The first test that I carried out was measuring the voltage drop across the components in the circuit. In doing this I found that the major voltage drops were across the the light bulbs and there were vey minimal drops across the wires and switch. These voltage drops added up to the supply voltage. Because the 2 light bulbs were the same size, they both used the same amount of voltage.
The next test was to measure the current flowing through the circuit. Connecting the ammeter around different points of the circuit I found that the current remained consistant, which it does in a series circuit.
I then used Ohms Law to calculate the resistance and power output of the circuit. Total resistance I found to be was 23.4 ohms and the power output of each bulb I found to be 6.6 watts.
The bulbs are putting out 6.6 watts because with 2 bulbs in series, there is more resistance, which means there is less current flow, which means there is less power output.
The next part was to construct a circuit with 3 bulbs in series. Voltage drop was measured again and was found that the major voltage drops were across the bulbs and because the bulbs were all the same size, they all had the same voltage drops. The current was again measure around the circuit and the current was again consistant at .44 amps, which was higher than the circuit with 2 bulbs. Resistance and power output were then calculated and I found total resistance to be 28.4 ohms, which is larger than the circuit with 2 bulbs, and power output of each bulb I calculated to be 5.5 watts, which is slightly less than the 2 bulb circuit. This is because the circuit with 3 bulbs has more resistance, which means less current, which means less power out put.
We were then asked about the difference between available voltage and voltage drop.
3 bulbs in series |
voltage drop is a measure of how much voltage being used by a componant-- not lost
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