Curry
  • Curry
Question with circuits
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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Curry
  • Curry
1 Attachment
Curry
  • Curry
@IrishBoy123
Curry
  • Curry
@jim_thompson5910

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Curry
  • Curry
@amistre64
Curry
  • Curry
@campbell_st
campbell_st
  • campbell_st
sorry not good with circuits
Curry
  • Curry
@radar
Curry
  • Curry
@inkyvoyd
perl
  • perl
@Michele_Laino
sidsiddhartha
  • sidsiddhartha
ok u should try first calculate V_c(0-) then V(infity ) right?
sidsiddhartha
  • sidsiddhartha
at t<0 switch was open|dw:1433646718816:dw|
sidsiddhartha
  • sidsiddhartha
and at t<0 capacitor will act as an open circuit
sidsiddhartha
  • sidsiddhartha
so use nodal at point A|dw:1433646864478:dw| makes sense?
sidsiddhartha
  • 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?
radar
  • radar
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|
Curry
  • 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.
radar
  • radar
Sorry @Curry I was on a previous problem lol.
Curry
  • Curry
I understand, I'm looking at your posts in relevance to my previous post. Let me read, one second.
Curry
  • Curry
@radar but that still doesn't give us a value of 0.8mF for the capacitor. @sidsiddhartha did you see me previous post?
Curry
  • 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.
radar
  • radar
You could "theviens" theorem (if you have studied it (spelling may be wrong)). The circuit would then look like this|dw:1433648004142:dw|
radar
  • radar
|dw:1433648176592:dw|
Curry
  • Curry
well ofc i've studied thevenin's and Norton's. But um, let's see. there seems to be a small flaw.
Curry
  • Curry
So when siwtch is open, the initial voltage is 5v.
Curry
  • Curry
and then, the switch is closed, so (5v/Rt)[40 + Zc),
Curry
  • Curry
but we can't do that, because we can't calculate inpedance, so we just calculate voltage of capacitor. ok. so then,
radar
  • radar
|dw:1433648508210:dw|
Curry
  • Curry
the Vc = 5 + (0 - 5)e^(-t/(48*0.1)
radar
  • radar
After 240.5 seconds (5 time constants 5 X 4.8) it would build up in an expotential manner to 5 volts.
Curry
  • Curry
yes, i too got 5. But the second part of the equation is -5/6e^(-t/RtC)
Curry
  • 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...
radar
  • radar
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)
radar
  • radar
|dw:1433649839635:dw|
Curry
  • Curry
So did you use the formula I(t) = cdv/dt?
radar
  • radar
No, I used the formulae:\[i _{c}(t)=(E/R)e ^{-t/\tau}\]
Curry
  • Curry
oh ok ok, i just checked it with the formula i posted, and it works too. so for future reference, :)
Curry
  • Curry
and why are we subtracting 8i?
radar
  • radar
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.
Curry
  • Curry
sorry, i meant, why are we subtracting?
radar
  • radar
The voltage drop across that resistor is subtracted from the thevinized 5 volt source that gives you the V(t).
Curry
  • Curry
Hmmm, i figured. But don't we have to account for the inital value for 5v also?
radar
  • radar
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
Curry
  • 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?
Curry
  • Curry
Wait actually, shuoldn't the inital value be 5v, not 4.167?
Curry
  • 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.
Curry
  • Curry
so we can just avoid that 40ohm R. And so the inital voltage would be (25/40)(10)
Curry
  • Curry
i mean, 25/50 * 10
radar
  • radar
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.
radar
  • radar
Or as you say back to (25/40) (10) which is 5 volts.
radar
  • radar
|dw:1433651558519:dw|Kind of like this.
radar
  • radar
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.
Curry
  • 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!
radar
  • radar
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.
sidsiddhartha
  • sidsiddhartha
yeah sorry i made a mistake took that 10 ohm as 40 yeah v(0)=5
sidsiddhartha
  • sidsiddhartha
i guess the problem is solved then, my internet connection got stuck sorry :(
Curry
  • Curry
So, if that capacitor was replaced with a inductor, would I treat the intial as if there's an open circuit there?
radar
  • radar
Yes, an inductor would behave as initially as an open at switch closure, however the current would increase to a value as if the opposition was only the DC resistance of the inductor (wire resistance). In other words inductor oppose any change in current. The change from 0 current (switch open) to the initial attempt of current condition (switch closed) is maximum as that is the greatest change (di/dt). Hard to explain......as they say "it gets complicated" lol.

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