capacitor
Hello, I made the calculations with the table of the link that you passed me and I put it into practice and it loads me in double the time of the calculations taken with the table, why will it be?
If you read what happened to you, you would already know that the time constant that is calculated by multiplying the farads by the ohms gives us the time it takes to charge the capacitor to 63.2% of the voltage and that the total charge of the consensus is actually that constant multiplied by 5.
The resulting voltage curve in terminals for the condenser is not linear, it could be compared with a water tank emptying through a hole in the lower part, at the beginning more water comes out and as it empties it goes out less and less Because the pressure decreases.
On the other hand, it is possible that you are causing an error by having the voltimetor connected to the capacitor all the time, which means putting a resistor in parallel with the capacitor that slows down the charging process (internal voltmeter resistance.
Whenever we make a measurement in a process we generate an error in the measured parameters so to minimize this we must know very well what is done and choose a methodology as appropriate as possible so that it hardly generates changes in the magnitudes.
¡Aprende a reparar con estos cursos YoReparo!
