Experiment: Capacitor charging/discharging

Purpose: Learn the capacitor are components that storing the energy via the electric field, and we will consider charging and discharging a capacitor.

The circuit have the charge side with the voltage source connect in the begining. After the charging, switching to the discharging circuit.

Charging:
Vc(t)=Vo(1-e^(-t/RC))
Ic(t)=(Vo/R)*e^(-t/RC)
Discharging:
Vc(t)=Vo*e^(-t/RC)
Ic(t)=-Io*e^(-t/RC)

Problems:
QThe Vs is 9V, and we want charging interval of about 20s storing 2mJ in the capacitor, and then discharging the capacitor in 2s.
1.for the W=V^2*C/2, C=2W/V^2=5m/81=61.73μF
2a.We want Vc(t) charging to 99% of the Vo, when 1-e^(-5)=99.3%
so -5=-20/R*61.73μ, and R =64.8kΩ
2b.peak current= Vo/R= 9/64.8k=0.1389mA
and peak power =VcIc=9*0.1389=1.25mW
3a. -5=-2/R*61.73μ, R=6.48kΩ
3b. peak current= Io= Vo/R=9/6.48k=1.389mA
P=IV=1.389m*9=12.5mW

Measure data.
Rcharge = 65kΩ
Rdischarge = 6.47kΩ
C=63.6 uF
The maximum Voltage is different with the ideal data.

Charging

The V=8.476, 8.476=9*Rleak/(Rleak+65k), Rleak=1.23MΩ

Discharging