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Studying for the PE and am going through all the problems in my circuit analysis book in college. I'm working problem 3.33 which asks to compute the total power created in the circuit. The answer in the back of the book is 1650 W, but I keep coming up with 1406.25 W. I'm not sure what I'm doing wrong, but I'm pretty sure it has to do with the power developed in the dependent source.

I've attached the circuit and my attempt. Can anyone see my mistake or offer a suggestion?

enter image description here

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  • \$\begingroup\$ The total power is coming from the +125 V supply and it really is 1650 W. The 50 V supply is actually dissipating (getting charged?) \$\endgroup\$
    – jonk
    Jul 1 '21 at 5:26
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    \$\begingroup\$ I don't want to work through a bunch of red ink. Worse, I don't want to have to draw up the schematic, but I feel someone has to do it. So I will in an answer I'll write, shortly. Next time you ask something here, please take a moment and use the schematic editor. It will number the parts for you. That helps avoid difficult and confusing discussions or having to write a lot more than we should have to, when discussing a point. And it also presents the schematic and any text you include with black ink (default) that can be read much more easily, too. \$\endgroup\$
    – jonk
    Jul 1 '21 at 5:56
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I hope this accurately reflects the problem:

schematic

simulate this circuit – Schematic created using CircuitLab

Here's the initial mesh work straight from head to paper as fast as I can write it out:

$$\begin{align*} 0\:\text{V} -R_4\cdot I_1-R_3\cdot\left(I_1-I_3\right)-R_1\cdot\left(I_1-I_2\right) &= 0\:\text{V} \\\\ 0\:\text{V} +V_1 -R_1\cdot \left(I_2-I_1\right)-R_2\cdot\left(I_2-I_3\right)-V_2 &= 0\:\text{V} \\\\ 0\:\text{V} +V_2 -R_2\cdot \left(I_3-I_2\right)-R_3\cdot\left(I_3-I_1\right)-V_{IA} &= 0\:\text{V} \\\\ I_3&=0.2\cdot R_1\cdot\left(I_2-I_1\right) \end{align*}$$

That's four equations and four unknowns. (Since you know that \$R_1=5\:\Omega\$, you could just go around the first three equations replacing \$I_3\$ with \$I_2-I_1\$ and reduce it to three equations and three unknowns.)

These solve out in the following way (using a free tool called sympy):

var( 'r1 r2 r3 r4 i1 i2 i3 via v1 v2' )
eq1 = Eq( 0 - r4*i1 - r3*(i1-i3) - r1*(i1-i2), 0 )
eq2 = Eq( 0 + v1 - r1*(i2-i1) - r2*(i2-i3) - v2, 0 )
eq3 = Eq( 0 + v2 - r2*(i3-i2) - r3*(i3-i1) - via, 0 )
eq4 = Eq( i3, 0.2*r1*(i2-i1) )
ans = solve( [eq1, eq2, eq3, eq4], [i1, i2, i3, via] )
for v in ans: v, ans[v].subs( {r1:5, r2:7.5, r3:2.5, r4:17.5, v1:125, v2:50} )
(i1, 3.60000000000000)
(i2, 13.2000000000000)
(i3, 9.60000000000000)
(via, 62.0000000000000)

So, given the direction of \$I_2\$ and \$V_2\$'s polarity, it's pretty clear that the power dissipated by \$V_1\$ is \$-V_1\cdot I_2=-125\:\text{V}\cdot 13.2\:\text{A}=-1650\:\text{W}\$. It's supplying power to the rest of the circuit. And the number matches up with the book value (at least in magnitude, if not sign.)

Just as a double-check, if the current \$V_2\$ is taken as exiting its positive terminal, then \$I_{V_2}=I_3-I_2=-3.6\:\text{A}\$. So the power is positive and therefore \$V_2\$ is dissipating. So you do not need to add it in.

Likewise, the current in \$I_A=I_3=9.6\:\text{A}\$ and voltage across it is \$62\:\text{V}\$. Given the polarity and direction this is also a dissipating element.

So there really is only one circuit element powering things, \$V_1\$. And you have the answer.

Please feel free to go through what I wrote, and what you did, and find your own error. It was less work for me to write this up than to parse your red script.

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    \$\begingroup\$ Excellent, thank you for this write up. Apologize for the red ink! And I will use the schematic editor next time (I wasn't aware). I found my mistake, it was when I developed the constraint equation for current i3. I did a KCL at that node (i2+i1 = i3) but should have just set i3 = .2Vd . That fixed the problem! Thank you \$\endgroup\$ Jul 1 '21 at 12:58
  • \$\begingroup\$ @elimenohpee Thanks so much for letting me know it helped! Best wishes, too, on studying for the PE. The PEs I've known have all been exceptional people. It's my privilege. \$\endgroup\$
    – jonk
    Jul 1 '21 at 16:26

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