Electricity Circuits: Individual, Series, Parallel and Series - Parallel

5th April - 9th April (2011)

Working in pairs and using circuit boards, we worked our way through the given booklet, building various circuits and measuring voltage and amps in variations throughout. This was to enable that each of us was aware of the basic principles of electricity in a circuit.

Definition:

Voltage - practical unit of electrical pressure. The force or pressure that is required to move the electrons in a circuit.
Amps - unit of current. When electrons move along a conductor, current is flowing through the circuit.

We started by building an individual circuits with a small bulb, which consists of a power source, a fuse, a switch and a bulb. With this circuit we had to ensure that each of these would follow on from the other before returning to the earth of the power source so it was a full circle. So each terminal had a wire input and a wire output.
With the switch then on, we measured the avaliable voltage (supply) at the positive power supply (12.64v), the terminal before the switch (12.71v), the termainal after the switch, before the light bulb, after the light bulb and the negative of the power supply (all 0v). This showed me that the fuse in the circuit (after the positive power supply and before the switch) added voltage to the circuit and also that the bulb consumed all the voltage in the circuit.
Also, voltage drop (decrease in voltage at different component of the circuit) was measured. From the positive of the power supply to the input of the switch (0v), from the input of the switch to the output of the switch (0v), from the output of the switch to the input of the bulb (0v), from the input of the bulb to the output of the bulb (12.4v) and from the output of the bulb to the negative of the power supply. The bulb of course has the greatest value of voltage drop which is due to the bulb consuming all the voltage in the circuit to power itself. However, even thought the voltage was entirely consumed, after measuring the amperage of trhe circuit it was seen to remain the same throughout as current flows the entirety of the circuit regardless.
According to ohms law, the resistance of the bulb was 43.1 ohms (volts: 13.8/amps: 0.32), and according to the power law the watts produced by the bulb was 4.4w (volts:13.8*amps: 0.32).

We then changed the consumer in the circuit from the small bulb to a larger bulb, and measured the voltage drop across the different components: wire before the switch (0v), Switch (0v), wire before the light bulb (0v), Light bulb (12.4v), wire after the light bulb (0v). There was obviously no difference in the results of the voltage drop between the two as even though the bulb is larger the power source is still the same and producing the same amount of voltage from both to be consumed by both.
However, the amps in this circuit increased from the small bulb (0.32a) to the larger bulb (1.83a), I believe the current needed to increase to generate more amps to power the larger bulb.
According to Ohm Law, the resistance of the bulb was 7.6 ohms (volts: 13.8/amps: 1.83). The larger bulb will need more current to power the bulb, for this reason the resistance is significantly lower to ensure better flow throughout the circuit.
The Power Law (volts:13.8*amps: 1.83) gave a total of 25.3 watts, a great deal higher than that of the smaller bulb. This being so because the amps are higher due to lower resistance.


We then changed the circuits from individual to series by adding the smaller bulb to the already existent large bulb circuit.

SERIES CIRCUITS:
The voltage drop of the components all remained at 0v, however light bulb 1 had a voltage drop of 12.6v, wre between light bulb 1 & 2 was 0v and light bulb 2 had a voltage drop of 0.2v. Being that the bulbs were different sizes and light bulb 1 was the smaller bulb, having a higer resistance it consumed the majority of the voltage and light bulb 2 could only use what was left over. This meant that although light bulb 1 was as bright as it could be, light bulb 2 was only just on, barely seen.
After measuring the wire before the switch, wire before light bulb 1, wire between light bulb 1&2 and the wire after light bulb 2, it became apparent that no matter where you measure the amperage it will be the same. The total in this circuit being 0.33A. By adding the small bulb to the large bulb circuit, the amperage greatly decreased, from 1.83A to 0.33A. The total resistance of the circuit also changed, increasing to 41.8ohms.
Since the voltage of the circuit was almost entirely consumed by light bulb 1, the watts generated by each bulb was entirely different. Bulb 1: 4.2watts (volts: 12.6*amps: 0.33) Bulb 2: 0.1watts (volts: 0.2*amps: 0.33), meaning the total watts in the enitre circuit was 4.3w.
By changing the series circuit, to a three bulb affair (each bulb with a similar resistance), we were able to see the difference than to that of one bulb with a significantly higher resistance. Bulb 1 voltage drop was 3.95v, bulb 2 had a voltage drop of 4.35v and bulb 3 had a voltage drop of 4.34v. So the results show that the voltage was better shared across the circuit with no one consumer using more than another. The amps in this circuit also increased slightly as the overall resistance was lower (28.7ohms), leaving the total amps at 0.44A.
Light bulb 1 generated 1.7w and light bulb 2 and 3 generated 1.9w. This tells me that each bulbs brightness is approx. the same as there is less resistance throughout the entire circuit but spread evenly across the resistors. The 2 bulb circuit had one lighter bulb but all voltage was only supplied to that one.
Also, avaliable voltage in the circuit was measured. Battery Positive (12.77v), Input to the switch (12.77v), Output of the switch (12.69v), supply to light bulb 1 (12.68v), Output of light bulb 1 (8.75v), Input to light bulb 2 (8.75v), output of light bulb 2 (4.4v), Input to light bulb 3 (4.4v), output of light bulb 3 (0v) and negative of battery supply (0v). There is a difference in the readings of voltage drop and avaliable voltage. The volatge drop readings tell me what components in the circuit are using how much voltage and the avaliable voltage readings are telling me how much voltage each component is using and how it is being shared out.

REPORT:
Explain what the different readings are telling you when using Voltage Drop compared with Avaliable Voltage

Voltage drop readings tell us the value/amount of voltage each component will be using, however avaliable voltage tells us the voltage that each component is able to use.

