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#562 – Electroboom!

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Kalvin:
Here is the circuit from Lewin's video with 1V voltage source.

If the KVL holds, it possible to split the original circuit in half at D and A by inserting two antiparallel 1mA current sources between nodes D and A. After split, the currents and the voltages of the original circuit remain unchanged.

If the KVL holds, it possible to split the original circuit in half at D and A by inserting a voltage source between nodes D and A with value V(D,A) because there is no current flowing through the new voltage source. After split, the currents and the voltages of the original circuit remain unchanged.

It can be seen that KVL holds and it is possible to spilt the circuit in half at D and A.

The 1V voltage source can be located some other point in the original circuit, which will affect the equivalent circuit values, but KVL will still hold.

Here is the circuit from Lewin's video when the solenoid is generating 1mA current flowing in the circuit. It can be seen that KVL fails when the circuit contains of non-conservative fields, and thus it is not possible to split the circuit at D and A.

Kalvin:
The simple circuit analysis above suggests that Dr. Lewin had it right, and there were no probing errors involved, even if the results look counterintuitive. KVL fails when there are non-conservative fields present.

jesuscf:

--- Quote from: Kalvin on December 03, 2021, 12:53:16 pm ---Here is the circuit from Lewin's video with 1V voltage source.

If the KVL holds, it possible to split the original circuit in half at D and A by inserting two antiparallel 1mA current sources between nodes D and A. After split, the currents and the voltages of the original circuit remain unchanged.

If the KVL holds, it possible to split the original circuit in half at D and A by inserting a voltage source between nodes D and A with value V(D,A) because there is no current flowing through the new voltage source. After split, the currents and the voltages of the original circuit remain unchanged.

It can be seen that KVL holds and it is possible to spilt the circuit in half at D and A.

The 1V voltage source can be located some other point in the original circuit, which will affect the equivalent circuit values, but KVL will still hold.

Here is the circuit from Lewin's video when the solenoid is generating 1mA current flowing in the circuit. It can be seen that KVL fails when the circuit contains of non-conservative fields, and thus it is not possible to split the circuit at D and A.

--- End quote ---

I see, you too, are confused with the 'non-conservative fields' statement.  Although it may appear that in the circuit the electric field is non-conservative because it is changing over time, so it is the magnetic field.   What happens is that the the total energy in the circuit is conservative: all energy added to the circuit is consumed by the circuit.  From the circuital point of view is easier to work with instantaneous power than energy.  For example at time t when the EMF is 1V the calculated current is 1mA, then: EMF*I=I2(R1+R2).  The external varying magnetic field is adding 1mW to the circuit and the circuit is consuming 1mW.  The fields in the circuit are conservative.

thinkfat:

--- Quote from: jesuscf on December 03, 2021, 03:07:21 pm ---I see, you too, are confused with the 'non-conservative fields' statement.  Although it may appear that in the circuit the electric field is non-conservative because it is changing over time, so it is the magnetic field.   What happens is that the the total energy in the circuit is conservative: all energy added to the circuit is consumed by the circuit.  From the circuital point of view is easier to work with instantaneous power than energy.  For example at time t when the EMF is 1V the calculated current is 1mA, then: EMF*I=I2(R1+R2).  The external varying magnetic field is adding 1mW to the circuit and the circuit is consuming 1mW.  The fields in the circuit are conservative.

--- End quote ---
Emphasis mine.

The "fields in the circuit" are not to be confused with the fields outside of the circuit. The "non-conservative electric field" described by the Maxwell-Faraday equation is not in the circuit.

PS: "change over time" is not a sign of non-conservativeness. The "curl" is.

bsfeechannel:

--- Quote from: jesuscf on December 03, 2021, 03:07:21 pm ---I see, you too, are confused with the 'non-conservative fields' statement.  Although it may appear that in the circuit the electric field is non-conservative because it is changing over time, so it is the magnetic field.   What happens is that the the total energy in the circuit is conservative: all energy added to the circuit is consumed by the circuit.  From the circuital point of view is easier to work with instantaneous power than energy.  For example at time t when the EMF is 1V the calculated current is 1mA, then: EMF*I=I2(R1+R2).  The external varying magnetic field is adding 1mW to the circuit and the circuit is consuming 1mW.  The fields in the circuit are conservative.

--- End quote ---

He understands what a non conservative field is. You don't. That's why you think he's confused.