Sunday, 15 May 2011

Testing Diodes

First of all, we had to identify the direction of flow through the diode. This is done by 1: Visual inspection of the diode itself shows that one end is painted which shows that it is the cathode end and 2: By measuring the resistance on each side it shows that when it is one way round it will read infinity as there is no flow through, and the other will have a high reading which means it is the anode end.
We had to then measure the resistance of the diode in both directions using the 2K ohms position on the meter:
Anode to Cathode: Infinity
Cathode to Anode: Infinity
It was said that both of these readings were going to read infinity as there should be no resistance within the diode. The voltage supplied was 0.05v which means theoretically there is not enough voltage to push through the bindery layer of the diode and get an accurate reading.

We then had to measure the diode in both directions:
Anode to Cathode: 0.683v
Cathode to Anode: Infinity
Explain what the Diode Test Position readings means when you test the diode in both directions and describe whether the diode was good or bad:
The diode is in good condition as it shows the amount of voltage allowed through in one direction. The other however has no reading, which is as it should be.

After building a circuit with a 1K ohm diode and a resistor, using a 12 volt supply.
Voltage drop across resistor: 13.22v
Voltage drop across the diode: 0.57v
Measure amp flow through the diode: 0.01A
Avaliable voltage at voltage supply: 13.8v
Voltage drop (sum) across the resistor and diode: 13.79v

Apply the rules of Electricity to these readings above and describe how these readings demonstrate the rules of electricity in action:
As voltage is quite high and so is the resistance, there appears to be verylow to minimal amps in the circuit. The resistor, being large allows very little voltage through, so the remainder of voltage in the circuit is below 1v. The diode had a voltage drop of 0.57v which is the remainder of voltage in the circuit. No voltage left as it grounds.

After changing the resistor in the circuit to a 4000 ohm resistor:
Voltage drop across resistor: 13.28v
Voltage drop across diode: 0.5v
Amp flow through D: 0.01A
Describe how this change of resistance lead to changes in your volt and amp readings. Discuss how this demonstrates how the rules of electricity work.
The readings did not change much, if at all,

After testing an LED (Light Emitting Diode):
Anode to Cathode: 1.834v
Cathode to Anode: Infinity
Compare the voltage drop of a normal diode and an LED. What does this tell you?
1.88v of LED, 0.5v for normal resistor. LED needs more voltage to energise it.

We then had to build a circuit with an LED in the diode position with a 1K ohm resistor and a 12v power supply.
Voltage drop across the resistor: 11.77v
Voltage drop across the diode: 1.88v
Amp flow through LED: 0.01A
Avaliable voltage at voltage supply: 13.5v
Voltage drop (sum) of resistor + diode: 13.65v
Apply the rules of electricity to these readings and compare how these readings are different than the readings for the diode above. In other words, how does the difference between a diode and an LED result in different readings for each part? Explain:
The voltage drop of the resistor decreases as more voltage is needed for the LED. The amp reading is still very low though.

Tuesday, 10 May 2011

Identifying, Testing and Combining Resistors

10th May 2011

We were given 6 resistors and a colour code identification sheet to allow calculation of each specific resistor. The basic principle of the coloured bands is that the last band is the tolerance percentage of the resistor, the 2nd to last the multiplier and the other bands will be the numbers taken down to represent the value of the resistor.
There are 4-Band resistors and 5-Band resistors. This does not necessarily mean that the 5-Band resistor has a greater resistance than that of a 4-Band however. The intensity of the resistor is entirely dependent on the multiplier.

The activity that we were given meant we had to write down each colour in the order it was on each of the resistors and give the value of each based on the identification chart. We then had to figure out the low tolerance value and the high tolerance level of each. From there we were to measure the value with the multi-meter to see if the value was correct and the resistor passed. 4/6 passed and the other 2/6 were unreadable on the multi-meter as they were too high a value.
From that chart we were to choose two resistors from the list we had just made and write the values of these down. From there we had to put the resistors in a series circuit (end to end, one right after the other) and measure the combined value, the result of which was 2,070 ohms. Following that, the resistors then needed to be put in a parallel circuit (both sets of ends connected, almost a circle), the result of which was 78.3 ohms.

