Electrical Engineering Stack Exchange is a question and answer site for electronics and electrical engineering professionals, students, and enthusiasts. It only takes a minute to sign up.
Sign up to join this community
Hello, I just need a little guidance on if I am analyzing this circuit correctly using Node Voltage.
I just want to make sure that I am allowed to do that instead of working with the dependent source and the super node.
Thank you!!
I usually like to redraw schematics before attempting to analyze them. Some simplifications can arrive. Even if you are not allowed to use those simplifications directly in the answer, you can certainly use them to simplify cross-checking your results.
Here's your diagram as I see it:
simulate this circuit – Schematic created using CircuitLab
I've inserted \$V_1\$ so as to make it clear what the CCVS operates on: \$I_{\text{V}_1}\$.
The direction of the current I used for \$I_{\text{V}_1}\$ has various conventions. It really doesn't matter which convention you choose, though. Everything still works.
For some, they want anything that consumes power to compute out as a positive value and anything that generates power to compute out as a negative value. In that case, my direction would be considered wrong!
For others, like me, it's more comforting to imagine the direction to be consistent with the direction you might choose for a simple series loop in mesh analysis with a voltage source and a resistor, for example.
I'm sticking by my convention. But you can use whatever you prefer, too.
A simplification that pops to mind is that \$R_3\$ is completely irrelevant and can be shorted out. Current sources have infinite impedance, so \$R_3\$ washes out completely. You have no interest in, nor need to know, the voltage at node d. When setting up the nodal equation for node a, all you need to know is that there will be \$9\:\text{A}\$ arriving from that branch. End of story.
Just set up a nodal equation for node a, another nodal equation for node b, and then add \$V_a=V_b\$ as your third equation. You will be solving for three "unknowns." (Technically, anyway.) These are \$V_a\$, \$V_b\$, and \$I_{\text{V}_1}\$. (Of course, you should expect the solution to produce the same value for \$V_a\$ as for \$V_b\$.)
You really don't need to bother with node c because it is synonymous with the equation for the CCVS: \$4\cdot I_{\text{V}_1}\$.
(Or, put another way, by treating the voltage at node c as \$4\cdot I_{\text{V}_1}\$ you effectively cut away the CCVS from the schematic. It's no longer needed.)
So, when writing out the nodal equation for node a, you just substitute that equation in where you would normally place voltage or voltage difference (depending on how you like to write out your nodal equations.)
With those solutions in hand, it's very easy to work out \$V_y\$, as you know that you have \$9\:\text{A}\$ arriving into node a and \$I_{\text{V}_1}\$ leaving node a. The remainder must be supplied by \$R_1\$. So it is trivial to get. Just multiply \$R_1\$ by the remainder current. Done.
If you need more help than that, let me know. But hopefully that will move you forward.
Nodal analysis is really easy if you just arrange your equations so that inflowing and outflowing currents are separated by the equal sign. It's almost impossible to mess it up, then.
Before I dive in, please note that I modified the above schematics to show my own personal choice for the direction of \$I_{\text{V}_1}\$. The equations below will be consistent to my preference. This may not be yours.
$$\begin{align*} \begin{array}{r} {\text{Node } a:}\vphantom{\frac{V_a}{R_1}}\\\\ {\text{Node } b:}\vphantom{\frac{V_b}{R_2}}\\\\ {\text{Assignment}:}\vphantom{V_b} \end{array} && \overbrace{ \begin{array}{r} \frac{V_a}{R_1=12\:\Omega} + I_{V_{1}}\\\\ \frac{V_b}{R_2=4\:\Omega}\\\\ V_a \end{array} }^{\text{outflowing currents}} & \begin{array}{c} &\quad{=}\vphantom{\frac{V_a}{R_1}}\\\\ &\quad{=}\vphantom{\frac{V_b}{R_2}}\\\\ &\quad{=}\vphantom{V_b} \end{array} & \overbrace{ \begin{array}{l} \left(I_1=9\:\text{A}\right)+\frac{V_c\,=\left(4\:\Omega\,\cdot\, I_{\text{V}_1}\right)}{R_1=12\:\Omega}\\\\ \left(I_2=2\:\text{A}\right)+I_{\text{V}_1}\\\\ V_b \end{array} }^{\text{inflowing currents}} \end{align*}$$
(See KCL Addendum.)
