# I need help redrawing a circuit!

1. Feb 18, 2014

### Magnawolf

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

2. Relevant equations

I just need help redrawing the circuit

3. The attempt at a solution

I don't understand how to draw the power sources in correctly. I'm getting mixed on on whether they're in series or parallel with the resistors and such. Any tips would be appreciated.

2. Feb 18, 2014

### phinds

What's the point? There are no load nodes indicated, so I can't see that you are trying for a Thevenin equivalent and there is nothing wrong with the way the circuit is drawn now. What is it you are trying to do?

Edit: by the way, if you can't determine the current I1, you really need to go back and study the basics some more. There just is no way to create a more simple example of when to use V=IR

3. Feb 18, 2014

### Magnawolf

I'm trying to study the basics right now lol.

I understand I1 = 6, because the voltage is 24 V and the resistor is 4.

What throws me off is the -8V on the other side. How does it behave in the circuit?

I2 is apparently 16A. Is it because the voltage through the resistor is 32V?

Now what confuses me is I3. I3 is apparently 0.8A, which is 8V/10Ω. Does that mean the 24V has no effect on that resistor? If so, why? Does the ground play a role in this circuit?

Also, I is 22V which I don't understand at all. I'm assuming I have to find R-total and E-net. Are all the resistors in parallel? That -8V is throwing me off. Would E-net be 24+8 = 32V?

4. Feb 18, 2014

### phinds

good.

. Yes, 24 minus a minus 8 adds up to 32.

How could it be anything else? It has an ideal voltage source across it that is 8v. That makes it irrelevant what the rest of the circuit is doing.

It gives you a reference point for ALL of the voltages in the circuit.

Forget the voltage. You've already FOUND all of the currents in that node except for I, so just do a simple node analysis. 16+6=22.

5. Feb 18, 2014

thanks bruh