Battery Testing

The first part of battery testing was inspecting for battery specifications. The make of our battery was a Lucas, battery number was 128HD. It has a Cold Cranking Ampere (CCA) of 400. It was a conventional battery and the electrolyte was easily accessable by removing the caps and using the tip of the hyrometer to check the level of the electrolyte.

We carried out visual checks and found that the terminals were clean and tight, there was no signs of swelling caused by overcharging.

In case of an accident, the nearest supply of water had to be identified, and was behind us about 2 meters away.

Next we had to check the electrolyte levels. To do this we dipped the hydrometer in until the tip touched the plates. We then removed it and the level can be seen on the rubber tip. The level was about 10 to 12mm above the plates. This was a bit high as the recomended level is 4 to 5mm above the plates.

The next test was to measure the open circuit voltage. With the meter range set on 20V, we got a result of 6.61v, which was due to the battery not being charged. The state of charge was recorded as under 25%.

To continue with the testing, we needed a state of charge of at least 50%, which equates to 12.4v, so our action was to put the battery on charge.

Using another battery, we got a set of hydrometer results. These results had a specific gravity variation of 100 points. The allowable specific gravity variation of a battery is between 25 and 50 points, so the result we got was a fail.

Next was to do a high rate discharge test, using a carbon pile. This was a different battery with a CCA of 310A. The load that we were to apply was 155A and the voltage that had to be held while the load was being applied is 9.5v. Our result was that the battery held a voltage of 10.6v while the loaad was being applied.

Next we used a digital battery tester to test the battery. The tester gave a reading of "PASS" which means the battery has adequate charge. The tester also said that the battery had and OCV of 13.46v. It also displayed a CCA reading of 300. The state of charge of the battery was 100%

The battery was in good condition. Although there was a surface charge on it, so putting a load on the battery for a short amount of time would have been appropriate.

Alternators - On car testing

Before we start testing the alternator, we need to carry out visual checks. This includes the connections, mountings, and tension of the drive belt. All of these were good.
The first tests that we do are no load tests, which means that there are no lights or other accessories on that will be causing the alternator to work harder.
The first test was the base battery voltage. This measured 12.42v, which is less than the 12.6v it should be.
The next test was the regulator voltage. It shoul have read between 14.5 and 14.6v. I measured 12.11v which is a fair bit less than it should be.
Finally with the no load tests was the amperage output. The carburetted output specification is between 5 and 12amps and I got a reading of .1A. I figured this was because the battery was fully charged and did not require charging from the battery. So I switched off the engine and let the fan run for a few minutes to drain the battery. I tried again and still only found that the output was .1A so I came to the conclusion that the alternator is at fault.
Next was to carry out a load test on the alternator. Since we only have have an engine, we cannot put a load on it using the accessories of a car, so we have to use a carbon pile to create a load.
since I found my alternator was not working properly, I joined the group next to me and found the output amps under load to be 29.1A and the charging voltage under load to be 12v.
The final tests were voltage drop tests. The first one was between the battery positve post and the output of the alternator. The voltage drop should be less than .2v, we measured .13v, so this is a pass. This is showing us if there is a drop in voltage from the alternator to the battery.
The second test was between the battery negative post and the body of the alternator. This should have also read less than .2v and ours read .16v which is a pass, so this is also a pass. This is showing us if there is a voltage drop through the earth part of the circuit.

Alternators - Off car testing report.

After we removed the rear cover, the regulator and the brush holder on the alternator, we were able to carry out tests for the rotor winding to ground and the rotor winding internal resistance test. In the rotor winding to ground test, we are testing to see if there is a circuit butween the rotor shaft and the slip ring. There should not be a circuit here and the meter should read infinity, when the meter is set to 2k. If there is a circuit here, it means the rotor winding is shorting to earth, and will need to be replaced.
On the rotor winding internal resistance test, we are testing to see if there is a circuit between the two slip rings. There should be a circuit between the two rings, and with the ohms meter set to 200ohms, we should get a reading between 2 and 6 ohms. In my test we got 3 ohms which is a pass. The ohms meter did not have any internal resistance I did not have to subtract it from the reading. If there is no circuit between the slip rings, they need to be replaced.
To carry out the next tests, I had to remove the rectifier, and the housing of the alternator so that we could access the stator winding terminals. We tested the internal resistance in these by connecting the negative of the meter to the common stator winding terminal. The specifications of the resistance should be between 0 and .2 ohms. I tested them all and found they had a resistance of .2 ohms each which is a pass.
The next test was the stator winding to ground test. This is where we test to see if there is a circuit between the stator winding and ground. We test this by connecting the positve of the meter to the common stator terminal and the negative of the meter to the body of the alternator. There should not be a circuit here so the meter, set on 2k, should read infinity. My test read infinity so that is a pass.
Next we tested the rectifier positive diodes. The rectifier is what turns the AC current produced by the alternator into DC current. To do this we put the negative lead onto the main terminal and then the positive lead onto each of the four p-terminals. The readings for the meter should be between .5 and .7VD My readings were between .491 and .497, so that is a fail. The next test we put the positive lead onto the main terminal and the negative lead on the p-terminals. The meter should now read infinity, and it did so that is a pass.

