# Potential difference across parallel circuits

1. Jul 5, 2007

### truewt

I am currently having some difficulty in recalling/understanding the potential difference across parallel circuits. Why is the potential difference across individual components of the parallel circuit the same?

Let's for example say 2 resistors of different resistance are connected in parallel. As p.d. is defined as the amount of energy dissipated by the charges as they pass through the resistor (am I wrong in the definition?), why is it so that for different resistances, the energy dissipated by the charges are the same?

2. Jul 5, 2007

### mjsd

i am not sure what you mean by "individual components" but remember in a circuit diagram, all wires are assumed to be ideal and so if you elements share the same set of terminals (ie. parallel to each other) than the potential difference across them will be the same (by definition)

well the resistor with less resistance will let more CURRENT through and the same potential difference means that more power will be dissipated on this device compare to the other one.

3. Jul 5, 2007

### truewt

So what is the explanation for saying the potential difference is equal?

4. Jul 5, 2007

### ranger

An alternate picture of viewing parallel connected components in this case (2 resistors) is to picture each of the resistors connect [individually] directly across the terminals of the voltage source. When you think of it this way, its roughly like a series circuit; hence the voltage drop across a single resistor is equal to that of the voltage source. Taking this thinking further, its obvious that the p.d across both resistors would be equal, however the current would not be [given unequal resistances].

5. Jul 5, 2007

### truewt

@ranger: That could be a way, but is there a better explanation for the potential differences being equal?

I'm trying to find the best argument for this concept. I can hardly remember what my teacher explained (or did he?) a few years back.

Basically what I'm saying is why are the charges in the circuit so smart as to be able to dissipate the same amount of energy?

Let's take another example. 3 resistors are in a circuit. 2 are in parallel, and the last resistor is in series with the 2 parallel resistors. Why are the charges (which split up into seperate branches in the parallel component) able to 'use up' the same amount of energy (ie the pd drop) and then join back into the main branch at the same potential?

6. Jul 5, 2007

### ranger

V = W/q; where W is work (joules) and q is unit of charge [moved]. As you know the amount of charge will vary depending on the resistance in the circuit. Now if we have a 10V source, so thats 10 joules of work per coulomb of charge. If we have 3 coulomb of charge flowing in one branch:
V = 30J/3q = 10J/C = 10V
[C = coulomb which is the unit of charge; J/C = V]
If in that very same circuit, on a different branch we have 7 units of charge. This will give an electric potential of:
V = W/q = 70J/7q = 10J/C = 10V.

Makes sense?

7. Jul 5, 2007

### youknowme

Hmm, I have wondered about this too and I thought of an explanation that might work. Think about this, the electrons in the main branch tries to move forward and when it comes to the split point, the electrons in the wire will exert an equal force to the electrons in each parallel circuit. So the potential difference for each circuit is the same.

8. Jul 5, 2007

### truewt

Why is it so that the ratio between the work done by the charges and the amount of charges (ie the pd) always the same in the branches of the parallel component?

9. Jul 5, 2007

### Staff: Mentor

First of all, remember that real wires have some small but finite resistance. So depending on how you connect up the parallel components, they may see slightly different voltages across them depending on how much voltage drop there is in the interconnections. This is what you should use to answer your original question -- ask yourself what the resistance of the interconnections is, compared to the resistance of the components that are connected in parallel. If you use 30AWG (tiny) wire to wire up a bunch of 1 Ohm power resistors "in parallel", with 100cm space between each resistor, and the first resistor adjacent to the power supply, then you will not measure the full power supply voltage 9m away at the end resistor....

And in your quoted question here, I think you are asking about the ratio of the power in a branch to the current in the branch? When you say "charges", I think you mean the flow of the charges with respect to time, right? If so, P=VI should answer the question, right. P/I=VI/I=V, which is the same across all branches....

10. Jul 5, 2007

### truewt

Hmmm, to the first part of your answer, it is right. But usually we assume ideal conditions and the wires have no resistance..

As for the second part of your reply, the point i'm asking is that why is the voltage the same? why is the work done per unit charge across each resistors in the parallel circuits the same?

