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Curry
 one year ago
Question with circuits
Curry
 one year ago
Question with circuits

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campbell_st
 one year ago
Best ResponseYou've already chosen the best response.0sorry not good with circuits

sidsiddhartha
 one year ago
Best ResponseYou've already chosen the best response.0ok u should try first calculate V_c(0) then V(infity ) right?

sidsiddhartha
 one year ago
Best ResponseYou've already chosen the best response.0at t<0 switch was opendw:1433646718816:dw

sidsiddhartha
 one year ago
Best ResponseYou've already chosen the best response.0and at t<0 capacitor will act as an open circuit

sidsiddhartha
 one year ago
Best ResponseYou've already chosen the best response.0so use nodal at point Adw:1433646864478:dw makes sense?

sidsiddhartha
 one year ago
Best ResponseYou've already chosen the best response.0that 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
 one year ago
Best ResponseYou've already chosen the best response.1It 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:dwThis 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 capacitordw:1433646874269:dw

Curry
 one year ago
Best ResponseYou've already chosen the best response.0Well 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
 one year ago
Best ResponseYou've already chosen the best response.1Sorry @Curry I was on a previous problem lol.

Curry
 one year ago
Best ResponseYou've already chosen the best response.0I understand, I'm looking at your posts in relevance to my previous post. Let me read, one second.

Curry
 one year ago
Best ResponseYou've already chosen the best response.0@radar but that still doesn't give us a value of 0.8mF for the capacitor. @sidsiddhartha did you see me previous post?

Curry
 one year ago
Best ResponseYou've already chosen the best response.0@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
 one year ago
Best ResponseYou've already chosen the best response.1You could "theviens" theorem (if you have studied it (spelling may be wrong)). The circuit would then look like thisdw:1433648004142:dw

Curry
 one year ago
Best ResponseYou've already chosen the best response.0well ofc i've studied thevenin's and Norton's. But um, let's see. there seems to be a small flaw.

Curry
 one year ago
Best ResponseYou've already chosen the best response.0So when siwtch is open, the initial voltage is 5v.

Curry
 one year ago
Best ResponseYou've already chosen the best response.0and then, the switch is closed, so (5v/Rt)[40 + Zc),

Curry
 one year ago
Best ResponseYou've already chosen the best response.0but we can't do that, because we can't calculate inpedance, so we just calculate voltage of capacitor. ok. so then,

Curry
 one year ago
Best ResponseYou've already chosen the best response.0the Vc = 5 + (0  5)e^(t/(48*0.1)

radar
 one year ago
Best ResponseYou've already chosen the best response.1After 240.5 seconds (5 time constants 5 X 4.8) it would build up in an expotential manner to 5 volts.

Curry
 one year ago
Best ResponseYou've already chosen the best response.0yes, i too got 5. But the second part of the equation is 5/6e^(t/RtC)

Curry
 one year ago
Best ResponseYou've already chosen the best response.0the 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
 one year ago
Best ResponseYou've already chosen the best response.1Yes, 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)

Curry
 one year ago
Best ResponseYou've already chosen the best response.0So did you use the formula I(t) = cdv/dt?

radar
 one year ago
Best ResponseYou've already chosen the best response.1No, I used the formulae:\[i _{c}(t)=(E/R)e ^{t/\tau}\]

Curry
 one year ago
Best ResponseYou've already chosen the best response.0oh ok ok, i just checked it with the formula i posted, and it works too. so for future reference, :)

Curry
 one year ago
Best ResponseYou've already chosen the best response.0and why are we subtracting 8i?

radar
 one year ago
Best ResponseYou've already chosen the best response.1Refer 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
 one year ago
Best ResponseYou've already chosen the best response.0sorry, i meant, why are we subtracting?

radar
 one year ago
Best ResponseYou've already chosen the best response.1The voltage drop across that resistor is subtracted from the thevinized 5 volt source that gives you the V(t).

Curry
 one year ago
Best ResponseYou've already chosen the best response.0Hmmm, i figured. But don't we have to account for the inital value for 5v also?

radar
 one year ago
Best ResponseYou've already chosen the best response.1We 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
 one year ago
Best ResponseYou've already chosen the best response.0so 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
 one year ago
Best ResponseYou've already chosen the best response.0Wait actually, shuoldn't the inital value be 5v, not 4.167?

Curry
 one year ago
Best ResponseYou've already chosen the best response.0in 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
 one year ago
Best ResponseYou've already chosen the best response.0so we can just avoid that 40ohm R. And so the inital voltage would be (25/40)(10)

radar
 one year ago
Best ResponseYou've already chosen the best response.1With 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
 one year ago
Best ResponseYou've already chosen the best response.1Or as you say back to (25/40) (10) which is 5 volts.

radar
 one year ago
Best ResponseYou've already chosen the best response.1dw:1433651558519:dwKind of like this.

radar
 one year ago
Best ResponseYou've already chosen the best response.1I 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
 one year ago
Best ResponseYou've already chosen the best response.0Yes, 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
 one year ago
Best ResponseYou've already chosen the best response.1Yes 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
 one year ago
Best ResponseYou've already chosen the best response.0yeah sorry i made a mistake took that 10 ohm as 40 yeah v(0)=5

sidsiddhartha
 one year ago
Best ResponseYou've already chosen the best response.0i guess the problem is solved then, my internet connection got stuck sorry :(

Curry
 one year ago
Best ResponseYou've already chosen the best response.0So, if that capacitor was replaced with a inductor, would I treat the intial as if there's an open circuit there?

radar
 one year ago
Best ResponseYou've already chosen the best response.1Yes, 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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