# Current Through a Battery

1. Oct 11, 2011

### roam

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

Determine the current through the 10-V battery.

[PLAIN]http://img69.imageshack.us/img69/8016/crct.jpg [Broken]

3. The attempt at a solution

This is very confusing. I know the equivalent resistance of the circuit is 15 since all resistors are in series. So, the current drawn from the 15V battery is I15V=15/15=1 A, and the current drawn from the 10V battery is I10V=10/15=0.66 A. What can I do next? The correct answer has to be 2.3 A, but I can't figure out how they've got that...

Last edited by a moderator: May 5, 2017
2. Oct 11, 2011

### vela

Staff Emeritus
The resistors aren't in series. To be in series, they need to be connected end to end with nothing else connected to the middle nodes.

Start by labeling the three currents in the circuit. Then apply Kirchoff's voltage law to two loops and Kirchoff's current law to one of the nodes. This will give you three equations and three unknowns, which you can solve.

3. Oct 11, 2011

### itaischles

I think you can also use the superposition principle here (correct me if I'm wrong...)
First short one of the sources, solving for the voltages and currents, then do it for the other one. Each time you have only one source and the resistors are connected in parallel or in series. The answer is a superposition of the different results you got for each branch.

4. Oct 19, 2011

### roam

Okay, so I labled the currents:

[PLAIN]http://img812.imageshack.us/img812/1738/picturebl.png [Broken]

This Kirchhoff's voltage law gives the equations:

5I1+5I2=25
5I1+5I2=10
I2=1.5

Kirchhoff's junction law gives:

I1+I2-I3=0

So from the very first equation we get I1=3.5, and from the junction law we get that I3=5.

If the current through the 10V battery is the I1, why is it that my answer is wrong? It should be 2.3 not 3.5...

Last edited by a moderator: May 5, 2017
5. Oct 19, 2011

### lewando

These make no sense. This could be part of your problem.

OK.

Recommend you refresh your mind with the correct method of doing KVL mesh analysis, and give it another try.

6. Oct 19, 2011

### vela

Staff Emeritus
One suggestion: Now that you've assumed directions for the currents, for each resistor, put a plus sign where a current enters and a minus sign where the current leaves to indicate the sense of potential difference across the resistor. As you go around a loop, when you go across a resistor from plus to minus, that means the voltage drops, so that term goes into the equation with a minus sign. If you go from minus to plus, that means the voltage increases, so that term goes into the equation with a plus sign.

7. Oct 20, 2011

### roam

Okay, I revised Kirchhoff's voltage law, I tried it again:

Left Inside Loop:

10-5I1-5(I1-I2)=0 ..........(1)

Right Inside Loop:

15-5I2-5(I2-I1)=0 ..........(2)

Outside Loop (looking at the contribution of the 15V battery to the brach with 10V battery in it):

15-5I+10-5I=0
I=2.5

I rewrote the first equation in terms of I1 and substituted it in the second:

I1=1-0.5I2

Substituting:

15-5I2-5(I2-(1-0.5I2))=0

I2=0.8

And substituting this into (1) we get I1=0.6.

The only current that I think goes past through the battery is the 2.5. But I still don't get it, how did they get 2.3 Amps?

8. Oct 20, 2011

### vela

Staff Emeritus
These are correct.
This last equation is wrong, but it doesn't matter since it's not necessary.
You made an algebra error solving for I1.
You made another error solving for I2.

You really need to be more careful doing the algebra.

Last edited: Oct 20, 2011
9. Oct 20, 2011

### roam

Oops, thank you for all your guidance.