Help with capacitance calculation

[danrogers]

Hi all, Im trying to figure out how to work out the following...

A power supply has a 10,000μF capacitor that is charged up to 500V via a 200Ω resistor. Determine how long it will take for the capacitor to charge up to 450V.

Could someone please point me in the direction of the formula to use?

Many thanks

 

[tomas632]

try using google

Capacitor - Wikipedia, the free encyclopedia

 

[tomas632]

you would need to rearrange the equation to find the time

 

[danrogers]

I cant get my head around it

is there anywhere with more simple step by step help?

Thanks!

 

[MrAl]

Hello,

For a simple circuit like this where we have a voltage source charging a capacitor through a fixed resistor we can use the following formula:

Vc=Vs*(1-e^(-t/RC))
where
Vc is the capacitor voltage at time t,
Vs is the fixed source voltage,
t is time,
RC is the RC time constant equal to R*C.

The problem stated however is not clear. The cap is charged up to 500v but then it is charged up to 450v, so which is it or is it really discharging?
You'll have to clarify a bit here.

 

[tomas632]

i think they are saying that the cap is charged to 500V, but they want to know how long it takes to get to 450V, not all the way to 500V

 

[danrogers]

Hi, thanks for that equation. What does the ^ signify?

Tomas has got it correct I just need to work out the time that it takes to charge up to 450v

 

[Grossel]

Hi, thanks for that equation. What does the ^ signify?

Tomas has got it correct I just need to work out the time that it takes to charge up to 450v

Hi. example: 2^3 means 2x2x2.

 

[danrogers]

Ah thankyou, so its e to the power of (-t/RC) is that correct?

 

[legend killer]

Ah thankyou, so its e to the power of (-t/RC) is that correct?

Hi,it is e to the power of (-t/RC).
If you dont want to calculate the exact time,then you can make a rough estimate by calculating 2RC
Here RC is the time constant,in two time constants the capacitor will charge upto about 86% of the final valueIn your case it is 90% of the final value

 

[RCinFLA]

To solve for time, multiple boths sides by Ln (natural log). Ln(e^-t/RC) = -t/RC

 

[MrAl]

Hello again,


So you are saying that the supply voltage is 500v then?

Starting with:
Vc=Vs*(1-e^(-t/RC))

we can solve this for t...

Divide both sides by Vs:
Vc/Vs=1-e^(-t/RC)

Subtract 1 from both sides:
Vc/Vs-1=-e^(-t/RC)

Multiply both sides by -1:
1-Vc/Vs=e^(-t/RC)

take the ln() of both sides:
ln(1-Vc/Vs)=-t/RC

Multiply both sides by -RC:
-RC*ln(1-Vc/Vs)=t

Swap sides:
t=-RC*ln(1-Vc/Vs)

To get RC you just multiply R times C, and Vc is the voltage you are charging the cap up to, and Vs is the source voltage.