The next test was to test the rectifier negative diodes. To do this we set the meter to diode test mode. The negative lead gets put onto the E terminal and the positive lead onto each of the p terminals. The meter should read infinity, which it did, so that was a pass. The positive lead was then put onto the E terminal, with the negative lead onto each of the p terminals. The meter should have read between .5 and  .7VD. I got readings between .485 and .492, which are all fails.
The next test was to test the voltage regulator. The regulator controls how much current and voltage is sent away from the alternator. To test a regulator, we used a device called the Transpo Voltage Regulator Tester. After testing the regulator, I found that the set point voltage of the regulator was 12.1v, which does not meet the specification of 14.5v, so this tells us that the regulator needs to be replaced.
The bearing should also be replaced when servicing an alternator. The bearing is what allows the rotor shaft to spin. It turned smoothly and without any resistance.

The final check was to check the protrusion length of the brushes. On the brushes that I pulled of the alternater, the brushes were so worn that they popped out of the holder. I found new brushes and measured the protrusion length of these ones. The first brush measured 10.5mm and the second measured 11.5mm. The minimum length is 4mm, so these were both passes.

Compound Circuits

To start off compound circuits, I constucted a circuit consisting of a power supply, fuse, switch, 2 bulbs in parallel and a third bulb in series. The first tests I made was the available voltage at the different parts of the circuit. Before the 2 parallel bulbs there was full supply voltage of 11.82 volts, after the parallel bulbs there was 8.3v. This meant that before the series bulb there was 8.3v and after the series bulb there was practically 0v. The voltage drop across the parallel bulbs was 3.5v and across the series bulb was 8.27v
The amperage at the switch was .57A and at parallel bulb 1 it was .16A and at parallel bulb 2 it was .4A. This is telling us that bulb 2 was a bigger bulb with less resistance so more current could flow through it. The current at the series bulb was back to .57A which means it has combined again after the parallel part of the circuit. The watts used at each parallel bulb was .58w and the watts used at the series bul was 6.6w. I found that the parallel bulbs were not glowing because the series bulb creates alot of resistance for the parallel part of the circuit, therefore the do not have enough current flowing through them to make them glow brightly.
What I found is that the amperage splits at the parallel part and then combines again after the parallel part.
A small amount of voltage is used up by each parallel bulb and then a large amount of voltage is used up by the bulb that is in series.

Parallel Circuits

To start off parallel circuits, we constucted a circuit using, a power supply, fuse, switch, and 2 light bulbs in parallel. We then measured the available voltage at each light bulb. which was 11.82v. There is the full supply voltage at each bulb because the full voltage goes through each branch. We then measured the voltage drop across each bulb and this turned out to be 11.74v for each bulb. This means that each bulb is using up the full supply voltage.

The next test was the current. Current flow through the bulb 1 circuit was .71A and current flow through bulb 2 circuit was .67A. These 2 values then add together to give the total current flowing through the circuit which is1.38A. Current will split evenly through branches in a parallel circuit as long as there is equal resistanc in each branch. If resistances are not equal, the most current will flow through the branch with the least resistance.

I then calculated the resistances for each branch and the bulb 1 circuit came to 16.6 ohms and bulb 2 circuit came to 17.7 ohms. I then calculated the total resistance in the circuit using the formula 1/Rt=1/R1+1/R2 and the total resistance came to 8.57 ohms. The total resistance of a parallel circuit should be less than the lowest resistance of any branch.

Total power output for the circuit was then calculated to be 16.3 watts and then power out put for each bulb was calculated. Bulb 1 had a power output of 8.4 watts and bulb 2 had a power output of 7.9 watts.

The bulbs are alot brighter in parallel because there is alot less resistance with the bulbs in parallel compared to in series. Less resistance means more current which means more wattage.

