# Circuit with batteries facing the same direction

1. Dec 9, 2013

### cvsanchez

1. The problem statement, all variables and given/known data
Find the potential difference at point P in the diagram.

2. Relevant equations
V=IR
ΣiVi=0
ΣIin=ΣIout

3. The attempt at a solution
I understand that you need to use Kirchoff's circuit laws to find the current and voltage at each resistor, but I can't comprehend how a current could flow at all since the positive and negative terminals are essentially isolated.
I haven't been able to find an example in my notes or textbook and nothing online clarified it.
My main problem is that I can't picture how any current "loops" could form in a circuit such as this one.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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2. Dec 9, 2013

### cpscdave

When doing KCL you really can't be wrong so long as your consitant.
On the LHS of the the equation where you put voltages from batteries when the loop goes in the + side of the battery if you count that as a +voltage then when the loop enters into the - side you count it as a minus

You're using KCL as it isn't immediatly clear which directly the current will flow. The great thing about KCL is it doesnt matter. If you define a current loop in a direction that turns out to be wrong you'll simply end up with a negative current

3. Dec 9, 2013

### cvsanchez

So are you saying that if I wanted to start with the 12V battery, I would analyze the circuit as if the other two batteries were just wires?

4. Dec 9, 2013

### Staff: Mentor

Potential difference with respect to what? For there to be a "difference" there must be a second point of reference involved.

There are several loops. A loop is simply a closed path that traverses components only once (i.e. the path does retrace its steps).

Here are some loops in your circuit:

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5. Dec 9, 2013

### cpscdave

Well that is one way to do it. The laws are linear as such you could split it into 3 problems and then add up the currents along each of the paths from each of the solutions.
But thats extra work.

When going through a loop you just need to be consitant with your voltage rises and drops. If the loop direction enters the negative terminal it's a voltage rise so 0 = +V1 + .....
if it entered the postive terminal instead it'd be a voltage drop so 0 = -V2 + ....
If you had both you would get
0 = +V1 - V2 + .....

6. Dec 9, 2013

### cvsanchez

The question itself said to find the potential difference at the negative terminal of the 3V battery. I took that to mean that I needed to find the voltage after the current went through the 5Ω resistor.

The loops diagram you posted was helpful! So, is the reason loop ABED is clockwise because the 12V battery has a higher voltage than the 6V, so the net current progresses clockwise?

7. Dec 9, 2013

### cvsanchez

Ok, so if we look at Gneill's diagram (I'll attach it), when the 12V passes through the 6V battery (going from B to E) I should reduce the voltage by 6V. For loop ACFD when the 12V passes through the 3V battery the voltage should be reduced to 9V. Would I then not do any calculations for loop BCEF?

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8. Dec 9, 2013

### cpscdave

I just solved this using KCL to make sure that I wasn't making any errors :)
In this problem you have 2 unknowns I1 and I2 (the third current going through the middle is just I1+I2) so you only need two loops.
I solved it with the following loops
ABED
and
CBEF
I can't think of a reason why using the 2 loops you specified wouldn't work :) and yes it sounds like you have the crux of it.

I have my 2nd loop CBEF going the oppisate direction as whats in Gneill's diagram as I found it that it is easier to do the math if you have the loops going in the same direction when they overlap :) Otherwise you have to remember of the loops is negative :D

9. Dec 9, 2013

### cpscdave

I strongly strongly STRONGLY suggest you use it only as verification, a lot of people I went through Circuits 1 with used falstad to solve homework and then failed the exam as they didnt know how to do it. Where as I used it to tell me when I made errors and did quite well on the finals :D

10. Dec 9, 2013

### Staff: Mentor

Perhaps you should post the exact wording of the question. Potential is always measured with respect to some reference point. It's analogous to height -- what is the height of a building? It's the distance measured from the ground to the top of the building, the ground being the point of reference.

I paid no attention to the batteries or their relative potential differences. I simply chose clockwise arbitrarily. This is fine because when you write the KVL equations in the direction of the loop the math will automatically take care of things; The resulting currents may turn out to be negative, but that just means the "guess" for direction was "incorrect" based on that criterion (it's not really incorrect, because the negative value tells you that the actual direction of the current is opposite to your guess. The result will still be correct).

11. Dec 9, 2013

### cvsanchez

This is perfect! I can actually see how the current is flowing now! Plus I can actually check my work! Thanks!