Transients(RC) elements in the circuit after commutation?

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builder_user said:
Really?Great!
But if before commutation Uc was !=0 so It would look like long time or not?

If Uc != 0 at commutation, then you would have to deal with that "initial" voltage in the "new" circuit.

One way to do that is to replace the charged capacitor with an equivalent circuit that behaves in the same manner. This would be comprised of the same capacitor, but uncharged, in series with a voltage source equal in value to the Uc value. You can then treat this new circuit as you would any other: in the instant after commutation the uncharged capacitor looks like a short circuit and the new voltage source will remain. This should allow you to find initial circuit conditions such as initial voltages and currents.

Of course as time moves forward (t > 0) the capacitor starts to charge as current flows, just as in any other circuit. Keep in mind that the "equivalent capacitor" comprises both the capacitor and voltage source! To find the voltage across the real-life capacitor that it replaced, you need to find the voltage across both items in series.

After a very long time the original capacitor will still end up looking like an open circuit of course, with whatever end-to-end voltage the circuit demands. The "equivalent capacitor" will behave similarly.
 
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gneill said:
If Uc != 0 at commutation, then you would have to deal with that "initial" voltage in the "new" circuit.
One way to do that is to replace the charged capacitor with an equivalent circuit that behaves in the same manner. This would be comprised of the same capacitor, but uncharged, in series with a voltage source equal in value to the Uc value. You can then treat this new circuit as you would any other: in the instant after commutation the uncharged capacitor looks like a short circuit and the new voltage source will remain. This should allow you to find initial circuit conditions such as initial voltages and currents.
.

You mean that immediately it will be look like pic.1? But I can use Laplace replace
 
Attachments
  • Circuit schematic labelled immediately after commutation: voltage source E feeds R1 (1k) in series, then a junction with R3 (1k) to ground, and a branch through capacitor C1 (100n) to R4 (1k) returning to ground.
    jpg.JPG
    10.3 KB · Views: 484


builder_user said:
You mean that immediately it will be look like pic.1? But I can use Laplace replace

It would look more like the figure below. Note that for the first instant this new C1 will look like a short circuit, just as you're used to doing for analysis.

The new voltage source needs to have the polarity that the voltage on the capacitor had just before the switch commutated.
 
Attachments
  • Circuit schematic labelled immediately after commutation: voltage source E feeds R1 (1k) and R3 (1k) in series, with C1 (100n) in series with a battery labelled Uc(t=0-) and R4 (1k) on the right branch.
    Fig1.gif
    6.7 KB · Views: 591


Hm..I think I'll don't like conductors. Inductor is in the circuit(as a wire) or is not in the circuit but conductor... I still don't know all cases. Or there is no sth(resistor for example) in parallel connection with it. Or there is no current if it series connection. Or it has a voltage source and etc...