RL Switching Theorm

[Chezma]

Any one knows about the 'RL Switching theorem',where a complicated RL switching circuit can be reduced to a single loop of resistance and inductance in series ?
i'm trying to use this theorem to find the circuit's time constant using step input function.
Any reference available?
Thanks for help

 

[VEBill]

What's the application?

 

[IRstuff]

Is this for school?

TTFN

FAQ731-376

 

[Chezma]

I have a very complicated magnetic device and i want to simpify the circuit to find its transient time response for step input and calculate its time constant.

 

[IRstuff]

I suspect there is no magic formula, just brute force and ignorance simplification of the network in question. At the end of the day (literally), you'd get a real term and an imaginary term, and the imaginary term can be associated with an equivalent inductor, and the real term would be the resistor.

TTFN

FAQ731-376

 

[Chezma]

I think thevenin's theorem can be used in very special way for step input to simpify the circuit and end up with inductance an resistor in series.

 

[Chezma]

I simplified the device to a resistor R1 in parallel with inductor L1 and this combination in series with another resistor R2 ,the input is step funtion.

But how this arrangement can be reduced furthe to one resistor in seires with one inductor ?

 

[electricpete]

Have you heard of Norton and Thevinin equivalent circuits?

Every linear system has one. It gives the same sinusoidal steady state response, by superposition must also provide the same transient response (provided initial conditions are properly translated).

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(2B)+(2B)' ?

 

[Chezma]

So,can you prove that the two above circuits are equivalent for step input and have the same time constant.
1)Circuit one:
R1 in parallel with L1 and the combination in series with R2.
2)Circuit two:
R3 in series with L2.

 

[electricpete]

I apologize, the Thevinin equivalent applies for sinusoid, but does not apply in general for transient analysis.

In the specific case you cited, if we tried to solve
R1*s*L1/(R1+s*L1) + R2 = R3 + s*L2
We cannot find R2 and L2 to satify the relation for all values of s.

=====================================
(2B)+(2B)' ?

 

[VEBill]

The brute force numerical approach would be to fire up Spice, and start a manual search (using binary algoithm) for the circuits that match you expectations. Allow a couple of days and it should be very easy.

 

[electricpete]

.
1)Circuit one:
R1 in parallel with L1 and the combination in series with R2.
2)Circuit two:
R3 in series with L2.

We could analyse by Laplace transform as shown attached to determine the analytical form of the step response for systems 1 and 2. The Laplace form is not the same function of s and so the time form is not the same function of t. We cannot determine any R3 and L2 parameters for circuit 2 that will match general R1, L1, R2 (only in special cases like L1-> infinity could we get an exact match).

So, I guess I have not heard of an R-L switching theorem that allows reducing a general R/L circuit into a simple R-L circuit which performs the same under transients like step response. I don't think such a theorem exists if the requirement is exact match. I'll be interested to hear what you find.

=====================================
(2B)+(2B)' ?