# Calculating current in five resistor/two battery circuit

## Homework Statement

The circuit above has five resistors. Calculate the current through each.

## Homework Equations

I1 = I2 + I3 (Node Law)
∑Vi = 0 (Loop Law)

## The Attempt at a Solution

I tried to equate the currents using the node law, with I3 = I1+I2+I4+I5

But I see a flaw here, because the currents seem to oppose each other at the node. Anyway, I tried to solve a system of equations using the above relation of currents, and seemed to get nowhere. Any advice?

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## Answers and Replies

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Homework Helper
Gold Member
The correct equation is ## I_2=I_1+I_3 ##. Do you see why? With that correction, it should work. Also ## I_1=I_4 ##, and ## I_2=I_5 ##. You only need 3 currents.

Is it because I1 and I3 meet at the node at the top, which then turns into I2? Almost forgot series resistors have equal current, that simplifies a lot.

Homework Helper
Gold Member
Note: You are basically solving a system of 3 equations and 3 unknowns (## I_1, I_2, ## and ## I_3 ##). Since you already have one equation, ## I_2=I_1+I_3 ##, you only need 2 voltage-loop equations. (The 3rd voltage loop equation would simply follow from the other two).

ddobre
CWatters
Homework Helper
Gold Member
In my opinion you should always apply KCL like this...

1) Define currents going into (or out of) a node as positive and write down your decision!
2) Then apply KCL and sum the currents to zero.

So with reference to the top node in your diagram...

1) I define into a node as +ve. So therefore...

I1 and I3 are +ve
I2 is -ve

2) The sum is...

(+I1) + (-I2) + (+I3) = 0

Obvious some symbols and brackets are unnecessary but I have included them to make the point about you "summing to zero". If you do it this way you are much less likely to make a sign error. You can now rearrange it how you like, for example my equation becomes....

I1 + I3 -I2 = 0
or
I1 + I3 = I2