# Circuit Analysis 1 - Find power of an independent current source in a complex (to me) circuit

#### TimothyG

Joined Dec 2, 2020
2
Problem:
Find the power absorbed by the 2 A current source in this circuit (HINT: CHOOSE A REFERENCE)

I chose the reference point to be the node where the two independent current sources and the independent voltage source meet.

The obvious equations appear to be:
V_1 - V_2 = 5V_delta
V_1 - V_5 = V_delta
V_1 - V_4 = V_x
V_4 - V_6 = 4i_delta
(V_2 - V_5) / 4 = i_delta
V_2 - V_5 = V_4 - V_6 = 4i_delta

The circuit to be analyzed is in the png file and my work is in the attached pdf. The first page is X-ed out because I felt like I was getting nowhere and needed a fresh start.

Nodal analysis seems to be the way to go, but I'm struggling to see how I can solve this quickly without making a large number of equations for a matrix. I was given this problem and really, really struggled with it. It's nothing like the homework that we've done, so maybe there's a leap of logic that I'm missing. Thanks in advance!

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#### WBahn

Joined Mar 31, 2012
26,398
It's hard to follow your work because I can't tell what your objective is from one line to the next. So I have to reverse engineer your equations to figure that out, and I'm only willing to spend so much time doing that.

You need to work on communicating your work to others in a clear and transparent way. That will not only make it much easier for people like us and, far more importantly, the person grading it to follow what you are doing and see where you are going right and going wrong, but it will make YOU much more organized so that you can see what you have covered and what you have missed.

I would strongly recommend breaking the work into four sections.

First, annotate your diagram with everything that you will use in your equations.

Second, set up the equations that describe the circuit. Don't simply them or anything else, just set them up. ALL of the EE and physics is captured by these equations, so it is worth your time to make them as obviously correct as possible because mistakes that go uncaught here are likely impossible to catch later since you are merely solving a different problem that you thought you were.

Third, manipulate the equations so that they are in a standard form.

Fourth, solve them.

Notice that the final two steps are pure math -- no EE and no physics at all. By separating the EE from the math, it allows you to focus on each one in turn without being distracted by the other.

#### WBahn

Joined Mar 31, 2012
26,398
Another thing that can sometimes be used to good advantage is looking for ways in which your circuit can be simplified before you get going.

For instance, voltage source insulates each of the parallel branches connected to it. This is because the voltage source will produce whatever current is needed by each of the parallel branches independently. So a resistor that is connected directly across a voltage source might as well not exist as far as the rest of the circuit is concerned.

Somewhat analogously, anything in series with a current source is insulated from the rest of the circuit. This is because a current source will produce whatever voltage is needed by the rest of the branch it is in to maintain the current level.

So in your diagram four of the components, including one of the dependent sources, can be removed before you even start.

#### The Electrician

Joined Oct 9, 2007
2,814
Another thing that can sometimes be used to good advantage is looking for ways in which your circuit can be simplified before you get going.

For instance, voltage source insulates each of the parallel branches connected to it. This is because the voltage source will produce whatever current is needed by each of the parallel branches independently. So a resistor that is connected directly across a voltage source might as well not exist as far as the rest of the circuit is concerned.

Somewhat analogously, anything in series with a current source is insulated from the rest of the circuit. This is because a current source will produce whatever voltage is needed by the rest of the branch it is in to maintain the current level.

So in your diagram four of the components, including one of the dependent sources, can be removed before you even start.
Five of the components can be removed.

#### WBahn

Joined Mar 31, 2012
26,398
Five of the components can be removed.
Correct. I don't know how I miscounted because the five are the exact components that I meant. Just gotta learn how to count, I guess.

#### RBR1317

Joined Nov 13, 2010
591
Nodal analysis seems to be the way to go
There is a more formal method to solve this problem than removing components: using supernodes with nodal analysis (which effectively eliminates some components because only the interfaces between supernodes are analyzed with node equations.) A supernode is formed when parts of a circuit are connected by a voltage source(s), either fixed or dependent. It would be difficult to compose a general rule for weird stuff, like a voltage source in series with a current source, but ultimately all currents flowing in & out of the supernode must add up to zero. So this circuit has 4 nodes/supernodes for which you can write 4 node equations. (Supernodes have been color-coded on the diagram)
Question: since there is no ground node shown, can you reduce the complexity of the calculations by designating a ground node, because a node equation is not necessary for the ground node?

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#### WBahn

Joined Mar 31, 2012
26,398
The supernodes don't reduce the number of equations needed since for every supernode you have to come up with the constraint equations that relate the boundary nodes to each other. In addition, you have the control equations for the dependent sources. So in this circuit that's nine equations in nine unknowns. This is fine, particularly if you are using an automated tool to solve the equations. But if you aren't, then reducing the number of equations can make things go not only much quicker, but make it much less likely that a mistake will get made.

In this case, a minor rearrangement of the circuit layout after removing the extraneous components yields a circuit with five meshes, two of which are determined by the two current sources, leaving three equations. You still have two control equations that have to be considered, but the third as also been eliminated.

But even more to the point, being able to apply fundamental concepts to simply circuits is a good skill to develop in its own right as opposed to blindly following a bunch of steps that, more often than not, are not understood, just followed blindly. You touch on this in the question you ask. While the choice of ground node is completely arbitrary, making a good choice can sometimes make a significant difference in the complexity of the ensuing work. But the ability to make a good choice is wrapped up with the ability to evaluate the impacts of the the various options available, which many people never develop because they don't want to spend the time and effort learning how to apply the fundamentals.

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#### The Electrician

Joined Oct 9, 2007
2,814
TimothyG, are you still looking for help with this?

#### TimothyG

Joined Dec 2, 2020
2
I'm so sorry! I got lost in the sea of finals and other complexities in my personal life!

tldr: I took the advice presented here and was able to solve the problem. Thank you!

I went back and looked at the problem using what WBahn said about breaking the problem down systematically. I reworked the problem from the ground up and was able to find all of the equations necessary to solve. The hardest part was actually the math. Figuring out what I needed to manipulate to get to the answer took the longest.