PARALLEL CIRCUITS:
Building a parallel circuit with 2 bulbs is much like that of series except that  each of the bulbs in the circuit has their own power source seperate from one another. When measuring rhe avaliable voltage of each of the bulbs, it became apparent that the bulbs have the same voltage. Light bulb 1 had a result of 13.11v and light bulb 2 had a result of 13.10v. The voltage drop of each light bulb was 13.03v.

Describe what happens to the volts in the parallel circuit (what rule of electricity applies here and how is it different from a series circuit.)
The voltage is shared amoungst the consumers evenly is divided amoungst the different pathways evenly. The consumers will be using the same voltage as they are both connected to the power source seperately.

After measuring the current flow of each of the bulbs, the results showed that each of the bulbs had a current flow of 0.75 amps. Which means the total amps of the bulbs is equal to the total amps of the circuit.

Describe what happens to the amps in a parallel circuit. (What rule of electricity applies here and how is it different from a series circuit.)
The current of the entire circuit is divided between the two bulbs. This means the sum of each bulbs amps will be equal to the total amps of the circuit. In series, each bulb simply has the same amps as each other component.

By using ohms law, i calculated the resistance of each bulb in the circuit. Light bulb 1: Volts (13.11v)/Amps (0.75) = 17.48 ohms. Light bulb 2: By using these calculations it was then possible to calculate the total resistance of the circuit. This was achieved by using the formula: 1/RT = 1/R1 + 1/R2
The final answer of total resistance was 8.73 ohms.
We then had to calculate the total watts used in the parallel circuit while both the lights were on. I used the power law, total volts and total amps to calculate the total watts. Volts (11.29v)*Amps (1.5A) = 16.9w. From there, using the same power law, i was to calculate the watts for each individual bulb. Light bulb 1: 13.11*0.75 = 9.8w, Light bulb 2: 13.10*0.75 = 9.8w.

How does these watts compare with watts used when the bulbs were in series? Why has this happened?
In series, the watts of different bulbs change as voltage avaliable to each will be different. However in parallel each bulb has their own link to the power source which gives each the ability to light and achieve the same power.

We then had to add a third bulb to the parallel circuit. One of the bulbs however was smaller than the other two and the resistance of which is higher than the others. The amps measured now for light bulb 1: 0.31A, light bulb 2: 0.7A, light bulb 3: 0.7A. Making the total current flow of the circuit 1.71A

What happened to the total amps flowing through the circuit when the third light bulb was added? (Did the current flow through each of the other light bulbs change? if so why?)
The amps of the two bulbs already in the circuit reduce slightly. The third bulb had a lower current as it is a smaller bulb with a higher resistance. However the sum of each bulbs current still adds to the totsl amps of the entire circuit.

What rule or concept does this show about amps in a parallel circuit?
The current is diverted down different pathways to energise each bulb seperately. If all the same size bulbs were used the amps would be evenly divided amoungst the 3 as opposed to two.

The avaliable voltage then needed to be measured of each bulb, light bulb 1: 12.18v, light bulb 2: 12.17v, light bulb 3: 12.17v. The voltage drop also needed to be measured, light bulb 1: 12.14v, light bulb 2: 12.14v, light bulb 3: 12.13v.

What happened to the avaliable voltage in the circuit when the third light bulb was added? why did this happen?
Avaliable voltage is the same between each bulb as in parallel circuits, each is even.

What happened to the voltage drops in the circuit when the third light bulb was added? Why did this happen?
The voltage drop used almost all of the voltage avaliable to each of the bulbs.

Again, resistance of each bulb needed to be calculated so the total resistance of the circuit could also be calculated. Light bulb 1: 39.3 ohms, Light bulb 2: 17.5 ohms, light bulb 3: 17.5 ohms. From this, the total resistance was 7.2ohms. Also, the total watts of the circuit needed to be calculated with the three bulbs on. As the same as before, i used the total volts of the circuit and also the total amps of the circuit. 12.39v*1.71A = 21.9w.

COMPOUND:
This consists of 2 bulbs in a parallel circuit but also another bulb in series. The avaliable voltage of each component was to be measured:
Switch - 12.82v
Before parallel light bulb 1 - 12.80v
Before parallel light bulb 2 - 12.80v
After parallel light bulb 1 - 5.6v
After parallel light bulb 2 - 5.6v
Before series light bulb - 5.6v
After series light bulb - 0v
Each bulb of the circuit's voltage drop then needed to be measured. Parallel light bulb 1's voltage drop was 7.2v, parallel light bulb 2 was 7.2v and the series bulb had a volts drop of 5.59v.
The measured amps of the circuit were all the same, except parallel light bulb 1 had a current flow which measured 0.25A, where as al the rest measured 0.5A. This means the watts of parallel light bulb 1 is 3.2w, parallel light bulb 2 is 6.4w and series bulb was 2.8w.

How does this compare with the series and parallel circuit watts that you calculated in the series worksheet and parallel worksheet?
A compound circuit consists of both of the circuits, series and parallel. The voltage is the same in the series and in parallel. Bulb is dim as voltage is shared.

How does the series light bulb affect the brightness of the light bulbs in the parallel circuit?  amd why?
It affects the brightness of the parallel bulbs as the voltage is shared in parallel. However, the consumer with the greater resistance will receive voltage first, then continues to others. Series use the one power source to power all, but in parallel each bulb has their own.

Explain what happens to the amperage in the series/parallel circuit:
Current drops as it goes to parallel bulb one and increased to normal at parallel bulb two.

Explain what happens to the voltage in this series/parallel circuit:
The voltage begins at over 12 volts but as it travels through the circuit it becomes consumed by the bulbs/consumers, starting from highest resistance to lowest then is grounded.