This project consisted of constructing circuits in a University of Colorado PHET Simulation. We were instructed to create two circuits in the program.
The first circuit consisted of a 50 Volt battery, an 80 Ohm resistor, a 50 Ohm light bulb, a 55 Ohm resistor, a 20 Ohm resistor, and an ammeter. The ammeter measured the current at 0.24 amps. The total power for the circuit, found by multiplying the current by the voltage, is 12 watts. The total resistance, found in a series circuit by adding all of the resistance, is 205 ohms. A voltmeter measured across the 80 Ohm resistor, measured a loss of 19.512 volts. The current calculated by hand, found by dividing the voltage by resistance, is 0.244 amps. |
The only criterion for the second circuit was that it must contain a parallel resistance construct. The circuit that I created contained a light bulb with 25 Ohms of resistance, a 50 Ohm resistor, a 30 Ohm resistor, a 15 volt battery, and an ammeter measuring the current at 1.4 amps. The power, again found by multiplying the current times the voltage, is 21 watts. In a parallel circuit, the resistance is found by taking the reciprocal of the sum of reciprocal values of resistance. 1 + 1 + 1 = 1 |
So, the total resistance for the second circuit is 10.718 Ohms. The current can be found by hand by dividing the voltage by resistance, so the calculated current would be 1.399 amps. In a parallel circuit, the voltage is all equal and consistent.