Physics capacitor discharging/charging equations help

12

When looking at the charging equations and discharging equations for capacitors for current, voltage and charge. E.G. V=Vo E^-T/RC for discharging.
Why is it the current equation stays the same for discharging and charging and can someone explain a little where these equations from and why voltage and charge have different equations for charging vs discharging, is this to do with the 37% and 63% idea? Can someone explain all this please?
Thank you

Reply 1

Nitrotoluene

12

Original post

by LisaPerera

When looking at the charging equations and discharging equations for capacitors for current, voltage and charge. E.G. V=Vo E^-T/RC for discharging.
Why is it the current equation stays the same for discharging and charging and can someone explain a little where these equations from and why voltage and charge have different equations for charging vs discharging, is this to do with the 37% and 63% idea? Can someone explain all this please?
Thank you

The key difference is in the sign and form of the equation.

The essential concept revolves around the form of the equation:
- Charging: V = Vo (1 - e^(-t/RC)) - The voltage increases from 0 to Vo.
- Discharging: V = Vo e^(-t/RC) - The voltage decreases from Vo to 0.
-Charge is a similar concept: Q = CV, but with coulombs instead of volts.
Thus, the charge increases with the function (1 - e^(-t/RC)), and the discharge is done with the function e^(-t/RC). The function RC determines the rate at which the discharge is.

Ciao,
Sandro

Reply 2

Fiana2727

7

Original post

by LisaPerera

When looking at the charging equations and discharging equations for capacitors for current, voltage and charge. E.G. V=Vo E^-T/RC for discharging.
Why is it the current equation stays the same for discharging and charging and can someone explain a little where these equations from and why voltage and charge have different equations for charging vs discharging, is this to do with the 37% and 63% idea? Can someone explain all this please?
Thank you

Hey,

I think about the p.d. between the capacitor and the other component.

So when charging the capacitor it initially has p.d. of 0 and the power supply has a high p.d. So initially there will be a large current as there is a maximum possible p.d. between the power supply and capacitor. As the capacitor charges it’s p.d. Increases so the p.d. between it and the supply decreases so current decreases until V(power supply) = V(capacitor) at which point I=0 as there is no p.d between them. If looking at charge or p.d. across the capacitor it will be increasing.

When discharging initially p.d. across the capacitor will be maximum and as it discharges its p.d decreases so the p.d of the circuit decreases so current decreases. If looking at charge or p.d in capacitor it will be decreasing.

I hope that helped!

Physics capacitor discharging/charging equations help

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