Transients(RC) elements in the circuit after commutation?

[builder_user]

Аh! I see.

Didn't the phasor result have a angle?

I forgot formula for amplitude. arcsin(x)=f <- phase

 

[gneill]

For x = a + jb, amplitude is sqrt(a² + b²). The phase angle is atan(b/a), but beware of the quadrant... the trig functions on a calculator deal with "priciple values" over a limited domain. If using MathCad, the atan2 function is available. Or, if you're working directly with complex numbers, |x| is the amplitude and arg(x) will give the correct angle.

 

[builder_user]

U=100*(0.72+j*0.69)

f=arcsin(0.72)=arcos(0.69)=46*

so for t=0
U=100*sin(0+46)=72?

 

[gneill]

atan(0.69/0.72) = 43.8°

I calculated 44° ...

 

[builder_user]

Oh.
43.9=44*

я used arcsin(0.72) but sin(x)=0.69 so arcsin(0.69) is needed

 

[builder_user]

last question.
How this circuit looks before & after commutation?
Is my circuits correct?

 

[gneill]

In the 'after commutation' circuit, immediately after the switch opens the capacitor will have a voltage on it that will want to discharge through the resistors R3 and R4. After a long time, the capacitor should look like an open circuit, not a short circuit.

 

[builder_user]

In the 'after commutation' circuit, immediately after the switch opens the capacitor will have a voltage on it that will want to discharge through the resistors R3 and R4. After a long time, the capacitor should look like an open circuit, not a sort circuit.

This circuits?

Long time after commutation I only need to find voltage on R3?Is R4 in the circuit?

Immediately after commutation I need to find current on resistor R4?

 

[gneill]

The circuit on the left looks fine for a long time after commutation. Note that there will be no current through R4.

For the circuit on the right, representing the instant immediately after commutation, if the capacitor had zero voltage on it from before, then the circuit is correct.

 

[builder_user]

The circuit on the left looks fine for a long time after commutation. Note that there will be no current through R4.

For the circuit on the right, representing the instant immediately after commutation, if the capacitor had zero voltage on it from before, then the circuit is correct.

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

 

[gneill]

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.

 

[builder_user]

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

 

[gneill]

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.

 

[builder_user]

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...