We were in pairs and had been assigned a battery for each pair. We were told and shown how to test the battery and had to work our way through the practical booklet we were given.
The battery Thomas and I were assigned was a Lucas conventional type battery with a 410 CCA rating. The battery number of which was 46G.
The first thing to do when testing any battery is to carry out visual checks. This consists of, battery terminals, surface, possible swelling of the battery etc. From this check it seemed as if the battery termials were clean and tight, there was no swelling of the battery evident. However, there was minor fluid apparent on the surface of the battery around the cells which could indicate leakage. Recommended further inspection.
We then had to remove the cell covers to check the electrolyte levels of each cell. Apparently if they are at the appropriate level, you will barely be able to see the cells at first glance as they will be covered by the electrolyte, however the level must not be too high. All our electrolyte levels appeared to be okay.
We then had to perform a battery open circuit voltage test. In this test it is important that surface charge is removed as it can give false readings. This can be achieved by turning on the headlights for approx 1-2 mins. When they are turned off the meter reading will drop and then build itself back up, where it stops is the reading needed. (The meter needs to be set to 20 DC volts) The result of this was 12.7v, which meant the state of charge on our battery was 100%. The battery must be over 50% charged before any other testing could be performed which meant ours was fine as it was over 12.4v. If it is lower than 50%, it must be charger slowly so as not to over exert itself.
As our battery was sufficient to continue testing we then began to test the specific gravity of the electrolyte. When this test is performed it is important that whoever is doing this test is wearing safety glasses and gloves as the electrolyte is an acid. Each cell needs to be tested individually with a hydrometer. This creates a vacuum which pulls the electrolyte into the hydrometer and makes the float float. First the fluid needs to be checked, if it is clear or murky. Then the reading, the variations of each of these cells should be no more than 50. In this regard our battery failed as our readings were: Cell 1: 1305; Cell 2: 1295; Cell 3: 1300; Cell 4: 1300; Cell 5: 1295; Cell 6: 1250. The specific gravity variation of this battery was 55.
We then carried out a high rate discharge test, using a load tester. The positive clip to be attached to the positive terminal and the negative to the earth terminal. This gave the load tester power. We then had to dial the amp on the load tester to half of the CCA rating which in this case was 205A. As it reaches the desired voltage, we must watch the voltage meter. The load should be held at 9.5v or over. Our battery held at 10.1v which means it was capable of holding its necessary load and is working.
When the engine is off, there are often things in a vehicle that can draw power from the battery. These at any given time should not be above 0.03Mv. If it is any more than this it will cause the battery to die very quickly. This is tested by adding an ammeter to the circuit as one of the components. (Series). Negative to negative terminal, positive to negative wire now unattached. The result of this test was 0 as we were working on engines off car, therefore there was no radio memory or anything to draw from the battery.
Report:
I recommend the battery terminals be cleaned as when the connection was removed for the draw test there was minor corrosion. This to be cleaned with hot water and baking soda until gone and then sealed with grease or petroleum jelly.
If the battery needed charging it should be done with a battery charger on slow charge.
If the amp draw were too high on the battery, the problem would be easily tracked down by keeping the circuit avaliable but removing possible draws until the amp drops. This would show what is drawing too much current while the engine is off.
When testing with a digital meter, it will simply tell you if your battery is pass or fail. The meter needs to be set to SAE and the CCA rating set. To test, the test button should be pressed and the screen will show a pass or fail. Our meter told us our battery failed which means that the battery needs to be replaced. Then press test button again to get O.C.V which was said to be 12.91v. From these results i would recommend cleaning the battery terminals and retesting in case corrosion caused false results. If it fails again, dispose of the battery.
Tuesday, 19 April 2011
Logic Probe Construction
We were given a list of parts that would be needed to build this probe:
Brass Rod 150mm long
Red LED
Green LED
Black wire 2 meters long
Red wire 2 meters long
2 resistors 1 K ohm
Red alligator clip
Black alligator clip
100mm plastic tube 7mm Id
Shrink tubing:
*Black 2.4mm diameter, about 300mm long
*Red 6.4mm diameter, about 175mm long
*Black 12.7mm diameter, about 125mm long
We were given this diagram to follow and a list of step-by-step instructions.
Had to solder the resistors to the LED light. One to the positve leg of the Green LED and the other to the negative leg of the Red LED. Then, twisting the wires of the ends of the black and red wires, so they were the same thickness as the covered parts of the wires. Then twisted the other leg of the green LED to the red wire and the other leg of the red LED to the black wire, symbolising red positive (green LED) and black negative (Red LED). Then to ensure insulation of the circuit, we covered the exposed wires with shrink tubing and heated to fit.
