# How to find currents of all nodes using superposition theorem

Given this circuit:

simulate this circuit – Schematic created using CircuitLab

Using the superposition theorem, I need to find the current of all the 3 nodes of the circuit. Using superposition I've found that the current from the source:

1. V1 is 6.32A = i_1
2. V2 is 1.05A = i_2
3. V3 is 1.89A = i_3

So is it correct to say that at the node of V1 the current is i_1 - i_3 ?
I am really not sure how to proceed.

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"So is it correct to say that at the node of V1 the current is i_1 - i_3 ?" do you mean "i_1=i_2-i_3"? – Davorak Mar 6 '13 at 21:27
@Davorak: No, I meant that the current of the node containing the V1 source is equal to i_1 - i_3 – Chris Mar 6 '13 at 21:40
The current through V1 is i_1 right? So you are saying i_1=i_1-i_3 then? – Davorak Mar 6 '13 at 21:41
Do you mean i_1 is the current through V1 when the other voltage sources are shorted out, in line with the first step of solving the problem with super position? – Davorak Mar 6 '13 at 21:51
@Davorak: Yes! The 2nd one is exactly what I mean. I've shorted it out finally, I hadn't calculated all the other currents at the nodes when I was applying superposition. I did now, and all the currents are correct. – Chris Mar 6 '13 at 23:28

## 1 Answer

You have found the total current when you short out any two voltage sources, but you also need the current going the other branches of the circuit.

your i_1 is i11, your i_2 is i22 in the diagrams above.

After you have solved the circuit for each source individually you can add together the currents of each branch to get the total current of the final circuit with all sources.

So the current through V1 with all sources would be i11+i21+i31. You can then use the total current to find any voltage drops you need.

Pervious:

I it is not possible, with out solving the circuit in some fashion to know the directions of current you are taking about in your questions. It would be a good idea to include a diagram like on fo the below or add an up/down, left/right annotation to your current where appropriate.

You can draw your current arrows in any direction you want, if the current is actually flowing in the opposite direction of your arrow then then it will just end up being a negative current in your solution.

example diagrams:

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