REPORT:
What principles of electricity have you demonstrated with this? Explain:

Showed the difference between resistors in series circuit and in parallel circuit. The total resistance in the series circuit is the sum of both the resistors added, however the total resistance in a parallel circuit is divided amongst the different pathways. The total resistance will always be lower than the lowest resistor.

Sunday, 8 May 2011

Relays

6th May 2011

Before we started to do any practical exercises on relays, we were told to read through the notes provided at the beginning of our practical workbooks and show on the 'Sample Wiring Diagram' how each relay is turned on to turn on the fuel pump in the circuit. I found this very effective as it showed the different paths of current and simultaneous switching to eventually lead to the same path.
We then did a few exercises to understand the basic principles of a relay, followed by an exercise involving wiring up a relay in a circuit with 3 bulbs in a parallel circuit (control circuit of the relay being negatively switched) and drawing a diagram of this. It took a while to figure out how this set up worked because in past circuits, we have had to wire the switch directly after the power supply, whereas in this case the switch had to be wired after the bulbs from the relay. From this circuit, we measured the avaliable voltage from the circuit when it was off and when it was on, and compare the difference.

REPORT:

When the circuit is off, each components avaliable voltage is approx. the same as no component is using any voltage from the power supply. However, when the circuit is on, some voltage is being used by the relay to energise the internal magnet, therefore they drop slightly. Terminal 85 however, has 0.14v avaliable. This is because terminal 85 is connected after the bulbs/consumers which are connected in parallel consuming the same amount of energy each. Therfore there is none to minimal voltage left.

We then had to draw a diagram showing how to wire up a relay to switch between two lights, as if switching from high beam to low beam. This was achieved by physically wiring up the board to show a better understanding of the diagram.

Wednesday, 4 May 2011

Starter Motor On Car Testing

4 May 2011

These tests were completed on a starter motor on an engine set up outside of a vehicle.
Before carrying out any tests however, the ignition or fuel injection system must be deactivated (which we did by removing the fuse), the battery to be checked for serviceability (after removing surface charge, checking voltage - result 100% - 12.7v), the right range has been selected on the digital meter (20DC volts) and that the vehicle is in neutral, which in this case does not apply.
The avaliable voltage across the battery while cranking needed to be checked, which is specified 9.5v minimum. The result was 11v which is plenty.
We then had to perform a voltage drop test (while cranking) between battery positive and solenoid starter input, across solenod main input and output, and between battery negative and starter motor body. The specifications of these were all very low. The first was specified less than 0.2v, however the result was 0.4v (well over and therefore failed). The second was specified less than 0.1v, and the result was 0.17 which is not too much over but still a fail. The third was specified less than 0.2v and the result was 0.91v which again is too high and failed. The maximum allowable voltage drop across the component is 0.5v and the total turned out to be 1.48v, almost 3 times that value.
The starter motor current draw to be tested using a clamp meter. The specified current draw is 125-150 amps. The test result was 133A which is a pass.

REPORT:

The voltage drop showed increased figures than specified which means there is resistance within the circuitfrom the battery to the starter motor and within the solenoid. Amps however were within specified range. The resistance source needs to be minimised for starter motor efficiency.
Starter circuit test failed.

Starter Motor Bench Testing and Repair

3 March 2011

In our group we had a Mitsubishi starter motor (Model Number M3T 49381) to pull apart and test various components to see if it was all working correctly. (All multimeter reading were measured using a 200ohms scale.)
The booklets we were given had a basic summary of all the tests we were about to perform, the possible results and why this would be so that we were actually understanding what we were doing.

To start, we did a no load test on the starter motor to make sure that the solenoid was operating. The result of this test was 11.5v and 35.2A. We then followed a strict procedure to dismantle the motor, marking sections so they could be realigned when put back together.

