## Curry one year ago Question with circuits

1. Curry

2. Curry

@IrishBoy123

3. Curry

@jim_thompson5910

4. Curry

@amistre64

5. Curry

@campbell_st

6. campbell_st

sorry not good with circuits

7. Curry

8. Curry

@inkyvoyd

9. perl

@Michele_Laino

10. sidsiddhartha

ok u should try first calculate V_c(0-) then V(infity ) right?

11. sidsiddhartha

at t<0 switch was open|dw:1433646718816:dw|

12. sidsiddhartha

and at t<0 capacitor will act as an open circuit

13. sidsiddhartha

so use nodal at point A|dw:1433646864478:dw| makes sense?

14. sidsiddhartha

that gives Vc(0)=25/3 now at steady state the capacitor will act as an open circuit again $V(\infty)=0\\ \Large v(t)=v(\infty)+[v(0)-v(\infty)]*e^{\frac{- t }{ R_{th}C }}$ is that making any sense?

It appears that the load real value is 30 Ohms with a parallel " Z matching capacitor" which we are to determine the value of capacitance. In a previous post, I mentioned that the internal impedance (10 + j20) is equivalent to an impedance of 22.36 Ohm at 63 degrees inductive. The capacitor needs to be a value such that combined in parallel with the load results in a value of 22.36 Ohms at 63 degrees capacitive.|dw:1433646405417:dw|This results in a capacitance reactance of 82.5 Ohms ( I realize that i did the math as if both of the values were real creating an error) Now calculate the value of the capacitor|dw:1433646874269:dw|

16. Curry

Well when I calculate the voltage across teh capacitor, i get the right answer. But, when I calculate the voltage across the resistor, i'm not able to do it.

Sorry @Curry I was on a previous problem lol.

18. Curry

I understand, I'm looking at your posts in relevance to my previous post. Let me read, one second.

19. Curry

@radar but that still doesn't give us a value of 0.8mF for the capacitor. @sidsiddhartha did you see me previous post?

20. Curry

@sidsiddhartha , the V(0) isn't 25/3 actually. It is 5. Because if the siwtch is open, then there is no current going through the 40ohm resistor, and therefore it is just 2 resistors acting on that circuit.

You could "theviens" theorem (if you have studied it (spelling may be wrong)). The circuit would then look like this|dw:1433648004142:dw|

|dw:1433648176592:dw|

23. Curry

well ofc i've studied thevenin's and Norton's. But um, let's see. there seems to be a small flaw.

24. Curry

So when siwtch is open, the initial voltage is 5v.

25. Curry

and then, the switch is closed, so (5v/Rt)[40 + Zc),

26. Curry

but we can't do that, because we can't calculate inpedance, so we just calculate voltage of capacitor. ok. so then,

|dw:1433648508210:dw|

28. Curry

the Vc = 5 + (0 - 5)e^(-t/(48*0.1)

After 240.5 seconds (5 time constants 5 X 4.8) it would build up in an expotential manner to 5 volts.

30. Curry

yes, i too got 5. But the second part of the equation is -5/6e^(-t/RtC)

31. Curry

the -5/6 is the problem i'm having. Ofc, solving the voltage across the capacitor gives me taht value. but they're asking for voltage across resistor...

Yes, the voltage across the resistor would vary with t until the capacitor becomes fully charged, which would occur practically in 5 RC seconds. Calculate the current and V(t) would equal 5V - 8I(t)

|dw:1433649839635:dw|

34. Curry

So did you use the formula I(t) = cdv/dt?

No, I used the formulae:$i _{c}(t)=(E/R)e ^{-t/\tau}$

36. Curry

oh ok ok, i just checked it with the formula i posted, and it works too. so for future reference, :)

37. Curry

and why are we subtracting 8i?

Refer to the "thevenized" circuit and note that the first two resistors (40 and 10 Ohm) become parallel in the thenize process and become 8 Ohm . See the 12th post above this one for the drawing.

39. Curry

sorry, i meant, why are we subtracting?

The voltage drop across that resistor is subtracted from the thevinized 5 volt source that gives you the V(t).

41. Curry

Hmmm, i figured. But don't we have to account for the inital value for 5v also?

We do, and when t=0 (switch first closed) you will see that the equation will give you a value of 5 - 40/48 = 4.167 whic is v(t) when t = 0

43. Curry

so initial value is 5v. and then as t fgoes to inifinty it's 5 - 5/6e^(something). And then we should be adding these values right?

44. Curry

Wait actually, shuoldn't the inital value be 5v, not 4.167?

45. Curry

in that circuit, when switch is open, the current going trhough that 40 ohm resistor is 0. which means there is no voltage trhough that resistor.

46. Curry

so we can just avoid that 40ohm R. And so the inital voltage would be (25/40)(10)

47. Curry

i mean, 25/50 * 10

With the switch open v(t)= 5 volts yes......but at t=0, the uncharged capacitor is connected to the circuit by the switch closure and now v(t) is 4.167 volts and increasing with t. As t increases to 5 RC (5 time constants) and more the capacitor becomes fully charged for all practical purposes and it will be fully charged and current will cease thru the rightmost 40 ohm resistor and v(t) will be back to 5 volts.

Or as you say back to (25/40) (10) which is 5 volts.

|dw:1433651558519:dw|Kind of like this.

I am still not totally satisfied with the previous problem regarding maximum power transfer - impedance matching. But I have to call it a day and retire for the night.

52. Curry

Yes, that problem is a stickler. I'll figure oout how to solve it, as i'm meeting with my friends. I'll be sure to let you know tomorrow!

Yes please do, and I believe you will get it, I am 76 yrs old and gotten rusty with these things, but you stick with it and you will overcome. Good luck in your studies. good night.

54. sidsiddhartha

yeah sorry i made a mistake took that 10 ohm as 40 yeah v(0)=5

55. sidsiddhartha

i guess the problem is solved then, my internet connection got stuck sorry :(

56. Curry

So, if that capacitor was replaced with a inductor, would I treat the intial as if there's an open circuit there?