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Suppose I have a following circuit:

enter image description here

I do not understand, why the potential difference between the points $c$ and $d$ is equal to the potential difference between the points $b$ and $a$? That is, why $\Delta V _{cd}=\Delta V_{ba}$? As I understand, potential is the potential energy per unit charge, and the potential difference tells us how much work shall we do in order to bring a test charge from one point to another. If so, why the potential differences mentioned above are the same? The distances aren't the same, so it seems obvious for me that the work should also be different.

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Possible duplicate: physics.stackexchange.com/q/8675/2451 and links therein. –  Qmechanic Jan 11 at 16:13
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2 Answers 2

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The crucial fact about these idealized circuits and electric potential differences that leads to the assertion you want to justify is

Wires are modeled as perfect conductors (Ohmic resistors with negligible resistance) for which there is zero potential difference between any two points. (This was edited from "perfect conductors are equipotentials.")

If we assume that the wires are perfect conductors (zero resistance), then Ohm's Law immediately gives the result above.

Having this fact in hand, we first notice that we have the following mathematical identity (which is basically the "loop rule"): $$ V_a + (V_d-V_a) + (V_c - V_d) + (V_b-V_c) + (V_a - V_b) = V_a $$ which we can rewrite as $$ \Delta V_{da} + \Delta V_{cd} + \Delta V_{bc} + \Delta V_{ab} = 0 $$ Now using the fact above, we note that the potentials at any two points connected solely by a wire are the same, so that $$ \Delta V_{da} = 0, \qquad \Delta V_{bc} = 0 $$ which gives $$ \Delta V_{cd} + \Delta V_{ab} = 0 $$ and therefore since $$ \Delta V_{ab} =- \Delta V_{ba} $$ we get the desired result $$ \Delta V_{cd} = \Delta V_{ba} $$

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Right, in other words, $a=d$ and $b=c$ for all the purposes of voltage. The work done on a charge is simply the potential difference times charge. The distances (mentioned by the OP) are completely irrelevant when we actually know the voltage. The same voltage may be divided to longer or shorter distances but we know the voltages and they're manifestly the same, $V_d-V_c = V_a-V_b$ because $V_d=V_a$ and $V_c=V_b$ because of the direct conection between the points. –  Luboš Motl Jun 10 '13 at 17:46
    
+1, I see that I wrote a similar answer simultaneously. –  Wouter Jun 10 '13 at 17:50
    
@joshphysics - Wow, thanks for the detailed answer. I know my question is quite newbie, but no one really cared to explain this thing in detail. However, I still have a question left - you emphasized "any two points connected solely by a wire" - can't a resistor be considered a "part of the wire"? I've told that they are both conductors (although the resistor's conductivity is lower), hence they should be equipotential as well (like one unified conductor). –  stuck_with_problem Jun 10 '13 at 18:06
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@joshphysics - Thank you. One more important thing that was an obstacle for me in understanding the subject - you've mentioned that conductors are equipotential. This is true for electrostatics. However, when we talk about currents, we usually think of another constant electric field applied inside the wire, that forces charges to have some net flow, drift velocity. That means we have a non-zero field inside the wire, and that means that the potential difference should be some non-zero value as well. However you say that it should be zero. What am I missing? Sorry, this is really confusing. –  stuck_with_problem Jun 10 '13 at 18:15
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@WalterWhite That's actually a great question. In fact, precisely as I was writing my response, I thought of that and realized that I had never thought of how to justify that step. Moreover, I turned to two of the grad students sitting next to me, and they had never thought about it either. In short; I don't know the answer to that one. In fact, I am thinking of posting that as a new physics.SE question if you don't mind. –  joshphysics Jun 10 '13 at 18:19
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First let's establish a feel for why they can be the same and then work out why they are indeed the same. The distances are not equal, but the resistance the test charge would face is also not the same, so the work might well be the same. This should show that the voltages can be the same.

To find that they are indeed the same, we need not consider the elaborate description of the work done on a test charge. It is sufficient to note that the potential difference must be zero if we complete the entire circuit. Seeing as the potential difference between the positive and the negative side of the source is some negative amount $-V$, we know the total voltage over the entire rest of the circuit must be $V$.

Taking the wires of the circuit to be perfectly conducting, we only encounter two obstacles: $R$ and $r$. Call the voltages (potential differences) over these $V_R$ and $V_r$, respectively. Then it is clear that we must have $V_R+V_r=V$.

But the potential difference between $b$ and $a$ is simply the sum of all the potential differences inbetween. So $\Delta V_{ba} = (-V) + V_r = V_r-V = -V_R$ which is obviously the same as $\Delta V_{cd}$ since $\Delta V_{dc} = V_R$.

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+1: as a courtesy due to simultaneity. –  joshphysics Jun 10 '13 at 17:52
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+1 from me, sir. I appreciate your answer and it helped me! –  stuck_with_problem Jun 10 '13 at 18:30
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