For resistors in series, the total resistance, RT, is given by:
RT = R1 + R2
This can be proved by the fact that when resistors are joined in series the same current I flows through them all. The potential differences across individual resistors are:
V1 = IR1 V2 = IR2 V3 = IR3
The total drop in potential V = IR, were R is the total equivalent resistance.
Since V = V1 + V2 + V3
IR = IR1 + IR2 + IR3
RT = R1 + R2
For resistors in parallel, the total resistance is given by:
When resistors are in parallel the current in the main circuit divides, part passing through each resistor. The potential difference across all resistors is the same.
Therefore: I1 = V I2 = V I3 = V
R1 R2 R3
Since I = I1 + I2 + I3
V = V + V + V
R = R1 R2 R3
When resistors are in parallel, the reciprocal of the equivalent resistance is equal to the sum of the reciprocals of the individual resistances.
As a preliminary experiment in class we set up a simple situation to prove our predictions. We created a simple version of the circuit that we will use for our full practical. Using a selection of bulbs, wires and an ohmicmeter I was able to simply measure the resistance of bulbs in different layouts within the circuit.
1 Bulb in Series 2 Bulbs in Series 2 Bulbs in Parallel
On the whole this smaller experiment proves my prediction. As shown in the results table the two bulbs in series provides double the resistance of one bulb in series. While the circuit with the two bulbs in parallel is halve the resistance than that of the one bulb in series circuit.
Firstly I will set up the equipment for the experiments in series. I will start of with the voltage on the power pack set to twelve volts. I will then add one bulb to the circuit. I will use the variable resistor to find the reading of 0.6 amps flowing through the circuit. I will then record the voltage across the bulbs. Increasing the amperage each time till I reach 1.3 amps. I shall then redo the experiment three times so I can work out an average reading and also to ensure accurate results. I shall then increase the number of bulbs to two in series and do the experiment again. Then I will do the same for two bulbs connected in parallel.
The independent variables in this experiment will be:
- The current flowing through the circuit, which I will be able to change using a variable resistor.
- The number of bulbs used.
- Whether the resistors are in series or parallel.
The variables I will be keeping constant in this experiment are:
- A voltage of 12V flowing through the circuit.
- The type of wires used.
- The types of bulbs used
This is the raw data provided from my practical for analysis I will use the average of all the three trials. To do this I must first add together the three trials and then divide that number by three.
E.g. 0.48+0.52+0.47 = 0.49
1 Bulb in Series
2 Bulbs in Series
2 Bulbs in Parallel
Using my average results I compiled them onto a graph to compare the results.
From the graph above you can see that my prediction was right. The circuit with one bulb, which could be known as the control, lies almost halfway between the other circuits, with the line of best fit having a gradient of 4. This is good because I predicted that the circuit with two bulbs would be twice as resistant as the control, this is because resistance doubles if you add another resistor of equal value in a series circuit. The circuit with two bulbs is shown on the graph with having a gradient of 7.5, which is almost double the control. I think this is suitable proof to say my prediction was correct. Also I predicted that the resistance in the parallel circuit would halve, because the current is split between the wires, so the resistance is less. On my graph the gradient of the line of best fit has a gradient of 1.75, which is also almost halve of the gradient of the control, which is more evidence to back up my prediction. Overall my prediction was correct in stating that for two resistors in series would double the resistance the current is held through each one. Also I stated that two resistors in parallel should halve because the current is split between the resistors. As my results agree with my results to a very close degree of accuracy I am very happy. The slight difference in the gradients not being exactly halve or double that of the control could be explained by the bulbs used not being ohmic, which means the resistance line is curved at the top and bottom due to the bulb heating up and cooling down. Using an ohmic set of bulbs or using the straightest part of the line for the line of best fit could avoid this. From my experiment I also found out that as you increase the current flowing through the circuit, the voltage across the circuit also increases, which means an overall increase in resistance.
Overall I think the experiment was carried out exceptionally well with very few errors. Also I did not record any unusual readings due to the fact that I checked the circuit was connected properly and I let the results stabilise before I took a reading. However, because I used non-ohmic bulbs there was a slight curve on the graph, this could have an effect on the reliability of my results. This could be avoided next time by using a set of ohmic bulbs to avoid getting the heating up and cooling curves on the resistance graph. From the results I am able to state a definite conclusion, which supported my prediction. My prediction stated, two resistors in series would double the resistance the current is held through each one and that two resistors in parallel should halve because the current is split between the resistors, through my results this is fully supported. However, I could extend the investigation I could add more bulbs in series and parallel circuits to see if it still follows the pattern I discovered in this experiment. For example if I used three bulbs in series and parallel would it mean that they would be three times as big and a third of the control.