Next we constucted a circuit with 3 light bulbs. I measured the current flow and found that the current through the 2 smaller bulbs were about .68A and the current through the bigger bulb was 1.67A. This meant that there was less resistance through that bulb which meant more current flow, which meant a brighter light. This means that the current will get split up between the seperate circuits, with the most current going through the circuit with the least resistance.

I then measured the available voltage at each light bulb and had supply voltage of about 11.55v at each light bulb. I then measured the voltage drop over each light bulb and that came to about 11.40v for each bulb. This means that the supply voltage is the available voltage in each branch.

I then calculated the resistance of each bulb usiong Ohms law. Bulb 1 came out to be 16.72 ohms, bulb 2 came out to be 17.52 ohs and bulb 3 (bigger bulb) came out to be 6.916 ohms. I then calculated the total resistance resistance of the whole circuit which I calculated to be 3.82 ohms, which is less than the lowest value of the individual resistances. I then calculated the total power output of the circuit which I found to be 36.15 watts.

Series circuits

Moving on to series circuits, we constucted a circuit consisting of a 12V DC power supply, fuse, switch, 2 light bulbs connected in series and then earthed the circuit back to the negative of the power supply.
The first test that I carried out was measuring the voltage drop across the components in the circuit. In doing this I found that the major voltage drops were across the the light bulbs and there were vey minimal drops across the wires and switch. These voltage drops added up to the supply voltage. Because the 2 light bulbs were the same size, they both used the same amount of voltage.
The next test was to measure the current flowing through the circuit. Connecting the ammeter around different points of the circuit I found that the current remained consistant, which it does in a series circuit.
I then used Ohms Law to calculate the resistance and power output of the circuit. Total resistance I found to be was 23.4 ohms and the power output of each bulb I found to be 6.6 watts.
The bulbs are putting out 6.6 watts because with 2 bulbs in series, there is more resistance, which means there is less current flow, which means there is less power output.
The next part was to construct a circuit with 3 bulbs in series. Voltage drop was measured again and was found that the major voltage drops were across the bulbs and because the bulbs were all the same size, they all had the same voltage drops. The current was again measure around the circuit and the current was again consistant at .44 amps, which was higher than the circuit with 2 bulbs. Resistance and power output were then calculated and I found total resistance to be 28.4 ohms, which is larger than the circuit with 2 bulbs, and power output of each bulb I calculated to be 5.5 watts, which is slightly less than the 2 bulb circuit. This is because the circuit with 3 bulbs has more resistance, which means less current, which means less power out put.
We were then asked about the difference between available voltage and voltage drop.

3 bulbs in series

The voltage drop readings are telling us how much voltage is lost at that component. The voltage available readings tell us how much voltage is available at any given part of the circuit.

individual circuits

We first learnt about voltage, resistance, current and power. Voltage, V, is measured in volts, v, and is the energy or pressure behind the current. Resistance, R, is measured in ohms, Ω and is what makes it harder for the current to flow around a circuit. Current, I, is measured in amps, A, and is the amount of electricity flowing around the circuit. Power, P, is measured in watts, w, and is the power output of a component or circuit. The first part of the electrical unit was individual circuits. These are simple cirrcuits consisting of a DC power supply, fuse, switch and a lightbulb, and then the circuit was earthed back to the power supply.

We then began to carry out tests. The first test was the available voltge - which is the voltage available at any given point on the circuit. We tested the avaible voltage at the positive of the 12V DC power supply, at the terminals before and after the switch, at the terminals before and after the light bulb and at the negative of the power supply. Getting accurate measurements became difficult because of the fluxuating voltage of the power supplu. In doing this I found that the only major voltage drop was at the light bulb. There were minimal voltage drops at the switch and wires which is explainable by a small amount of resistance in these components. We then measured the current flowing through the circuit by connecting the ammeter in series before the light bulb. This turned out to be .36 amps. We then used Ohms laws to calculate the resistance and power of the light bulb. We then did the same circuit with a bigger bulb and found that there was a higher current draw, less resistance and a higher power out put in the circuit with the bigger bulb.

safety

On the day of our first practical class we were briefed on a bit about safety.
We were told where the fire extinguisher was located and where our emergency exits are. These are staight down the hall and the down the stairs or down the hall to the left and then down the stairs. And the fire extinguisher is located in the corner of the room next to the door.

We were also told about personal protection equipment. Steel cap boots and overalls or a coat are a must and anything else is optional. If we are not in the correct PPE we should not be turning up to practical class. Or if an accident occurs, unitec will be in alot of trouble and the students could potentially have missing toes etc.