# Comparing the current through multiple batteries

1. Mar 4, 2015

### acdurbin953

1. The problem statement, all variables and given/known data
Is the current through battery D greater than, less than, or equal to the current through battery C. Bulbs identical, and batteries ideal and identical. I took a picture of the circuit, hopefully you can see it - ignore the scribble!

2. Relevant equations
Kirchhoff's current rule

3. The attempt at a solution
Originally when I did the problem I though IB=IA, and got the paper back with it marked as wrong. I am reviewing for an exam now, and am hoping someone can let me know if I've resolved the problem.

IB < IA. We know that the current that goes into the top node must be the same as the current through the bottom node. Why B is less than A is because B is in series with a bulb, which increases the resistance of that branch, and thus lowers the current.

2. Mar 4, 2015

### haruspex

You mean at X and Z? Those are junctions, so it's not clear what you mean by the current 'through' them.
You could consider voltages there, but you have the problem that battery C might oppose the potential difference, so get the lower current.

3. Mar 4, 2015

### BiGyElLoWhAt

Another very relevant equation is Ohm's Law. You can use this to quantitatively check your results. Let the resistance of the bulbs be R.

4. Mar 4, 2015

R1 and R2?

5. Mar 4, 2015

### BiGyElLoWhAt

They're identical.

6. Mar 4, 2015

### haruspex

Ah yes, missed that in the text - thanks.

7. Mar 4, 2015

;)

8. Mar 4, 2015

### acdurbin953

Sorry, I meant that the current which enters X (our prof refers to them as nodes), must be the same the amount of current that leaves Z.

Is this correct?

VC=I(R/2), where R/2 is the equiv. resistance from the bulbs in parallel. So IC=2V/R
and
VD=IR, where R is the one bulb in its parallel branch. So ID=V/R

Current is proportional to voltage, so IC must be greater.

9. Mar 4, 2015

### BiGyElLoWhAt

Right answer, but slight misconception. Look at the sea of electrons model. An electron leaves battery c, what does it see resistance wise? Repeat the process for d. Hint* the results will differ by a factor of 4, not 2.

10. Mar 4, 2015

### Staff: Mentor

Redraw the circuit, varying the layout trying different or more conventional layouts. Do this a couple of times, more if necessary, until you experience that "Ah, ha!" moment.

11. Mar 4, 2015

### haruspex

Still not meaningful. What do you mean by "the current which enters X"? It has three links. There may be current entering on two and leaving on one. If you mean the net current then that must be zero. If you just mean the sum of the positive flows into X then I don't see why that should be equal (and opposite) to the sum of the negative flows into Z.
Certainly the flow entering X from Z equals the current leaving Z for X.

12. Mar 5, 2015

### Staff: Mentor

13. Mar 6, 2015

### acdurbin953

Thanks for all the help, you guys! I did finally figure it all out. Re-drawing the circuit helped a lot.

14. Mar 7, 2015

### Staff: Mentor

Care to share your discovery? What is the ratio of battery currents?

15. Mar 7, 2015

### acdurbin953

Well redrawing the circuit confirmed for me that D must have less current than C. For finding the actual ratio I used Ohm's law as suggested up-thread. I used a random value for the batteries (5 V) and a random resistance for the bulbs (10 ohm), and got that battery C has 1.5 times the current.

16. Mar 7, 2015

### Staff: Mentor

That ratio does sound right! C has more current.

17. Mar 7, 2015

### acdurbin953

18. Mar 7, 2015

### acdurbin953

My a-ha moment came in realizing that the current through each parallel branch must sum up to the current experienced by battery C. So I analyzed each branch separately as if it was just a series circuit. So, the left branch in series with battery C would have current 0.5 A. The branch on the right would have current 1 A. The total current then for battery C must be 1.5 A. Current ratio C/D = 1.5/1 aka 4/3.

Last edited: Mar 7, 2015