Timeline for Using KVL in a changing magnetic field
Current License: CC BY-SA 4.0
9 events
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| Sep 22 at 20:44 | comment | added | Ján Lalinský | @Peltio we "have to" if we want to explain standard circuit theory. You don't have to if instead of circuit theory and $LdI/dt$, you want to discuss IEC definition, relativity, or simple connected regions. $V=RI$ is not valid in copper in presence of induced EMF, or battery EMF, or other impressed force not due to potential difference. I repeat, you can check this experimentally by an oscilloscope. Simple Ohm's law $V=RI$ is too simple for many situations, including inductor, where approximately $V=LdI/dt$, and it makes no sense, and is not standard, to try to recover it there. | |
| Sep 22 at 9:34 | comment | added | Peltio | No we don't have to. This might have been a rational position in the late 1800, but now that we know the relationship between electric and magnetic fields, and relativity, we can appreciate how work per unit charge done by the electromagnetic field can be path dependent. V = L i' is valid between the terminals (and in all other paths inside the dB/dt-free simply connected region of space enveloping the circuit path), while V=Rcopper * i is valid inside the copper of the coil. I really need to make another answer with figures for the other question, but I do not have the time right now. | |
| Sep 21 at 23:43 | comment | added | Ján Lalinský | @Peltio We have to end up with $V=LdI/dt$ for a perfect inductor, so Ohm's law for $V$ cannot hold when current changes in time. | |
| Sep 21 at 23:43 | comment | added | Ján Lalinský | @Peltio I get here you want to recover, in the copper, Ohm's law in the form $V=RI$, but this is a misconceived goal - there is no need for the simple Ohm law to be valid in AC regime for any part of the coil. In this regime, we have generalized Ohm's law, taking into account induced EMF, instead. If you do it anyway, and define voltage this way all the way from one terminal to another, then when integrating voltage contributions from one terminal to another, you'll get a value that has nothing to do with voltage as used in circuit theory. | |
| Sep 21 at 22:46 | comment | added | Peltio | What I mean is measure the voltage with a voltmeter along a one quarter turn of a one coil transformer supplying 12V to a 12kohm load. If you use my definition of voltage you find a value that is nearly zero and agrees with ohms law in copper (say V = rho_copper * length / area * I), but if you call voltage potential difference you have V = 1/4 *12V = 3V with 1 mA and a resistor of a few milliohms in copper cable. I will make a few examples, including a partial coil, in my other answer in a few days. | |
| Sep 21 at 22:22 | comment | added | Ján Lalinský | @Peltio If by Ohm's law you mean the modern local relation $\mathbf j = \sigma \mathbf E^*$, then this does apply inside the conductor that makes the inductor, but this is also not using voltage in any way, and is thus independent of how we define the word "voltage". | |
| Sep 21 at 22:21 | comment | added | Ján Lalinský | @Peltio Voltage values used in KVL are electric potential differences. You can check that the path-restricted IEC definition you use gives exactly those values. Voltage in other contexts may be something different, like the IEC integral of net electric field - I'm fine with that. But the question is about KVL. There is no drawback, because Ohm's law $V=RI$ does not apply to inductors (as a whole, $V$ being voltage on inductor terminals), so why would we expect this law to apply inside it. Ohm's law was generalized already by Kirchhoff to include all kinds of EMFs. | |
| Sep 21 at 22:06 | comment | added | Peltio | just to say that what you call "voltage", both the IEC and ISO call "(scalar) potential difference". Your definition might have been a valid one in the late 1800s, but as of today, we know better. See Purcell, for example. One big drawback of using scalar potential difference instead of voltage is that Ohm's law ceases to work inside inductors. Apply it to two points inside the copper of a coil with a 12V terminal voltage supplying a current of 1mA to a load and see if it makes sense. | |
| Sep 21 at 21:54 | history | answered | Ján Lalinský | CC BY-SA 4.0 |