For this lab, we are given two capacitors that are 1 mF each. We are going to find the total capacitance in parallel and series. But first, we need to find the actual value for the two capacitors.
We get the value of 0.994μF and 0.984μF.
We then put them into series.
The capacitance is 0.489μF.
We then put the two capacitors in parallel.
The capacitance is 1.978 μF.
From these values we can find a relationship between capacitors in series and in parallel.
In series, the relationship of total capacitance of capacitors is 1/Ctotal=1/C1+1/C2.
In series, the charge is the same for the capacitors and so when looking at V=Q/C, the charge can be ignored and the only variable left is V.
In parallel, the relationship of total capacitance of capacitors is Ctotal=C1+C2.
In a parallel circuit, the voltage is the same for the capacitors and so when looking at the equation Q=CV, voltage can be ignored and only variable left is C.
Charge Build up and Decay in Capacitors
The graph of a circuit which is the same as our lab set up.
For this lab, a capacitor is first charge. Then, we will show how energy is used from it.
This is the experiment set up.
From the video, we can see that at first the light bulb lights up. Then, the light goes dimmer and dimmer, and eventually it goes out, because it is charging the capacitor now.
Then we connect the two wires together to build a closed circuit. The light bulb turns on again, and it goes dimmer and dimmer, and eventually turns off. When the wires are removed from the power supply and the circuit is closed, the bulb is being powered by the capacitor.
This lab shows us how the capacitor works, it is first charged and stores the power. Then it is discharged by charging the light bulb.
A Capacitance Puzzle
In this lab, we first charge the two capacitors, one is at 3.0 V and the other one is at 4.5 V. The charge will reach equilibrium and that can be found by finding the average charge of the two capacitors. The next step was to find the new voltage of two capacitors in parallels. Because the capacitors are in parallel, the capacitance adds up and we get a total capacitance of 1.12 F. The voltage is found by the formula V=Q/C and is calculated to be 3.75 V. Using the voltmeter, the capacitance was calculated to be 3.67 V. This amount of error can be due to the capacitors we use. The capacitance may not be the same as what we see on the cover. Also, the charges may discharge somewhere. Overall, our calculated value is very close to our theoretical value.
Quantitative Measurements on an RC System
For this lab, we do more quantitative measurements on a RC system.
This is the set up for this experiment.
Our goal is to find a mathematical relationship between voltage across a capacitor and time that describes how voltage changes as the capacitor discharges.
First, we measure the voltage across a charged capacitor . A current meter, LoggerPro and a resistor are also attached to the circuit. From Logger Pro, we get a Potential vs. Time graph and a Current vs. Time graph is made. As seen in the LoggerPro graphs below, there are four graphs; potential and current graphs for with voltage and potential and current graphs for without voltage.
A best fit line is made for each of the graphs with the equation Ae^(-ct) + B. It can be concluded that the relationship between potential/voltage and time for a charged capacitor is VC=Vmax (1-e*(-t/RC)) and for a discharged capacitor it is VD=Vmaxe^(-t/RC).
In our lab, the voltage initial is not correct because we do not start from 0. So when LoggerPro tries to do best fit line, it does not start at the right time.
In today's class, we learn how to calculate the total capacitance of capacitors in series and parallels. And we learn how a capacitance change in a capacitor as time changes.We also learn how a capacitor works.
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