Basic capacitor theory ?

[tron87]

Suppose i had a rc charging circuit consisting of resistor(rc) in series with the capacitor(c) and the following parameters.

rc = 47kohm
c = 1000uf
vs = 5v

I know so far that 1 time consent = 47sec
capacitor is fully charged at 5 x 47sec = 235sec

Using the following formula how would i find the voltage across the capacitor at 120 sec ?

Vc = V(1 - e-t/RC)
where

Vc is the voltage across the capacitor
Vs is the supply voltage
t is the elapsed time since the application of the supply voltage
RC is the time constant of the RC charging circuit


Also what formula do i use to find the time required for voltage across capacitor to reach say 6v ?

 

[Ratchit]

tron87,

Suppose i had a rc charging circuit consisting of resistor(rc) in series with the capacitor(c) and the following parameters.

rc = 47kohm
c = 1000uf
vs = 5v

I know so far that 1 time consent = 47sec
capacitor is fully charged at 5 x 47sec = 235sec

The capacitor energized to 99% of the supply voltage at that time.

Using the following formula how would i find the voltage across the capacitor at 120 sec ?

Vc = V(1 - e-t/RC)
where

Let's not get sloppy. The correct way to write that equation is Vc = Vs(1 - e^(-t/RC)) or Vc = Vs(1 - exp(-t/RC)) . Just plug in the values and solve for the unknown. It will be 92% of the supply voltage.

Also what formula do i use to find the time required for voltage across capacitor to reach say 6v ?

Is that a joke? How can the capacitor ever charge to 6 volts when the supply voltage is only 5 volts.

Ratch

 

[tron87]

tron87,

Is that a joke? How can the capacitor ever charge to 6 volts when the supply voltage is only 5 volts.

Ratch

No joke ! typo should read 4.5v.

 

[BrownOut]

4.5V = 5(1-e^-t/RC)

4.5/5 - 1 = -e^-t/RC

e^t/RC = 1/(1 - 4.5/5)

t = RC*ln(1/(1 - 4.5/5))

t = 47*ln(10)

t = 47*2.3

t = 108seconds