# Help kirchof's rules,

1. May 23, 2012

This isn't really a problem, but more of a conceptual question. I have a circuit drawn (see attached), there are 2 batteries. I don't understand how to split currents at junctions with two batters, It seems like I cannot split the current from both batters at EVERY junction. It's confusing. If you look at a current from right to left on each battery, they run into a number of junctions such that the current from one battery is moving in opposite direction of current from the other if you split currents of BOTH batteries at EVERY junction.

For instance if I have current from battery 2 going from right to left, at A there is a junction, so I split the current with some going up towards B and some going towards b. As I go right to left with battery 1, I also hit the junction at B. Does the current go down there towards A? I don't believe you can have current going in opposite directions. So do I just take the current from Battery 1 to the junction at c without splitting it at B? Also, does the current from Battery 2, going right to left, split at b? but what about the current from battery 1 at c, is it going to split? If current from both batteries 1 and 2 split at the junctions c and b, respectively you would have current from battery 1 going opposite current from batter 2 between b and c. I don’t get it….

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2. May 23, 2012

### Staff: Mentor

You generally do not put voltage sources (like batteries) directly in parallel like that. If their voltages differ at all, they will fight each other.

If you put some resistance in series with each battery, you can just write the usual KCL or KVL equations to solve for the currents in each component.

3. May 23, 2012

### Staff: Mentor

Is that a wire between A and B? If so, if the batteries are not perfectly identical in potential it's going to cause an unrealistic situation: an infinite current will flow through the batteries as they both attempt to "have their own way" and make the potential THEIR potential, not the other guy's potential. It's not a situation that is encouraged in the world of circuit design and any circuit built that way will fail (sometimes spectacularly ) in short order.

If the batteries ARE identical, then that wire AB makes R2 and R3 into a parallel pair, and the batteries into a parallel pair; Replace them both with a single resistance and a single battery.

Kirchhoff's rules do in fact allow currents to flow in both directions through a given wire, at least mathematically speaking. The actual current through the wire is then the SUM of the calculated currents, and whatever that sum is is the actual current that flows and in the direction dictated by the sign of that sum.

You will discover that there are circuit analysis methods that take advantage of this mathematical trick and allow you to write the circuit equations automatically by a straightforward procedure. If you're curious, look up "mesh analysis".

4. May 23, 2012

Edit: I just saw the second reply...this was to the first guy,

Yes, I understand how to apply the kirchof rules in many cases where there aren't so many junctions. The problem I'm having looking at a circuit like the one I drew is I don't know how many unknowns there really are. If I have current from battery 1 traveling against current from battery 2 on the same wire, with no junction between them, that's impossible right? So how do you decide which currents split and juntionss and which don't? I split the current from Battery 2 at A, but couldn't I have split the current from battery 1 at 1 instead?

5. May 23, 2012

Yes, that is a wire, there's current through it right?

6. May 23, 2012

### azizlwl

If there is potential difference between A and B.
Normally when we connect batteries in parallelm,as in your diagram, we assume that they have same voltage. Thus no potential difference and not current flows.

In calculation we assume also there current flow in opposite ways but the net flow(actual flow) must be in normal direction, positive to negative(higher to lower potential).

Last edited: May 23, 2012
7. May 24, 2012

I'm looking at the case when the batteries are not equal. The circuit comes from a problem in my text; in many problems my teacher will draw the currents over the batteries much like I did i.e. the batteries are in parallel, and the currents are both from left to right. I'm just confused when it comes to dealing with junctions.

Perhaps someone could show me how they'd work with the currents and junctions...

8. May 24, 2012

### azizlwl

In the batteries loop circuit, there no resistance. Thus you can't apply Kirchoff Loop Rule(KLR).

9. May 24, 2012

### asgerbj

Just as a note, you can try to use LTSpice to help you solve these circuits, and also to create better drawings :-) Its free and easy to use. Just to avoid any questions about the circuit itself.