What is the instantaneous value of UL after commutation?

[builder_user]

l = 12

l=0.012

 

[gneill]

l=0.012

If you say so. In your first post you had:

K1=1.2
L = 10*K1

You did not indicate units (for anything!) so I assumed Ohms for resistances and Henries for inductance.

 

[builder_user]

Thevenin equivalent circuit and wrote KVL around the loop.

This loop?

 

[builder_user]

You did not indicate units (for anything!) so I assumed Ohms for resistances and Henries for inductance.

I forgot.In every task I have millihenries and so I've not written it there.

 

[gneill]

This loop?

No, this loop. All of the source and resistor network is replaced by its Thevenin equivalent. Makes life simple.

 

[gneill]

I forgot.In every task I have millihenries and so I've not written it there.

Okay. Perhaps in future you should just state the component values. All the K multipliers and assumed units is confusing.

 

[builder_user]

No, this loop. All of the source and resistor network is replaced by its Thevenin equivalent. Makes life simple.

Thevenin equivalent after commutation?Or at the moment of commutation?One of the equivalents that was found before?Or not found in this topic?

 

[gneill]

Thevenin equivalent after commutation?Or at the moment of commutation?One of the equivalents that was found before?Or not found in this topic?

After the switch closes, so yes, after commutation.

The circuit at the moment the switch closes remains the same thereafter.

It's only the initial conditions (the existing current in the inductor) that makes the moment of commutation special.

 

[builder_user]

How to do inverse Laplace transform in MathCAD?

 

[gneill]

How to do inverse Laplace transform in MathCAD?

Select the 's' variable and then Symbolics --> Transform --> Inverse Laplace.

 

[builder_user]

Thanks.