Ladder Circuit Paradox: Equivalent Resistance Analysis

  • Thread starter Thread starter GPhab
  • Start date Start date
  • Tags Tags
    Circuit Paradox
AI Thread Summary
The discussion centers on calculating the equivalent resistance of an infinite ladder circuit by equating the total resistance to that of a segment combined with the existing ladder. Participants explore whether the derived formula for resistance holds true across different ladder models, with specific attention to configurations involving resistors in series and parallel. The conversation highlights the need for circuit diagrams to clarify complex relationships and calculations. A query arises regarding the impact of adding an additional resistor directly connecting segments on the overall resistance. The analysis emphasizes the importance of understanding both the mathematical process and the underlying logic in circuit design.
GPhab
Messages
24
Reaction score
0
We find the equivalent resistance(let us denote it by R) of an infinitely long ladder circuit by considering that it is equal to the same resistance in combination with one segment of the circuit. But when we equate "R" to "R" combined with two segments, will we get the previous result for all possible ladder circuits(I got it for two models. Are these the only possible?)?
 
Last edited:
Physics news on Phys.org
It's hard to talk specifics w/o looking at a circuit diagram, but in general ladder networks are modeled as an (infinite) series of circuit elements, each consisting of (a) a resistor R1 in series and (B) a resistor R2 in parallel w/ the "existing" ladder.

Call the (effective) resistance of the "existing" ladder R.

Now add another "link" to the ladder.

This link will consist of R1 in series + (R2 in parallel w/ R).

Set this equal to R, the effective resistence of the ladder.

You can now solve for R in terms of R1 and R2.
 
Hi GPhab,

Provide us with the diagram of at least one model.
 
Good explanation... but what if there is an additional resistor connecting them directly. I understand the process but not the logic behind it; how would this addition change the resistance?
 
Hello! Let's say I have a cavity resonant at 10 GHz with a Q factor of 1000. Given the Lorentzian shape of the cavity, I can also drive the cavity at, say 100 MHz. Of course the response will be very very weak, but non-zero given that the Loretzian shape never really reaches zero. I am trying to understand how are the magnetic and electric field distributions of the field at 100 MHz relative to the ones at 10 GHz? In particular, if inside the cavity I have some structure, such as 2 plates...
Back
Top