19-Tanzanite

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Electrical Engineering Challenge #3

Forum|Forum|2 years ago
April 26, 2024
4 replies
6369 views

Hi,
For below circuit:

1. Find the expression of the voltage across C2 (Vb) at t>0
2. Find the voltage across C2 at t=100ns when C1=100nF
3. Find the voltage across C2 at t=100ns when C1=10nF

Also let C1 and C2 voltage as t-> infinity to be Vb.

Best answer by Werner_E

Here in Prime using correct units (one of the very few advantages of Prime over real Mathcad)

EDIT:

Yes, the definition of C1 at the top can be deleted as the solve block is parametrized.

Furthermore, too retain full formatting capabilities in the plots, you should rather define the range t as follows:

There is hardly a need to use more than 50000 point for a single trace anyway 😉

Prime10 Worksheet attached

EDIT; I made a mistake with the initial condition for i(0). This affects the behaviour in the first 4 ns.

Here is the correct condition:

P

Perez

14-Alexandrite

For C1 = 100nF:

𝑉𝑏(t = 100𝑛𝑠) ≈ 120.34 V

For C1 = 10nF:

𝑉𝑏(t = 100𝑛𝑠) ≈ 16.97 V

C

Cornel

Author

19-Tanzanite

Not correct for sure... It gives me around for C1=100nF Vb(t=100ns) =4.45V, and for C1=10nF Vb(t=100ns)=3.46V, but not through math calculation. But lets see other opinions...
With your answer like that you are qualified to obtain Nobel prize 😉

P

Perez

14-Alexandrite

I double checked and now I get:

For C1 = 100nF:

𝑉𝑏(t = 100𝑛𝑠) ≈ 9.28 V

For C1 = 10nF:

𝑉𝑏(t = 100𝑛𝑠) ≈ 0.38 V

But I'm not the expert in this field so let's see other results

W

Werner_E

25-Diamond I

Here numerically in real Mathcad (without units)

C

Cornel

Author

19-Tanzanite

2_Ode_solve_MCP9.zip

W

Werner_E

25-Diamond I

The syntax of odesolve has changed from real Mathcad to Prime. See the Prime version I posted.

Furthermore you can't numerically evaluate u2! u2 is a function with a single (time) argument and you can use u2 to evaluate it at a specific time value.

W

Werner_E

25-Diamond I

The symbolic in Prime 10 is able to solve the simple simple system symbolically, but only if we omit i(t):

!!!!

It was just now when I compared the plots for i(t) from the solve block and the symbolic solution that i realized that my initial condition for i(0) was wrong!! This failure is responsible for the unnatural behaviour in the first 4 ns.

The correct condition would be i(0s)= U1.0/R which would yield 0,5 A with the given values

!!!!

But Prime fails when it comes to the simple limit t-> infinity. Quite disappointing!

C

Cornel

Author

19-Tanzanite

That's great that symbolliycal solution can be obtained as well. I was wondering for this.

I saw also that strange shape at the beginning in the current plot, and i wanted to ask why looks so, but i see that You figured ouț already that inițial condition for the current was set wrong.

Regarding that second limit where Prime 10 cannot do the calculation I see this also in Prime 9, and my thinking on this is that assume cannot recognize that factor lambda is > 0 when lambda is in the denominație or assume keyword does not work well in this situation. If You put -t/1 then prime will give 0 as the result for that limit. But indeed, i would expected that prime to be able to do also such limit calculation with parameters and with assumptions on parameters, but seems that prime cannot do such thing, indeed bad thing.

W

Werner_E

25-Diamond I

Actually its just a system of two simple ODEs which also could be solved by hand. But its convenient that P10 is able to do so symbolically.

According the limit its not understandable that Prime would respect the assume modifier when I write -lambda*t, but not when we write -t/lambda. Tend to call it a bug.

Here is a very clumsy workaround - two steps are necessary, doesn't work in one go and you sure have to know which result you are striving for (so its quite a useless workaround).