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In a previous physics lab, we did a experiment on a discharging RC constant circuit. We used a double pole switch with a resistor (r) on the charging leg and resistor (R) on on the discharge leg. As we test for the RC constant with both resistors we computed t = 0.6, but when we tested with one resistor we computed t = 26.42. For the first portion we obviously flipped the switch to charge and then flipped it to discharge. In the second portion we removed resistor (R) and flipped the switch to charge and then opened the switch to discharge since resistor (R) was removed. The DC power supply was providing an initial 4 volts and the capacitor was 24*10^-6 F. The resistance for the first part was 25 ohms and the second part was 1100 ohms. Why I am not really concerned about the calculations, I am wondering the logic behind this. The lab states your first resistance should be higher since in part 2 resistor (R) was removed. The reasoning being part 1 is the combined resistance of the input impedance of the sampling circuit of the voltage sensor and resistor (R); while part 2 should just be the input impedance. I was thinking in the second part since we just opened the switch it acted as a open circuit and took allot longer to discharge, but I am not sure or possibly our second circuit was hooked up wrong. Because we completely removed the line that the resistor (R) was placed into in the diagram that is shown in the handout via the link provided.I know its hard to understand the circuit in words so I posted a link to the lab handout. Thanks. http://www.utc.edu/Faculty/Harold-Climer/PHYS%20281/281%20Lab%205%20RC%20Time%20Const.pdf
And got the following answer:
When the resistor R is in place, technically, the capacitor isn't just discharging through R, it's discharging through the parallel combination of R and the input impedance of the voltage sensor. Because the input impedance of the voltage sensor is very high, though, the equivalent resistance of that parallel combination is essentially the same as the resistance of R. But when the resistor R is removed, then the capacitor discharges through the voltage sensor. You know that the equivalent resistance of a parallel combination is always lower than the resistance of either resistor by itself. So when R is removed, the resistance through which the capacitor discharges increases. (So, no, the circuit isn't really open when you remove R. If that was the case, the capacitor wouldn't discharge at all. I can't know whether you built the circuit incorrectly, but since your time constant increased as expected, I wouldn't think so.)