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I'm having trouble connecting voltage drops and induced current. Imagine you have a triangular loop make up off three resistors. You place this loop in a constantly changing magnetic field. This changing magnetic field will cause an EMF and a resulting induced current in the loop.

Now imagine I place a voltmeter across one of the resistors. What does the voltmeter read? $I\times R$ or $I\times 2R$? If you look at the situation one way, you are measuring the voltage drop across one resistor, so V would equal $I\times R$. But if you look at in another way, you're also measuring the voltage drop across the other half of the circuit which contains TWO resistors (hence a voltage drop of $I\times 2R$).

I have tried to model this situation in my schools physics lab and I do see that there is a voltage drop across the resistor so the answer is not "there is no voltage drop". My model wasn't perfect, but it seemed to imply that the voltmeter read $I\times 3R$? How could that be?

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When you measured the voltage, did you ensure the voltmeter circuit DOESN'T see the magnetic field?? I mean that there is no way an EMF was generated in the circuit of voltmeter, because if that happened the reading of voltmeter will be irrelevant. Please double check it –  Gotaquestion Nov 3 '13 at 21:31
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What the voltmeter reads in this case is not the classical voltage drop as the concept of voltage itself, as a scalar function whose gradient gives the field, becomes redundant. What it displays is, indirectly, the current through it, which in the case of induced emfs, is not related to "potential drop" across its end. (it will display $V$ as its reading where $V=Ir$, $I$ the current through it, and $r$ the resistance of the voltmeter.

As an example, consider the case you present in the question. Three resistors of resistances $R=5$ each are connected to form a triangle. A voltmeter of resistance $r=100$ is connected across two vertices. Let the change in magnetic flux be confined to the triangular portion only, and be $1 weber/s$. Therefore, the line integral of the field along a line which completely includes the area with changing magnetic flux must be $1 volt$. Let the current through the voltmeter be $I_1$(from A to C) and across the two resistors (except the one across which the voltmeter is attached) be $I$(clockwise). The current through the resistance across which the voltmeter is connected is $I-I_1$(from A to C).

enter image description here First, consider, the line integral along the triangular portion in clockwise manner. The integral will give $$-2IR-(I-I_1)R=1$$ $$-15I+5I_1=1$$ Considering the line integral along $C\rightarrow B\rightarrow A\rightarrow \text{Voltmeter}\rightarrow C$ clockwise, we get $$-2IR-I_1r=1$$, since in both the case the change in magnetic flux (time rate of change) is same, as the change is confined to the triangular area which is bound completely by both the line integrals. $$-10I-100I_1=1$$ on solving we get $I=21/310=0.068$(anticlockwise) and $I_1=1/310=0.0032$(anticlockwise). Therefore the voltmeter will show $I_1r=0.32V$ which is equal to $(I-I_1)R$ but not $2IR$ since, in cosidering the voltmeter attached to only the resistance $AC$, we get a simple circuit without any changing magnetic flux, but considering the voltmeter as being attached to the resistances $CB$and $BA$, we get a circuit which has a changing magnetic flux, which needs to be accounted for, and hence reading will not be according to $2IR$, or the hypothetical "potential drop" across the two resistors.

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The series resistance of a voltmeter must typically be small compared to the other resistances in the circuit. So the reading of the voltmeter reads the voltage on a resistor in the circuit with a tiny error due to series resistance of the voltmeter itself. If the voltage drop across series resistance of voltmeter is comparable to voltage drop elsewhere, it can’t be used to measure the voltage. I don’t think that the series resistance of voltmeter is relevant here –  Gotaquestion Nov 4 '13 at 11:40
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@Gotaquestion Voltmeter resistance must be high compared to other circuit resistances since it has to attached in parallel and not in series. The resistance of an ammeter needs to be very low, as it has to be attached in series with other circuit elements. –  Satwik Pasani Nov 4 '13 at 14:42
    
I made a mistake, it should be "large" instead of "small" in the first sentence of my comment, thanks for bringing my attention to it. Anyway, the point of my comment was that the resistor of voltmeter should be chosen such that its effect is neglected. In your answer you seem to rely on voltmeter's resistor in explaining the strange reading the question states. From designing point of view the circuit shouldn't see the resistor of voltmeter. It is sees it, then the voltmeter's reading is completely irrelevant to what one is trying to measure. I suppose you agree on that, don't you? @Satwik –  Gotaquestion Nov 4 '13 at 15:26
    
Satwik: Thanks very much for that answer. Very clear. But let me ask another conceptual question based your answer. In your diagram, take the voltage meter and flip it over to the other side of the circuit. So the loop that contains the voltmeter and the single resistor AC now has the changing flux through it. If you redo the calculations, you get different numbers for the current and most specifically, a different number for the volt meter reading. It's now 0.64V Does that make sense? Conceptually? For the record, I didn't that change in my lab, but there could have been other problems. –  Jeff Knapowski Nov 4 '13 at 22:33
    
@Gotaquestion You are right, for an ideal voltmeter, the resistance must be infinite and hence in an ideal circuit, resistance of the voltmeter makes no sense. But in circuits with changing magnetic fluxes (non-conservative fields), an ideal voltmeter cannot exist, since if it does, it will show the value of potential drop without affecting the circuit, but potential drops in such circuits are not well defined and hence an ideal voltmeter gives a paradoxical result. Therefore, to have a meaningful reading, the voltmeter must have a finite resistance. –  Satwik Pasani Nov 5 '13 at 1:56
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