11. Jul 5, 2007

### Staff: Mentor

Why is the voltage the same? We already answered that. Because the interconnection resistance is negligible compared to the parallel component resistances. To get a voltage drop, you need current flowing through a resistance. V=IR, so if R is negligible, the V is too.

I also already answerd the second question. P/I=VI/I=V, which is the same for all branches.

Let's try the second one a different way to see if it helps. Let's say that you have a single resistor of value R across the power supply, so obviously I=V/R, and P=VI=V^2/R. Now split that resistor into two parallel resistors of value 2R each, so the total I=V/R still, but now I/2 flows through 2R in each of the two parallel branches. Do you see where I'm going with this? What do you do next in this exercise to show the property that you are asking about? What is the current in each of the 2 parallel legs? What is the power in each of the two split legs compared to the original power in the single R before the split? Does that make it more intuitive?

12. Jul 5, 2007

Imagine a water pump hooked up to two pipes, in a configuration as like the one below with my beautiful ASCII art.

---W------________======B=======
|--------|___A___//_______________\\
|--------|======__________________=====D===
|--------|_______\\_______________//
----------________======C=======

If the water pump is W, and B and C are equal pipes. What would be the water pressure through A? What about the pressure at B and C. How would the flow of water change through B and C? What would the flow through A be compared to D?

13. Jul 5, 2007

### truewt

Yes I get what you are trying to explain. But aren't we following a circular argument since we try to explain with equations, which are derived from certain concepts?

I just can't seem to understand why is the voltage drop across each resistor
the same? Why can't it be different? We argue that voltage drops because of resistance, but in parallel component, why must the drop be the same?

@Frogpad: Yea this explanation is what most of my teachers use when I was taught a few years back. But it seems that I'm still stucked at the problem with why can't the 'pressure' be different in the parallel component.

Thanks everyone for your replies but it seems that my brain has become 'more retarded'.

14. Jul 5, 2007

The pressure is the same, but the RATE at which the water is flowing is different.

Maybe we can think of it like this (this is not a 100% accurate analogy).

When you hook up one pipe there is water moving through. Imagine it moving really fast through the pipe.

Now when you hook up two pipes, the water is now split, so the flow must decrease accordingly in each pipe. Imagine the water slowing down in each of these pipes.

Now hook up 10 pipes. What would happen to the flow of water?

But the thing is, this whole time the PRESSURE on the top of these pipes has remained constant. You are applying the pressure to the top equally. Imagine this:

if you stand on a board with a spring underneath.
like
____________
[___________]
-------S
-------S
-------S
______S______

If S is the spring and you are standing on the board on the top you put a certain amount of force on the BOARD right?

Now imagine putting two springs underneath. When you stand on the board you put the same amount of force on the board right? Your weight hasn't changed. However, the system underneath has. Try to mentally think about the difference between current and voltage.

At least for me, I may not have a 100% correct mental image, but I have intuition from these models. Now back your intuition up with the math.

Maybe it would serve better to explain at an atomic level what is going on?

15. Jul 6, 2007

### Defennder

The following explanation assumes this experimental setup: Two resistors of unequal resistance R1 and R2 connected in parallel to a constant DC emf source. Let the current through R1 and R2 be I1 and I2. And pds across each of them be V1 and V2. U denotes the energy dissipated by the entire circuit in a given time t. U1 and U2 denotes the energy disspiated in R1 and R2 over the same time t.

We know that V = U/q right? Now we show that U = VIt . For a constant current, q = It. Hence U = VIt.

Now we make use of the principle of conservation of energy. This is essentially the explanation why pds across parallel circuits are the same.

You would agree that the energy dissipated by the entire circuit in a given time t would be equivalent to the sum of the energy disspated by both R1 and R2, right?

Hence U = U1 + U2, for a given time t.

VIt = V1(I1)t + V2(I2)t

Since by Kirchoff's current law, I = I2 + I1 (may be proven by conservation of charge),

V(I1 + I2)t = V1(I1)t + V2(I2)t

Dividing by t:

V(I1) + V(I2) = V1(I1) + V2(I2),

where by comparing coefficients, we see that the only way the above is true is when V=V1=V2.

Hope this helps.

16. Jul 7, 2007

### truewt

Thanks Defennnder, now I have come up with a very mathematical proof for what you mentioned. Some arguments involved which you fail to mention. But thanks for giving me a lead.

