Calculating current in five resistor/two battery circuit

[ddobre]

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?

 

[Charles Link]

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.

 

[ddobre]

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.

 

[Charles Link]

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).

 

[CWatters]

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