Calculating Current Through a Battery in a Series Circuit

[roam]

Homework Statement

Determine the current through the 10-V battery.

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

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 I_15V=15/15=1 A, and the current drawn from the 10V battery is I_10V=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...

 

[vela]

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.

 

[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.

 

[roam]

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.

Okay, so I labled the currents:

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

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...

 

[lewando]

This Kirchhoff's voltage law gives the equations:

5I₁+5I₂=25
5I₁+5I₂=10
I₂=1.5

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

Kirchhoff's junction law gives:

I₁+I₂-I₃=0

OK.

So from the very first equation [that makes no sense...] we get I₁=3.5, ...

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

 

[vela]

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.

 

[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?

 

[vela]

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)

These are correct.

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

This last equation is wrong, but it doesn't matter since it's not necessary.

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

I₁=1-0.5I₂

You made an algebra error solving for I₁.

Substituting:

15-5I₂-5(I₂-(1-0.5I₂))=0

I₂=0.8

You made another error solving for I₂.

You really need to be more careful doing the algebra.

 

[roam]

Oops, thank you for all your guidance.