We then sanded a point to one end of the brass rod and the middle so that the solder would take to the rod easily. Then on either side of the sanded middle, shrink tubing was applied to again keep the circuit insulated. the exposed leg of the LED's then are wrapped around the sanded part in the middle of the rod. they are then soldered into place. It is important that there is no sharp points on any of the solder so it doesnt create any holes in the shrink tubing. The red and black wires then need to be pulled back to the end of the rod (without the point) and shrink tubed together (with the rod) to hold it neatly together. Then the end of the plastic support tube needs to be cut so there is a rectangle shape at the end to fit in LED lights and fit ove the brass rod. A little extra room at the end should be given so shrink tubing can be attached to improve presentation. The plastic support tube slides on from the bottom of the wires up to the rod and LED's. Once fitted into place it then needs to be shrink wrapped over as much as possible to improve the general appearance.
The wires then to be plated around one another to also improve appearance leaving approx. 30cm at the end so the wires can be seperated. Finally, the alligator clips need to be attached to the end of the wires, red to red and black to black. They must be pinched onto the wire and soldered down to ensure they are properly secured.
ref: Logic Probe Construction Practical Workbook
This tester is only for 24 volts or less, DC volts only. Connect the red Clip to the battery positive and the black to the battery earth (negative). Both red and green LEDs will light. This tests the connections and that the tester is properly connected. Now touch the brass probe end to the positive battery terminal and the green LED goes out, and the red LED gets brighter. Touch the probe to the earth battery terminal and the red LED goes out, and the green LED gets brighter.
Both LED's go on when they are connected to the battery because the circuit is then a compete seires circuit, in which the current has to flow through positive to earth to be complete.
The green LED goes out when the probe contacts the battery positive because the current now flows through the probe as opposed to the first resistor which makes the circuit then shorter. The red LED then gets brighter as now there is only one resistor in the circuit as current can not flow back to the green LED.
FINISHED PRODUCT
Brass Rod 150mm long
Red LED
Green LED
Black wire 2 meters long
Red wire 2 meters long
2 resistors 1 K ohm
Red alligator clip
Black alligator clip
100mm plastic tube 7mm Id
Shrink tubing:
*Black 2.4mm diameter, about 300mm long
*Red 6.4mm diameter, about 175mm long
*Black 12.7mm diameter, about 125mm long
Had to solder the resistors to the LED light. One to the positve leg of the Green LED and the other to the negative leg of the Red LED. Then, twisting the wires of the ends of the black and red wires, so they were the same thickness as the covered parts of the wires. Then twisted the other leg of the green LED to the red wire and the other leg of the red LED to the black wire, symbolising red positive (green LED) and black negative (Red LED). Then to ensure insulation of the circuit, we covered the exposed wires with shrink tubing and heated to fit.
We then sanded a point to one end of the brass rod and the middle so that the solder would take to the rod easily. Then on either side of the sanded middle, shrink tubing was applied to again keep the circuit insulated. the exposed leg of the LED's then are wrapped around the sanded part in the middle of the rod. they are then soldered into place. It is important that there is no sharp points on any of the solder so it doesnt create any holes in the shrink tubing. The red and black wires then need to be pulled back to the end of the rod (without the point) and shrink tubed together (with the rod) to hold it neatly together. Then the end of the plastic support tube needs to be cut so there is a rectangle shape at the end to fit in LED lights and fit ove the brass rod. A little extra room at the end should be given so shrink tubing can be attached to improve presentation. The plastic support tube slides on from the bottom of the wires up to the rod and LED's. Once fitted into place it then needs to be shrink wrapped over as much as possible to improve the general appearance.
The wires then to be plated around one another to also improve appearance leaving approx. 30cm at the end so the wires can be seperated. Finally, the alligator clips need to be attached to the end of the wires, red to red and black to black. They must be pinched onto the wire and soldered down to ensure they are properly secured.
ref: Logic Probe Construction Practical Workbook
This tester is only for 24 volts or less, DC volts only. Connect the red Clip to the battery positive and the black to the battery earth (negative). Both red and green LEDs will light. This tests the connections and that the tester is properly connected. Now touch the brass probe end to the positive battery terminal and the green LED goes out, and the red LED gets brighter. Touch the probe to the earth battery terminal and the red LED goes out, and the green LED gets brighter.
Both LED's go on when they are connected to the battery because the circuit is then a compete seires circuit, in which the current has to flow through positive to earth to be complete.
The green LED goes out when the probe contacts the battery positive because the current now flows through the probe as opposed to the first resistor which makes the circuit then shorter. The red LED then gets brighter as now there is only one resistor in the circuit as current can not flow back to the green LED.
FINISHED PRODUCT
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.
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.
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