prove of Capacitor charging

[Highlander-SP]

laplace+capacitor+charging

How can be proved the formula of the capacitor charge in a RC series circuit:

Vc = Vin * [1 - e^(-t/(R*C))]

where:
Vc = capacitor voltage
Vin = power source
R = resistor
C = capacitor
t = tempo

and how about the discharce: Vc = Vcmax * [ e^(-t/(R*C))]

 

[Kral]

capacitor charging s domain equation solution

Treat the RC circuit as a voltage divider with R as the top element
Write transfer function Vo/Vin =Xc/(Xc+R).
Xc = 1/jwc (Capacitive Reactance)
Using the Laplace operator in place of jw,
Xc = 1/sC
So
Vo/Vin = (1/sc)/(R + (1/sc)) = 1/(RCs+1)
Laplace transform of a step input is 1/s
LaPlace transform of the output response to the step is the product
of the input times the transfer function (multiplication in s domain is
equivalent to convolution in the time domain)
vout(s)=[1/s] [1/RCs+1]
Expanding by partial fractions
vout(s) = 1 - 1/(s+(1/RC))
taking the inverse Laplace transorm:
vout(t) = 1-e^(-t/RC).
~
A similar approach can be taken for the discharge condition.
Regards,
Kral

 

[Tinamuline]

The proof is very easy. You can find the proof from any physics textbook, like 'Fundamentals of physics, by Halliday, 6 or 7th edition', or ' Physics by James walker, 2nd edition'.
I can help you to prove it, but I cannot type it on here because it requires integration symbols. Feel free to contact me should you have any physics problems

 

[azaz104]

U need to know that the capacitor doesn't really charge, the total charge in a capacitor is preserved and what happens is that one plate accumulates electrons which reppels the electrons on the other plate.
if u get this then u must know that this will genetrate an electrical field that will fade through time. the recombination rate of negative and positive charges is explained by maxwells equations.

 

[yousefsam]

The solution is very very very very easy :
from the relation q(t)=Q(1-exp(-t/RC)
you can substitude instead of charge Q=VC then you can get this relation :
VC=εC(1-exp(-t/RC)) , but the capacitance of capacitor doesn't change then you can get the relation V = ε(1-exp(-t/RC)) , where :
V : Voltage of capacitor
ε : electromotive force or the voltage of the source .
t : time
R: Resistor .
C: Capacitor ,
best wishes ,
Yousef Omran .

 

[janath]

you can draw the circuit with source resistor and capacitor all in series. then

voltage across the resistor is V1=IR, capacitor V2= C*integral(Idt);

If source is Vin then Vin=V1+V2; from this you ger a differential equation and hence get I; V2=Vin-IR get the voltage across the capacitor