Basic experiment
of the capacitorOne evening , when I take bath with my son, we played with cup and dipper. I poured the water on the cup to the dipper. Because the dipper is bigger than the cup, the level of the water on the dipper becomes lower than the level of the cup. I planned to experiment the same thing with electricity.
I went to the shack , and I made the circuit shown on Fig2. Actually 10V battery is the voltage variable DC supply. C1 is the Chemical capacitor of 47 micro (u) farads. C2 is the Chemical capacitor of 100 micro (u) farads. In order to measure the voltage , I connected the oscilloscope instead of the voltage meter. Switches are the lead wires actually.
1. At first, I turned on the switch1, and charged the capacitor1. Then Voltage meter indicated the voltage 10 volts. I decided to call the voltage of this timing as V1. I decided to call the quantity of the electron of this timing as Q1. Then Q1=V1*Q1=10V* 47uF=470uC (micro coulombs)
2. Next, I turned off the switch 1, and cut the power supply from the circuit.
3. Next, I turned the switch2, and charged the capacitor2 with the electron charged on the capacitor1. Then the voltage meter indicated the voltage of this two parallel capacitors as V2 ,3 volts. The quantity of this timing is calculated as Q2=(C1+C2)*V2=(47uF+100uF)*3V=447uC.
Because the capacitance of the Electric capacitor is not so correct, this experiment has many errors.
In this experiment , all of the electron charged on the capacitor1 is not poured on capacitor2. On the last timing the capacitor1 and the capacitor2 connected parallel share the electron of each other. The Fig3 indicates the image of this experiment.
Next, I made another experiment with changing the capacitor 1 and 2 each other. Then , V1=10V, V2=6V. In this case ,
Q1=C1*V1=100uF*10V=1000uC
Q2=(C1+C2)*V2=147uF*6V=882uC
You will say that this experiment is not so correct, but I could experience the relation between cup and dipper with capacitors.
******Description add on 7Feb2000************
I bought the digital bolt meter on the last of 1999. So I made this experiment with it.
1. C1=47uF,C2=47uF,V1=19.69V, V2=9.94V,V2 must be (V1)/2=19.69/2=9.84V,difference is 0.48%
2. C1=47uF,C2=94uF(two 47uF parallel),V1=19.69V, V2=6.55V,V2 must be (V1)/3=19.69/3=6.65V,difference is 0.56%
3. C1=47uF,C2=141uF(three 47uF parallel),V1=19.69V, V2=5.02V,V2 must be (V1)/4=19.69/4=4.92V,difference is0.56%
I used the same bag of capacitors in this experiment. Anyway I feel that the difference between the calculation and measurement is very small.I think that this experiment is good to educate what is capacitor.
******Description add on 17Mar2001**********
My friend Mr.Hashimoto taught me that this experiment can be done without any power sause if you have an analogue tester. Please charge the capacitor with "Ohm" renge with your tester! It nedds about 30 second to make full charge. After that you should change the tester to DC voltage range! And after that you can continue as I descrive in upper of this page.