Wednesday, May 20, 2015

Day 13

Capacitor Voltage-Current Relations

In this lab we measure the relationship between the voltage and current across a capacitor. To get a better understanding of the relationship we will use three different time-varying signals: sine waves, triangle waves, and square waves as voltages applied to the capacitor. Due to the fact that our measurement tool (Analog Discovery) doesn't measure current we will use the formula: I(c)=C(dv/dt)


Pre-Lab

Below is our guess for what the current across the capacitor will look like graphically when a sinusoidal voltage and a triangular voltage is applied to the capacitor. Since we know that the current is proportional to the derivative of the voltage we can make a pretty good assumption by aligning the slopes of the graph.


Lab

For this lab we needed a capacitor and a resistor connected in series. To simplify the procedure we use a 1uF capacitor with a 100 Ohm resistor. By using the oscilloscope function of our Analog Discovery we can measure the output voltage of both the resistor and capacitor. We also added a Math Channel of where we divided the voltage by the resistor (100 Ohms) so we can plot the current. Because we are very well educated individuals our predictions were correct.


We chose a frequency of 1kHz for our sinusoidal voltage and we saw that the current across the capacitor is indeed proportional to its derivative. The output voltage for the capacitor is a sine wave (as it should be since we are applying a sinusoidal voltage). And as expected the current is a cosine wave in line with the sinusoidal voltage.



Now we chose a frequency of 2kHz for our sinusoidal voltage. The results haven't deviated from the above experiment. The resulting current was a cosine wave with a 2kHz frequency.




Here is the 1uF capacitor in line with the 100 Ohm resistor because we all like sexy circuit pics.


To conclude this lab we were able to see the relationship between the voltage and current across a capacitor. We were successful at our analysis that I(c)=C(dv/dt) holds true. We also got more experience using the Math Channel to analyze current using Analog Discovery.

No comments:

Post a Comment