How Do You Handle KVL Limitations in High-Frequency AC Circuits?

Click For Summary
KVL (Kirchhoff's Voltage Law) is not applicable in high-frequency AC circuits due to the presence of fluctuating magnetic fields, which disrupt the assumption of a conservative electric field. In such cases, circuit analysis must account for electromagnetic wave propagation delays, making traditional lumped element models inadequate. Instead, circuit elements should be represented in schematics to include magnetic coupling, such as transformers. The discussion emphasizes that circuit analysis is an approximation and often relies on simplifying assumptions. Ultimately, when dealing with high-frequency circuits, alternative methods must be employed to accurately calculate voltage and current.
Muhammad Usman
Messages
52
Reaction score
3
Hi,

I was reading KVL and KCL and I read that KVL cannot be applicable to all the situations specially when :-

" KVL is applicable on the assumption that there is no fluctuating magnetic field linking the closed loop. While, in presence of changing magnetic field in a High Frequency but short wave length AC circuits, the electric field is not a conservative vector field. So, the electric field cannot be the gradient of any potential and the line integral of the electric field around the loop is not zero, directly contradicting KVL. That’s why KVL is not applicable in such a condition "

My Main question is how to deal with this inconsistency or in other words if we face such circuit where AC frequency is high then which method is used
for such kind of circuits to calculate the voltage and current. Many thanks
 
Engineering news on Phys.org
The PF Insights article https://www.physicsforums.com/insights/circuit-analysis-assumptions/ goes over the assumptions needed for KVL, and KCL, and circuit analysis in general. I think the key one for your question is:

The time scales of interest in CA are much larger than the end-to-end propagation delay of electromagnetic waves in the conductors.

When the frequencies are high enough so that is not true, then you can't use circuit analysis.
 
  • Like
Likes Fisherman199 and jim hardy
KVL and KCL are tools applied to "lumped element" circuit analysis. What this means in practice is that you only consider circuit elements that are drawn in the schematic. If magnetic coupling between circuit loops is to be considered, then that should be drawn in the schematics as a circuit element(s), like a transformer between the loops, for example.
Circuit analysis is always an approximation in a purely theoretical sense. Problems often have implicit simplifying assumptions to allow us to use tools like KVL.
So, for example, you probably don't have to worry about quantum mechanics or cosmic rays, unless someone says they matter.
 
  • Like
Likes jim hardy
Muhammad Usman said:
" KVL is applicable on the assumption that there is no fluctuating magnetic field linking the closed loop.

I still regard that claim as 99% sophistry.
Why cannot the induced emf be lumped and represented as just another voltage source?

Antennas , transmission lines and waveguides are a different story .
To me they're a separate field with its own textbooks and math. No place for beginners...

old jim
 
jim hardy said:
I still regard that claim as 99% sophistry.
Why cannot the induced emf be lumped and represented as just another voltage source?

It might be a semantic difference Jim, but an equivalent circuit is a different circuit. Once you have the equivalent circuit, it is analyzed using CA rules without flux and without charge. An ideal transformer in a transformer equivalent is a device that makes no reference to magnetic flux.

From the article.
The time rate of change of magnetic flux outside any conductor is zero. [##\frac{∂ϕ}{∂t}=0##]
If the initial flux at t=0 was zero, it remains zero. [So basically forget magnetic flux for CA. Forget Poynting vectors.]

The time rate of change of electric charge inside any conductor is zero. [##\frac{∂q}{∂t}=0##]
 
  • Like
Likes Fisherman199 and jim hardy
I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage. The suggested explanation that the current is equal to the power supply divided by the voltage, and hence higher voltage leads to lower current and as a result to a lower power loss on the conductives is very confusing me. I know that the current is determined by the voltage and the resistance, and not by a power capability - which defines a limit to the allowable...

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 25 ·
Replies
25
Views
3K
  • · Replies 4 ·
Replies
4
Views
6K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
6K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K