Calculating current in five resistor/two battery circuit

  • Thread starter ddobre
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  • #1
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Homework Statement


9sNwJTD.jpg

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

egThju4.jpg


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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  • #2
Charles Link
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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.
 
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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.
 
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  • #4
Charles Link
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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).
 
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  • #5
CWatters
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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
 
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