Week Six

A lot of week six was a majority of testing and putting the entire system together. We have our system set up now that our pump is set up in series with 3 10 ohm capacitors. This makes it so there is a total of about 33 ohms when it was tested. Since this circuit is run with 2 12 volt batteries in order to boil the water quickly and efficiently, the pump is being hooked up to one and it will receive a total of 1.09 Volts. This was found by doing the simple Voltage = Current*Resistance equation. By using the total voltage and the total resistance, we can divide the voltage by that resistance in order to get the current. Since resistors set up in parallel all get the same current, we can use the equation again, except by using the resistance of the pump itself. Since we know the pump is 3 ohms of resistance, we multiply that by the current and get the voltage of 1.09 volts total going into the pump. The picture for the circuit is shown below and to the left. This picture shows how the water is hooked straight up to the batteries, making sure that the water boils fast and efficiently. The other picture to the right is the picture of the pump. The pump will be hooked up to only one battery, and has the three resistors all leading to pump. The pump fills up the beaker in approximately 1.7 seconds, while the water boils in about 1 minute and 36 seconds. By the end of this project the pump will be glued to the outside of the bucket, pumping in water every time the button is held down, and the beaker will be in the center of the bucket boiling water that will condense for the plants roots in order to ensure they get the water that they need.