Timeline for Measuring voltage drop from induced current
Current License: CC BY-SA 3.0
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| Nov 9, 2013 at 11:55 | history | edited | stochastic13 | CC BY-SA 3.0 |
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| Nov 5, 2013 at 2:00 | comment | added | stochastic13 | @JeffKnapowski You are correct.In such cases, the reading (and the currents) depend on the spatial orientation of the circuit elements and the voltmeter. If it(loop of the voltmeter and resistor $AC$) includes a part or the whole of the changing magnetic flux, you will get a different answer, because the current distribution changes. It is conceptually sound, since Faraday's laws($\int E.dr=-d\phi/dt$) will be valid in all conditions. In short, the reading changes with the spatial position of the voltmeter, and hence use of a voltmeter in such circuits is meaningless. | |
| Nov 5, 2013 at 1:56 | comment | added | stochastic13 | @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. | |
| Nov 4, 2013 at 22:33 | comment | added | Jeff Knapowski | 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. | |
| Nov 4, 2013 at 15:26 | comment | added | Gotaquestion | 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 | |
| Nov 4, 2013 at 14:42 | comment | added | stochastic13 | @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. | |
| Nov 4, 2013 at 11:40 | comment | added | Gotaquestion | 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 | |
| Nov 4, 2013 at 10:22 | history | answered | stochastic13 | CC BY-SA 3.0 |