Saturday, April 23, 2016

04/14 - Lab 17: Passive RC Circuit Natural Response, and Lab 18: Passive RL Natural Circuit Response

Lab 17: Passive RC Circuit Natural Response

In this experiment, we explored how capacitors change a circuits response to a sudden change in voltage. Capacitors store energy in an electric field, which prevents sudden voltage drop throughout the circuit. We set up a simple RC circuit and test manual switching, as well as sending a square wave signal through it and measured the voltage drop across the capacitor.

Diagram of the circuit modeled

Before setting up the circuit, we ran some calculations to determine the theoretical time constant, Tao. We used R_1 = 1 kOhm, R_2 = 2.2 kOhm, and C = 20 uF. 


Actual resistor values used

Actual breadboard circuit

Charge voltage across the capacitor using trigger

Discharge voltage across the capacitor using trigger

Voltage over time across the capacitor when implementing a square wave

The discharge time of a capacitor is typically ~ 5 time constants, which we calculated to be 15 ms. In other words, the theoretical charge/discharge time should have been approximately 75 ms. The graph shows that the total charge/discharge time of the capacitor in the circuit was nearly spot on.


Lab 18: Passive RL Natural Circuit Response

In this experiment, we explored how inductors change a circuits response to a sudden change in current. We ran essentially the exact same lab as the one above, but we swapped the capacitor for an inductor and used Ohm's law to calculate the change in current across the circuit when the voltage supply is switched on and off.

Diagram of the circuit

Calculations to determine the time constant,Tao

Actual breadboard circuit

Inductor discharge using manual trigger

Voltage across inductor using a square wave

In the calculations above, the time constant was found to be 1.45 microseconds, and thus, the time for discharge should be ~ 6 - 7 microseconds. The manual trigger showed this result exactly, but the square wave did not, likely because the frequency was slightly too high.

In Class Examples

1. Given the circuit below, find Vc(T) and Ic(t) given that Vc(0) = 10V.



2. The switch in the circuit shown below has been closed for a long time. At t = 0, the switch is opened. Calculate i(t) for t > 0.



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