Then just use sympy:
var('va vb iv1 i1 i2 r1 r2')
eqa = Eq( va/r1 + iv1, i1 + (4*iv1)/r1 )
eqb = Eq( vb/4, i2 + iv1 )
eq0 = Eq( va, vb )
ans = solve( [ eqa, eqb, eq0 ], [ va, vb, iv1 ] )
for x in ans: x, ans[x].subs({ i1:9, i2:2, r1:12, r2:4 })
(va, 124/3)
(vb, 124/3)
(iv1, 25/3)
It's a good thing that \$V_a\$ has the same value as \$V_b\$! Otherwise, I'd know I'd screwed up somewhere. Having them be the same doesn't mean I'm safe. But at least it's some small comfort.
At this point, all you need to do is compute \$V_y=R_1\cdot\left(I_1-I_{\text{V}_1}\right)\$:
vy = ( r1*( i1 - ans[iv1] ) ).subs({ i1:9, i2:2, r1:12, r2:4 })
vy
8
That's not \$36\:\text{V}\$ as your comment suggests. But let's say you wanted to verify your answer. Just follow the logic below:
If \$V_y=V_{R_1}= 36\:\text{V}\$, then \$I_{R_1}= \frac{36\:\text{V}}{12\:\Omega}=3\:\text{A}\$ away from node a. Since there is \$9\:\text{A}\$ coming into node a, then it follows that \$I_{\text{V}_1}=9\:\text{A}-3\:\text{A}=6\:\text{A}\$ away from node a and into node b. Since there is also \$2\:\text{A}\$ coming into node b, this means that the \$I_{R_2}=6\:\text{A}+2\:\text{A}=8\:\text{A}\$. This means that \$V_b=8\:\text{A}\cdot 4\:\Omega=32\:\text{V}\$. We know \$V_a=V_b\$ so \$V_a=32\:\text{V}\$, too. And this must mean that \$V_c=V_a-V_y=-4\:\text{V}\$. Since \$4\:\Omega\cdot I_{\text{V}_1}=-4\:\text{V}\$ so also it must be that \$ I_{\text{V}_1}=-1\:\text{A}\$. But this is a contradiction!
So that's a way of testing your answer.
Now let's test sympy. (Hopefully it's good because I use it a lot.)
If \$V_y=V_{R_1}= 8\:\text{V}\$, then \$I_{R_1}= \frac{8\:\text{V}}{12\:\Omega}=\frac23\:\text{A}\$ away from node a. Since there is \$9\:\text{A}\$ coming into node a, then it follows that \$I_{\text{V}_1}=9\:\text{A}-\frac23\:\text{A}=8\frac13\:\text{A}\$ away from node a and into node b. Since there is also \$2\:\text{A}\$ coming into node b, this means that the \$I_{R_2}=8\frac13\:\text{A}+2\:\text{A}=10\frac13\:\text{A}\$. This means that \$V_b=10\frac13\:\text{A}\cdot 4\:\Omega=41\frac13\:\text{V}\$. We know \$V_a=V_b\$ so \$V_a=41\frac13\:\text{V}\$, too. And this must mean that \$V_c=V_a-V_y=33\frac13\:\text{V}\$. Since \$4\:\Omega\cdot I_{\text{V}_1}=33\frac13\:\text{V}\$ so also it must be that \$ I_{\text{V}_1}=8\frac13\:\text{A}\$. No contradiction!
Sympy didn't let me down. Good thing.
