=-16, so this means V4 must be 0v, fixed at 0v. You can't later determine V4 to be -18.1v.

Last edited by a moderator: May 6, 2017
3. Feb 18, 2014

### zealeth

Node 2 was just a typo I think, but nonetheless I got it (P dissipated = 389 Watts). V_4 being 0 was indeed the problem I was having. Thanks.

Last edited by a moderator: May 6, 2017