# Fundamental Circuit Analysis

1. May 23, 2005

### splitendz

Hi Guys,

I'm a first year Electrical Engineering student and we're just covering the basics of circuit analysis. Currently we are going over Nodal analysis.

I'm having some trouble correctly identifying the currents that are entering and leaving a node in a circuit. I can get most simple questions correct but when multiple sources are introduced and the complexity increases, especially when postive and negative signs are not allocated for each resistance, my answers are usually wrong because of an incorrect equation based on an errornous identification of current entering or leaving a node.

I'm wondering is there a general rule of thumb (or logical procedure to follow) for labeling the correct flow of current in/out of a node? I've got exams coming up soon and want to be confident when writing out my equations.

Any help would be appreciated :)

Thanks.

2. May 23, 2005

### learningphysics

Just remember X amps entering a node is the same as -X amps leaving a node.

If you've got 5A entering a node through particular branch... you can just as well think of that as -5A leaving the node. There's no difference. And -5A entering a node is just like 5A leaving the node.

You can label the currents any way you want. I'll draw a rough sketch and example here (ignore the ','s I put them in there for alignment):

,,,,,,,,,,,,,,,,,,,,|4A upward
,,,,,,,,,,,,,,,,,,,,|
3A leftward---O------- X leftward
,,,,,,,,,,,,,,,,,,,,|
,,,,,,,,,,,,,,,,,,,,|
,,,,,,,,,,,,,,,,,,,,|3A downward

Suppose 0 is a node... and we have four branches coming out. We want to solve for the right branch. We label the current X leftward. By KCL, the total current leaving the node is 0... so 4 + 3 + 3 + (-X) = 0. Do you see why I switched from X to -X? X amps are entering the node... so -X amps are leaving the node. So solving the equation we get X=10A. So we know that 10A are entering the node through the right branch, because we defined X as the current leftward (entering the node)

I could also just as easily solve for that current like the following:

,,,,,,,,,,,,,,,,,,,,|4A upward
,,,,,,,,,,,,,,,,,,,,|
3A leftward---O------- X rightward
,,,,,,,,,,,,,,,,,,,,|
,,,,,,,,,,,,,,,,,,,,|
,,,,,,,,,,,,,,,,,,,,|3A downward

This time I've labelled the current X rightward. Using KCL (currents leaving the node)... 4 + 3 + 3 + X = 0. Here X = -10. So we have -10A leaving the node (since we defined X as the current rightward... leaving the node). Hence we have 10A entering the node.

We get exactly the same result both ways. You can label the currents however you want. You just need to know how to interpret the '-' sign. In the above two examples I could have changed 4A upward, to -4A downward, and the circuit would have been exactly the same... And I could change the 3A leftward to -3A rightward.... etc.

It takes a little time to get used to this. But you'll get the hang of it.

I'm not sure what you meant by negative resistance... resistance is always positive.

Last edited: May 23, 2005
3. May 23, 2005

### splitendz

Thanks for your detailed response learningphysics. It's helped clear some things up.

Just another question related to current flow and probably a stupid one but can two currents produced by two indepandant batteries in a DC network flow in opposite directions through the same branch? I have attached a .jpg image to help illustrate my question. If not, why?

#### Attached Files:

• ###### example.JPG
File size:
11.7 KB
Views:
63
4. May 23, 2005

### learningphysics

There is only one current flowing through a particular branch. The contribution from each source to that current can be in opposite directions.

You can use superposition (if you wish) to find that current through a particular branch. First you turn off one source (replace a voltage source with a short) and find the current through that branch. Now bring back that voltage source, and replace the other one with a short, and find the current through that branch. Now, your actual current through that branch will be the sum of the two numbers that you found.

For example, in your particular circuit... I can directly find the through the 10 ohm resistor. It is (60 - 6)/10 = 5.4 A flowing from left to right through the branch.

Now let's use superposition to find the current through the same branch. First I'll turn off the 6V source, so point B is shorted to ground. The current through the 10 ohm resistor is just 60/10 = 6 A from left to right.

Now we'll leave the 6V source on and turn off the 60V source. Now the current through the 10 ohm resistor is (0-6)/10 = -0.6 A flowing from left to right (or 0.6A flowing right to left).

6 A + (-0.6 A) = 5.4 A flowing left to right. Just like I found before. Note that I took the current from "left to right" for both parts, and added them together. I could have added both currents "right to left" if I wanted. So this would have been -6 A + 0.6 A = -5.4 A flowing right to left. It's the same answer.

The following would be an incorrect total current: 6 A + 0.6 A = 6.6 A since the 6 A is left to right, and 0.6 A is right to left. We want both as left to right, or both as right to left... before we add them together.

To answer your original question... there is only one current through the branch. But you can say that the contribution due to each source is in opposite directions.

Hope this helps!