# Linear Equations - Currents

1. Jan 22, 2012

### AlegbraNoob

1. The problem statement, all variables and given/known data

"Determine the currents in the various branches"

2. Relevant equations

The one I am stuck on is 6.

3. The attempt at a solution

I know there is 3 equations, I believe one of them is I1 + I2= 3 and a second is 4= 2I3+I2 ??
but I am not sure if that is correct or what the 3rd one is. I know I can solve the equations if I know the 3 equations.

2. Jan 22, 2012

### Curious3141

You'd better show how you got those equations, because both of them seem wrong.

3. Jan 22, 2012

### AlegbraNoob

I know they seem wrong, that's why I am asking for help...

4. Jan 22, 2012

### Curious3141

Show us how you derive the equations systematically using Kirchoff's laws and we may be able to help critique them.

5. Jan 22, 2012

### AlegbraNoob

Well I know that the sum of all the voltage drops around a closed loop is equal to the total voltage in the loop so.I'm guessing here:

I1+I2+I3= 0
I2-2I3= 3
I1+ I2 = 4

6. Jan 22, 2012

### Curious3141

I don't think it matters. Clearly, familiarity with Kirchoff's laws is a prerequisite here.

There are two: the current law and the voltage law. You can find them easily on wiki. I suggest you read them first.

This looks like a current law statement. That's actually talking about current flows at a point (node in the circuit) rather than a loop. In this case, the directions of the currents I1, I2 and I3 have been indicated. If you solve the equations and find that one or more of them are negative, it just means that the current is actually flowing the other way than indicated, so don't worry.

Look at the node to the right of the positive terminal of the 3V cell. Identify which way the currents are flowing in and out of the node. I1 is flowing "upward" out of the node. I2 is flowing "rightward" into the node. I3 is flowing "downward" out of the node. Can you now set up an equation using Kirchoff's current law?

Here, you're trying to apply Kirchoff's voltage law, but not doing it correctly. A tip is to identify a closed loop and choose a direction (clockwise works fine here). When going across a resistor determine if you're going with the current in segment or against the current in that segment. If it's the former (with current), put a negative sign on your I*R term. This is because you're actually going from a higher potential to a lower potential (because that's the direction of current flow). If it's the latter, leave the I*R term as positive. When you're going across voltage sources (like cells), if you're going from the negative terminal to the positive, leave that as positive V (since you're going up in voltage). If you're going from the positive to the negative terminal, make that term negative (-V). Collect all the terms in a sum on the left hand side. The right hand side is always equal to zero.

With these rules, you should be able to generate the right equations.

You've neglected the 3V source completely here.