# A simple electric circuit question

1. Aug 7, 2010

### arkofnoah

1. The problem statement, all variables and given/known data
http://img19.imageshack.us/img19/7701/screenshot20100808at010.png [Broken]

2. Relevant equations

3. The attempt at a solution
Do add up the resistances like this:

$$R_{total} = (\frac{1}{R} + \frac{1}{r} + \frac{1}{r})^{-1}$$

or

$$R_{total} = R + (\frac{1}{r} + \frac{1}{r})^{-1}$$

Last edited by a moderator: May 4, 2017
2. Aug 7, 2010

### diazona

Is R in series or parallel with the power sources?

As a reminder, two circuit elements are in series if and only if any current that comes out of the first one must enter the second one and any current that enters the second one must come out of the first one.

Also, circuit elements A and B are in parallel if and only if all current that goes into both A and B comes from the same section of wire, and all current that comes out of A and B goes to the same section of wire.

Last edited: Aug 7, 2010
3. Aug 7, 2010

### arkofnoah

So does that mean that when taken alone, any one of the power source r is in series with the resistor R because the current of any one must enter R?

also imagine that if i remove one of the power source, the other power source is now in series with the resistor. however if i now put in the second power source again, the first two is no longer in series with the resistor anymore?

can there be a current flowing between two positive terminals with equal potentials? this means the current coming out from, say, the leftmost power source, will only have a single path to follow, won't it?

4. Aug 7, 2010

### diazona

Ah, thanks for asking that. I neglected to mention one condition. In order for two elements to be in series, it must also be the case that the current coming into the second one must come from the first one. I edited my previous post to include that.
So are you talking about the two power sources as a group, or individually? It's possible for a group of circuit elements (A,B,C) to be in series with another circuit element (D), even if individually A is not in series with D, B is not in series with D, and C is not in series with D.
Remember ΔV=IR. If two terminals have equal potentials, what's ΔV, and then what's I?
Yeah, that's correct.

5. Aug 7, 2010

### arkofnoah

Is the following an example of the above circuit:

http://img62.imageshack.us/img62/1310/screenshot20100808at014.png [Broken]

So ABC taken together is in series with D. A, B and C are in parallel with one another. But is A, taken individually, in series with D?

Anyway I will assume that they are all in parallel because after what you said I have a strong feeling that they are in parallel. So following that I take the formula

$$P = \frac{V^{2}}{R}$$

to get the power dissipated by a single power sources

$$P_{r} = \frac{V^{2}}{r}$$

and double that for the two power sources.

am I on the right track so far?

Last edited by a moderator: May 4, 2017
6. Aug 7, 2010

### arkofnoah

Last edited by a moderator: May 4, 2017
7. Aug 7, 2010

### diazona

That's exactly the case I was talking about.
Yep, that works.
Right, you can use this circuit as a model for how you add up the resistances in your original question.

Last edited by a moderator: May 4, 2017
8. Aug 7, 2010

### arkofnoah

ok after some calculations I got the reciprocal of the answer C.

I just divide

$$P_{r} = \frac{V^{2}}{\frac{r}{2}}$$

by

$$P_{total} = \frac{V^{2}}{R_{total}}$$

and ended up with

$$\frac{2R+r}{r}$$

Hmm... where did I go wrong?

9. Aug 7, 2010

### kuruman

Although it is a good exercise to actually derive the fraction of power loss, this multiple choice question can be correctly answered by eliminating the answers that cannot possibly be correct. You can do that by inspection, without math. For example, answer (A) cannot possibly be correct (Why?) Can you find the other two and justify their elimination? Think "physics".

10. Aug 7, 2010

### diazona

In this formula, you've plugged in the equivalent resistance of the whole circuit, but you're still only using the voltage drop over one of the resistors. That doesn't work. If you want to calculate the power dissipated by a resistor, you have to use the voltage drop over that resistor in P=V^2/R.

11. Aug 7, 2010

### arkofnoah

Okay interesting approach and a good practice in itself!! A is obviously wrong because the fraction is more than 1. I'm thinking B is wrong because of the negative sign in 2R-r but I'm not sure if means anything. And for D I'm suspecting it has something to do with the numerator being 2R but again I don't find this line of reasoning convincing.

Give me some hint :tongue:

Last edited: Aug 7, 2010
12. Aug 7, 2010

### arkofnoah

OH!! The voltage drop for the internal resistance is the difference between the emf and the terminal p.d, while that for R is the terminal p.d itself. I didn't notice that.

But anyway I think in this case it's easier to use the P = I^2 R formula because the current passing through each of the power source is exactly half of that of the resistor R.

I get the answer now Thanks!

13. Aug 7, 2010

### kuruman

Correct thinking for (A). For the rest, what is the internal resistance of an ideal battery and what should the ratio be when you have ideal batteries?

14. Aug 7, 2010

### arkofnoah

Zero for both? That's when the fraction of power dissipated by the sources is at the minimum.

So I conclude that the fraction of power dissipated by the sources is proportional to the sum of the resistance of the sources and inversely proportional to the sum of resistance for the entire circuit! Correct reasoning?

15. Aug 8, 2010

### kuruman

Correct. When r goes to zero, there should be no internal power dissipation. So when r goes to zero how many of the multiple choice ratios (A) - (D) go to zero and how many do not?

I don't see how you reached this conclusion merely on the basis that the ratio has to be zero when r goes to zero.

16. Aug 8, 2010

### arkofnoah

Wow!! Okay I completely did not think of that