Armature:

First, the Visual Inspection, we checked the armature for any possible signs of overheating, burning, physical damage or poling. There were no real major visual signs of distress, only minor scratches on the plates which still means that the armature was serviceable. We then performed a ground circuit test, placing one of the meter leads on the armature core and the other lead on between each of the commutator segments. This should read infinity as there shoul be no circuit. As this passed, it shows that the armature is serviceable. We then had to test the continuity of the armature by placing one probe on the commutator and moving the other around the commutator (not breaking contact) to see how much resistance is within the circuit. The result of this was 0 ohms which is within recommended specifications and is therefore still serviceable. Then we had to measure the diameter of the commutator and the mica undercut depth. The diameter of the commutator that we were working with, was 20.05mm which is below manufacturers specifications of minimum diameter (26.8mm - 31mm). Therefore in this regard it failed and was not serviceable, however, it was explained that the specifications were perhaps not for the particular model that we were working with. Also the mica undercut was suggested to be 0.7mm - 1mm, it was said that the measurement of this was impractical and to make an estimation of depth. The armature was then to be placed "V" blocks to perform a dial test to check for run out. The armature had to be turned 360 degrees to get an acurate reading. The specifications for this were 0mm - 0.2mm. Our test came back with 0.05mm and therefore is serviceable.
We then used an alternative method of testing the armature using a 48 volt test light, to test continuity and ground. For the continuity test, it is performed the same way but on the machine, this test applies greater pressure to the windings and should be used where avaliable, the light on the test glowed, which means that the continuity of the circuit was good and still serviceable. For the ground test, it is also performed the same way, but in this instance the light will stay off to show theere is no circuit, this means it is still serviceable.
Also, to test for short circuits in the armature, it is placed on the "V" of the growler and switched on. A hacksaw blade is held loosely on top of the armature, the blade did not vibrate which means that there are no short circuits.


Field Coil and Pole shoes:

First, the visual inspection, where we checked for and visible signs of overheating, burning, physical damage or poling. Again, no major visual faults, only minor such as covering of wires tearing slightly, and coil wires becoming loose and frayed. This means that it was still serviceable.
We then checked the fied coils for continuity, but first we had to check the internal resistance of the meter itself which turned out to be 0.4 ohms. We were given 2 options on the worksheet 'If Grounded' (specifications - Test N/A) or 'Not Grounded' (specifications - 0-0.02 ohms). The results of my test turned out with just the internal resistance of the meter, so there for the test was N/A, meaning it was grounded and serviceable. To then check if the field coils were grounded we had to place a positive probe on the field wire or brush, then the common probe on the body of the starter. Again, given the same 2 options 'If Grounded' (specifications 0-0.02 ohms) and 'Not Grounded' (specifications Infinity), and being that our meter read infinity means that the component is still serviceable.
The length of the brushes had to be measured with the vernier calliper and specified to be longer than 5mm. Also, we had to check for cracks or other damage of the brushes. There was no visible damage to the brushes and results showed the 2 brushes were 5mm each which passed. The brushes also needed to be checked for short circuits between insulated brushes. This is achieved by placing one probe on the brush and the other on the metal retaining plate. The result of this test showed infinity which means that the holder is insulated and it passes to remain serviceable.

Solenoid Magnetic Switch:

We first had to identify the terminals. The battery terminal of the solenoid is obvious as it is isolated from the rest, the starter motor supply is also obvious as it lead from the solenoid into the starter, therefore leaving the remaining terminal to be the ignition/starter switch supply. We then checked that the pull in windings of the solenoid were working by connecting a 9v power supply to the ignition terminal and the starter motor supply terminal. This is only applied for 5 seconds to prevent heat damage to the windings. The maufacturers specifications of the current draw of these windings should be 8-12 amps, the pull in windings of ours however had a current draw of 23A which is too much current to be activating the windings. When they are activated however, the plunger of the solenoid should be pulled in which it did. ALso in the solenoid are hold in windings which are activated after the plunger is pulled in to hold it in place. This time, the 9v power supply is connected to the ignition terminal and the solenoid body. The plunger must be pushed in by hand and released to see if it is held in place. The current draw of this should be 5-8 amps. The result was 7A which passes.

Pinion Gear and Overrunning or One Way Clutch:

This needs to be inspected for damage and smooth movement along the shaft. The bushes must be checked for wear and also the bush clearance in the appropriate end housong. Each were in good condition and serviceable.

