Electrostatic potential...in electrodynamic regime in the expression $P=U.I$

Question

Let us consider a two terminal electrical component. It is typical to calculate the power it consumes by considering the voltage drop $U$ and the current passing through it $I$ and to say it absorbs a power $P=U.I$ (of course voltage and current should be oriented in the appropriate manner to say that it absorbs this power). I emphasize on the fact that this formula is also used when varying-in-time fields are considered (the instantaneous power reads $P(t)=U(t)I(t)$)

Now, from electromagnetism theory, we know (among other things) that:

$$\mathbf{E}=-\mathbf{\nabla} V - \partial_t \mathbf{A}$$ $$\mathbf{B}=-\mathbf{\nabla} \wedge \mathbf{A}$$ $$\mathbf{\nabla} \wedge \mathbf{B}=\mu_0 \mathbf{j} + \mu_0 \epsilon_0 \partial_t \mathbf{E}$$

In electrostatic regime, the relationship between voltage drop around two points $A$ and $B$, and electric field is simply:

$$U=\int_A^B \mathbf{E}.\mathbf{dl} $$

This is "consistant" with the general expression of the electric field I gave in term of potentials, assuming $\partial_t=0$

Question1:

What happens in the varying regime ? To what the $U$ in $P=U.I$ corresponds to ? Is it still the line integral of the electrical field ? If so, the voltage drop now depends on the precise path where the integration has been performed because we also integrate the quantity $\partial_t \mathbf{A}$. Then it is not really well defined because we don't need to specify "a path" when we talk about the voltage around a two terminal electrical component in varying regime.

Or maybe the answer is to take in general the "path independant" quantity:

$$U=\int_A^B (\mathbf{E}+\partial_t \mathbf{A}).\mathbf{dl} $$

Question2:

If the voltage drop is still defined as $U=\int_A^B \mathbf{E}.\mathbf{dl} $, how to make sense of the fact such definition depends on the exact path of integration ? Is there a way to make sense of it in the lumped-approximation regime, which means that we don't see spatial variation of electrical waves around the size of the electrical element we are considering ?

Answer 1

Question1:

What happens in the varying regime ? To what the $U$ in $P=U.I$ corresponds to ? Is it still the line integral of the electrical field ?

Most usually in AC circuits, voltage is drop of electrostatic potential, the line integral of electrostatic part of the field (the Coulomb field that depends only on charge distribution). The power $P=UI$ is then rate of work of electrostatic forces only (due to electric charge present because of battery or other energy source). This work turns up as other form of energy (for example, as internal energy ("heat") in a resistor or as a magnetic energy in ideal inductor).

We can try to change definition of $U$ to integral of total electric field along the current path, but then (if there is significant induced electric field present in the current path), the power $P=UI$ will have different meaning and value: rate of work of total electric field, which turns up always as non-electromagnetic energy. For example, rate of work of total electric field in a real inductor with ohmic losses turns up as internal energy in the inductor. With good conductivity, this can be much lower than rate of work of electrostatic forces.

"Voltage" both in physics and engineering is unfortunately an overloaded term and confusion is ubiquituous. Usually it is best to stick with the meaning

1. voltage is drop of Coulomb potential (thus integral of the Coulomb field, a quantity that depends only on endpoints, not path).

However, sometimes when only total field is known or of interest (working of a voltmeter) term "voltage" is used to mean

2. voltage is integral of total electric field, which then is not function of endpoints only, but depends on the path choice; and the chosen path is often assumed to be the path of electric current.

These two definitions are at odds with each other in situations with induced electric field, so one should use only one definition at a time or find another more descriptive name for quantity 2), e.g. "total electric electromotive force".

In AC circuit theory, where currents change in time, the first definition is actually the standard one and is much less confusing.

For example, potential drop on an ideal inductor (no material core, no ohmic losses) in the pre-chosen positive direction is known to be

$$ p.d. = L\frac{dI}{dt}. $$ This is a unique value, self-inductance $L$ does not depend on choice of integration path in space, only on geometry of the ideal inductor. However, if we used the second definition of "voltage", on ideal inductor it would always be zero (because ideal inductor is made of ideal conductor for which total electric field vanishes inside). That would not be very useful as it would be inconsistent with the standard exposition of AC circuits.

Question2:

If the voltage drop is still defined as $U=\int_A^B \mathbf{E}.\mathbf{dl} $, how to make sense of the fact such definition depends on the exact path of integration ?

"Voltage drop" is a very confusing term for this, it is better to say just "voltage". The sense is made by the fact that some preferred path is implicitly assumed (path of electric current) so this "voltage" has unique value. This is just the definition 2) described above.

Is there a way to make sense of it in the lumped-approximation regime, which means that we don't see spatial variation of electrical waves around the size of the electrical element we are considering ?

One could try to formulate everything in terms of definition 2) where only total field is used and the integration path is always the path the current takes. But it is not standard, see the problem with ideal inductor described above.

The solution is much simpler. In lumped element model of AC circuits, use the definition 1). The definition 2) is not needed.