Anyhow, I still find it fascinating that charges are so smart as to be able to use the same amount of energy supplied to each individual charge to overcome the resistance when the circuit is set up in parallel branches.

And the same goes for the water pressure system. In parallel pipes (Frogpad's illustration), the water is able to maintain the same pressure under the correct circumstance (cross sectional area of pipe) after passing through the resistance. Or is it more correct to say that we are forcing the water pressure to remain constant by setting up the pipes to certain measurements?

Because from the water pipe analogy, I found out that in order for the water pressure to be the same in both parallel pipes after passing through the resistance, you need to 'configure' the pipe's cross sectional area in order to allow for that. Or am I wrong?

17. Jul 7, 2007

### guten

Charges do what they do. Why do they do things that follow any order at all? Why have they followed the same order for every experiment for a couple hundred years? Why can't we trick them? It is fascinating.

If you are looking for a answer that doesn't use math I would suggest finding a book about the history of science or a history of the early scientists. It is very likely your library would have a good one.

Look for something that reads similar to http://en.wikipedia.org/wiki/Georg_Ohm#_note-1

This will probably put charges into a different context for you to think about them. And should aid in figuring out why charges are so smart.

18. Jul 7, 2007

If I put my hand on your back and push, why doesn't my hand go through your body? You could argue philosophy, or maybe argue from an atomic standpoint. Lets pick the atomic standpoint and dramatically simplify the situation. Consider two protons atoms. One proton can represent my hand, the other your back. Lets completely disregard quantum effects. As I move two protons together a Coulombic force will exist between them; opposites attract, and likes' repel. Do you accept this?

From your argument, you should not. Why are the protons smart enough to obey Coulombs law? In fact why does any artifact of nature obey physical laws? I think a better way of phrasing it would to say, why do OUR models of the natural world characterize these systems so well? As you can see, these questions quickly digress, but I hope you see my point.

The resistance of a nearly uniform current density inside a conductor can be modeled as:

$$R = \frac{\rho L}{A}$$
Where A is the cross section of the conductor, L is the length of the conductor, and $\rho$ is the resistivity.

Notice the dependence on cross section though. When you increase the cross section, you decrease the resistance.

About the water analogy. You MUST remember that this is a CLOSED system! You can add all these pipes to our imaginary system, but in the end they all connect to one point, and end at one point (since they are in parallel). So when thinking of a potential difference, it is the the potential difference across the top and bottom of all the pipes.

19. Jul 8, 2007

### truewt

Thanks Frogpad. I get where you're going.

Talking about the two protons analogy you gave, coulombs' law is a hypothesis proven theory later on for an observation made? What about this parallel circuit issue? Was the 'law' a hypothesis proven theory after an observation was made first?

I can't understand the explanation you gave for the water analogy. Right now I think i'm digressing the topic, because my mind is some sort thinking more deeply into the water system instead of the water pipe system as an analogy for the electrical system.

Because we know water pressure is dependent on the speed of the water, and the speed of the water is dependent on the flow rate which in turn is dependent on the cross sectional area of the pipes.

That is why I am sort of thinking that in order to use the water pipe system as an analogy, we sort of need to configure the variables.

20. Jul 8, 2007

What math class are you up to?
What physics (or engineering class) are you up to?

The reason I ask, is I think you will be unsatisfied with an anwer until you take more uppder division classes. It would not be a such a bad idea to just accept that the potential across a parallel circuit is the same in the ideal case. I mean you are definitely NOT denying it. So first accept this as fact, and then keep asking questions. Just so you know, I am not trying to jump ship. You can keep this thread open until you die. Someone WILL be able to answer your question in a suitable fashion.

About the water analogy. Honestly, I only use that analogy to gain intution into the problem. I don't know if they map 1:1 (so to speak). When I get some time, I will look into water pressure and see how much that relates to your question.

I think I see more where you are coming from with this statement. I honestly don't know. All I can say is it just makes sense inuitively for me. I don't know if it did early in my studies or not; I guess I have blocked that out.

Answer the first two questions I asked in this thread. We can always go the route of jumping into it from a more rigorous (as rigourous as I can make it anyways (don't count on much)) treatement of what you are asking by jumping into it from a vector analysis approach.