# How do I find total current when there are two emfs?

1. Oct 28, 2014

### Sol_Engineered

The question is what it is. Its a general confusion which I am having.

Problem:

Q. Using kirchoffs laws to calculate branch currents and total current.

Now I can use Kirchoffs voltage law to calculate branch currents through individual resistors. But when it comes finding total current I am dumbfounded.

Attempt:

1k ohms=I1
2.2k ohms=I2
3.3k ohms=I3

Kirchoffs voltage law:

1000I1+2200I2=12
3300I3+2200I2=12

Kirchoffs current law:

I1+I3=I2

Simplification using above equation:

1000I1+2200I2=12 (1)
1100I2-3300I1=12 (2)

I1= -13/1900A, I2=179/20900A, and I3= I2-(-I1)=161/10450A

Where to go from here?

(1) By total current does it mean what the current is produced by "each of the emfs "or "both emfs collectively"?
(2) If I were to find the total current for each emf in the circuit, then how would I be using my branch currents to do that? Which resistor would I be choosing? For starters I think I1 would give me total current for 12V but then there is I2 in series with it. How do I know if am choosing the correct resistor?
(3) Should I break this circuit into two individual circuits? Like making it into network A and B and then make 2200ohms and 1000ohms resistors in series, and similarly 2200ohms and 3300ohms in series, to find total current by each emf respectively?

Last edited: Oct 28, 2014
2. Oct 28, 2014

### Staff: Mentor

(1) I can understand your confusion. The phrase "total current" is at best ambiguous in a circuit with multiple sources. It's possible that they mean the current in the branch with the 2200 Ω resistor. It could also mean the sum of the currents produced by all sources. If this is an assignment where you hand in your workings, I'd suggest explaining your confusion and provide both.

(2) The answer to (1) doesn't impact finding the individual branch currents, and you don't need any "total current" value(s) to determine them. You have three parallel branches and you want to find the current in each. A branch current isn't associated with any one resistor, it's associated with the branch itself. All components connected in series in a given branch share the same branch current.

(3) Don't break the circuit into separate circuits. When loops border each other as in this circuit, what happens in one loop can effect what happens in the other thanks to any shared components.

There are specific cases where you can break a circuit into subcircuits, for example if the only shared component is a fixed voltage source. But that's not the case here.

Oh, and you might want to re-do your calculations for the currents. Your two loop equations and your current equation look fine, but the result you obtained for the currents don't look right to me.

3. Oct 29, 2014

### Sol_Engineered

Thank you so much sir.

You mentioned my values are wrong: either (1) Solved wrong (2) Sign problem (positive or negative). I will look into that. I have seen that you have to substitute in the exact value (without taking off the negatives) into the equations if (2) is the case you are pointing towards?

This is my final question: can I ever sum the branch currents to find the total current? Likewise can I add Power dissipated for individual resistors to calculate total power dissipated?

Thanks this is my final query. You have explained it well sir. :)

4. Oct 29, 2014

### Staff: Mentor

Begin by checking your simplified equation (2) where you eliminated I3:

This is my final question: can I ever sum the branch currents to find the total current? Likewise can I add Power dissipated for individual resistors to calculate total power dissipated?
[/QUOTE]
If there's a single power source feeding several parallel branches then it makes sense that the total current (supplied by the source) would be the sum of the branch currents. When there are multiple sources things can be trickier. For example, it's possible that current can be forced to flow into the + terminal of a voltage source by another, larger source (this is how battery chargers work). Should that current then be subtracted from the total?

Regarding power, yes you can sum the individual powers dissipated by the resistors to find the total power dissipated.

5. Oct 30, 2014

### Sol_Engineered

Thank you so much sir for your help