04-14-2016

04-14-2016: Passive RC and RL Circuit Natural Response

I. Introduction:

Today, we spent most of our time to discuss the RC and RL. We also reviewed the first differential equation to find out the relationship between current and voltage across the inductor.

Given this simple circuit, our goal is to find out the equivalent inductor. For the inductor, a rule of thumb here is to follow the same as the equivalent resistors. For my notation, + indicated they are in series, not adding.

II. Formula Derivation:

Here is our derivation of the voltage across the capacitor over the course of time. Notice that when the switch is closed and DC current, the capacitor acted like an open circuit. If the switched is open, the capacitor will be fully charged and has its own voltage, ideally a straight line, but in real world, the voltage will be dropped over the course of time

Here is our derivation to find out what time the voltage across the capacitor would be left with 1% of the original voltage, which means it would take 5tau. tau=R*C

After we got the voltage, the current would be voltage divided by the resistor, the power is the multiplication of both voltage and current

Given a simple circuit, our goal is to find out the V(t) and i(t). First of all, we have V(0)=10 V already. We then had to find tau and got our V equation, I=V/R

Here we need to find the smallest tau and the largest tau, since tau is directly proportional to RC, we picked the lowest values for RC and largest values for RC respectively

II. Prelab:

Given a simple circuit,we had to tau, we first find the Rth in the circuit, then use that to find out tau

II. Voltage and current in capacitor and inductor:
1. The capacitor:
For this lab, we were trying to measure the current and voltage of capacitor and inductor over the course of time. We change the voltage by two ways: mechanically, which means that we had to basically take in take out the voltage supply power; or we can use the square wave voltage built in feature of the analog discovery

Here is how the graph looks like when we did it manually, the region we concerned about is the going downslope of the graph

Here is what it should look like if we the built-in feature of square wave voltage instead of doing it mechanically. 

Here we used triggering and single acquisition to find out the slope of the curve

Here is our data. For part A (mechanically), we found the percent to be 35.88%, which was not really favorable in terms of acceptable range. It could due to that the value we were trying measure to small therefore a small change of number could cause a great impact to our percent error

III. Some More derivation

Here we derived the formula to find the current across the inductor

After we got the formula of the current. We know that V=l*di//dt. We took the integral to find the formula for the voltage

Here is our setup for this lab, this setup applied for both the capacitor and the inductor 

The percent error is about 35% for the manual experiment

Here is the portion we need for the inductor. The percent error is about 28% in this case

here is the setup for the inductor circuit

III. Analysis:
Since we don't have an ideally inductor or capacitor, the current or voltage are not perfectly trapped in the circuit, over the course of time, those values would be dissipated into the resistors, causing a great percent error in our lab. Moreover, since we dealt with small value of time division, it could make our measurement really sensitive to the percent error
IV. Conclusion:
For this lab, we spent our time discussing the behaviors of voltage and current in capacitor and inductor circuit. We derived many equations using derivative and integral to the review of what we have learned in our 4b class. In the lab, we verified our knowledge about natural response of the passive RC and RL circuit. The percent error is great due to the time is too small to measure. For the rule of thumb, the capacitor acts like the open circuit with DC, and the inductor acts like short-circuit.