The starter motor then had to be reassembled, housing realligned to marks made. However, when we were putting the component back together, the frayed wiring of the field coil terminals became increasingly worse and one of the terminals snapped off completely. It was suggested that we continue to put it together without that terminal and complete a no load test to see if it would still work.
The final no load test was unable to be performed and the starter motor was unable to perform.


REPORT:

The initial no load test carried out gave results showing the starter was within manufacturers specifications. However, once taken apart it became apparent that the brush wires were frayed and therefore the brush came off completely causing the circuit to be incomplete. When final no load test was carried out after re-assembly of starter, the results were unavaliable as the windings within the solenoid were not able to generate anything. I would recommend changing/replacing brush holders.

Alternator Off Car Testing

In this exercise we had to dismantle an alternator and carry out different tests to see if the alternator was working properly and if not, what component would be the source of this failure.
Following the given list of instructions in the booklet, we took apart the alternator and began testing.
Starting on the Rotor windings, we tested it to ground. (Multimeter set to 2K on ohms scale.) To do this, we placed the black lead from the multimeter on on the centre of the rotor shaft and the red lead on the slip ring. The meter should end up with no reading. Also, the rotor winding internal resistance is tested. For this the leads are to be placed on the slip rings. The meter should read between 2-6 volts as specified. The reading achieved was 3.2 (minus the internal resistance of the multimeter) which gives a final result of 2.8 ohms. Results of these tests are within specs and pass.



Then further opening the alternator, the stator winding resistance is tested. This is achieved by placing the black lead of the meter on the common terminal (terminal with the most wires leading to it) and the red on each of the other terminals to gain readings of each. The spec for this was to be between 0 - 2 ohms. 0.2, 0.1 &0.1 were the readings of the terminals which makes them within specifications and therefore also pass. Then to make sure that the common terminal is grounded, the black lead is placed on the body of the alternator whilst the red is on the common terminal. To pass, there will be no reading on the meter. This was the case which means that there is no circuit between the stator winding and ground.

The meter then needs to be chnaged to diode test mode to test the rectifier positive diodes. Testing each of these, Black lead on a common terminal and the red on each of the rest one at a time. There were 3 terminals which gave relatively similar readings (0.580, 0.590, 0.579) and the other gave no rading at all. This is so as one of these terminals is the common ground and as such should have no circuit. Therefore all of these passed. This test is then reversed and all results should read infinity (no reading at all). All results were as required.
Testing the rectifier negative diodes is slightly different as the red lead is placed on a body terminal and the black is placed on each of the 4 other terminals one at a time. The results were the same however.

Testing the voltage regulator, we first had to find out the make of it, this was done (personally) by comparing the regulator to the diagrams in the booklet and finding a match. Carrying out a Transpo regulator test, we had to make sure the short circuit light was of (Pass), The warning light came on and stayed on (Pass), the field light should flash continuosly (Pass) and the reading (12.5v) was what it needed to be (Pass). This means that the regualtor is not short circuited and is doing its job correctly.
Then using a vernier caliper we checked the length of the brush protrusion. This is very important as if they are too short the brush springs can not apply enough pressure to maintain contact and in turn reduces output of the alternator. The length we got was 10.2mm and 10mm. This means it was well above the minimum length of 4mm.


REPORT:

Overall, the alternator passed, meaning that it operates as its supposed to and each component is working well.
Testing the rotor winding internal resistance gave a measurement of 2.8 ohms which is between recommended specifications of 2 and 6 ohms. Being that the rotor is an electromagnet if resistance is too high, this causes the magnetic field to become weaker which in turn decreases the field current of the circuit. If this happens the voltage regulator will have less field current to adapt to constant voltage at the stator output. This will also reduce the field current that will be induced into the stator windings and therefore have less output to the terminals of the coils with the same result of rectifier which will draw more energy from the battery to compensate for the fault, greatly decreasing the battery's state of charge over time.
It is very important that the rotor winding internal resistance is between specified values, if not they may need to be replaced.

Tuesday, 19 April 2011

Battery Testing

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.

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

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.