Capacitor Charging/Discharging

In the lab, we considered charging and discharging a capacitor. TO avoid the need for sophisticated electronics in a pluse-forming network, we will consider the ideal capacitor. Like the figure below, in the charge position, as long as Vs>Vcap, energy is transferred from the source to the capacitor. IN the discharge position, the energy is absorbed by the resistance Rdischarge. In either case, C,Rcharge, and Rdischage control the rate at which energy is transferd.

In the experiment, the two modes of operation are done and the circuit is shown in the figure below. We can compute the Thevenin voltage and resistance as seen by the capacitor in each case. We substituted this voltage for the source voltage in the charging mode and we replaced the charging or discharging resistance by the appropriate Thevenin resistance.

In the RC charging circuit, the voltage and current across capacitor can be express as V(t)=Vs(1-e^(-t/tau)) and I(t)=Vs/R*e^(-t/tau), where tau is time constant which is Rcharge*C. In addition, in the discharging circuit the voltage and current across the capacitor are V(t)=Vic*e^(-t/taud) and I(t)=-Vic/Rdiscahrge*e^(-t/taud), where taud is time constant in discharging circuit. However, in real capacitor model the capacitor connected to resistor Rleak in parallel, so we must apply Thevenin's theorem to find out the real value of resistance.

 Then the circuit was built using a 65k charge resistor, a 6.47k discharge resistor, three21 uF capacitors. The three capacitors were placed parallel to simulate a 66 uF capacitor.

Vs = 9.01 V

Vfinal = 8.562 V = Vs*Rleak/(Rleak+65k), so R(leak)=1.24 Mohms

The Thevenin equivalent voltage and resistance seen by the capacitor during charging are

Vth=9.01*1.24M/(1.24M+65k)=8.55V and Rth=1.24M and 65k in parallel which is 61.8k ohms, and the Thevenin equivalent voltage and resistance seen by the capacitor during discharging are

Vth=9.01*1.24M/(1.24M+65k)=8.55V and Rth=1.24M and 6.47k in parallel which is 6.44k ohms

Questions:

1. The capacitance can be found using the energy equation w = 0.5Cv^2= 1.42F.

2. The capacitors in series add up to equivalents of C/2. There are 4 parallel branches, so the total capacitance is 2C. 2C has to equal 1.42F, C=0.71F.