In this instance, there is a resistor in parallel with the two capacitors that are in parallel. This resistor has a very high resistance. Because it is in parallel with the two capacitors, it acts like a voltmeter, measuring the potential across the two capacitors. Measure the half times for the charging and discharging cycles of the two capacitors in parallel and fill in the rest of the table. Because the two capacitors are in parallel, you will need to calculate the total capacitance of the circuit with the equation: Ctatal = Cl + C2 Ctotal 12 Time (charge) 1/2 Time (discharge) Tcharge Tdischarge Taverage Ctotal based on time constant % Difference in Ctotal10pF 1 E012 H:2 ms/div V:2 V/div I: 69.5 ms Max=4.999 V ' me step = 5 us reSIstor, 1EO12 Q Min:-4.999 V