Resistors and Relays

Resistors:

We started off resistors by calculating values of different resistors. An example of this was a resistor with the colours red, red, orange and gold. The first red is a 2, as is the second. So we have 22. the orange means multiply by 1000, so we now have 22000Ω and the gold means a 5% tolerance. Measuring this on the multimeter, we got a reading of 21800Ω.

Next we got two resistors of 98.6Ω and 813Ω. We put these two resistors in series and got a reading of 910Ω. We then put the two in parallel and got a reading of 88Ω.

When 2 resistors are placed in series, the total resistance is the sum of the two, therefore the total resistance will be higher than the highest.

When they are in parallel, the inverses of the values are added up, then the inverse of that value is taken to get the total resistance. This value will be lower than the value of the lowest resistor.

Relays:

To start off relays, we measured the resistance through the different terminals. 86 to 85 had 75Ω of resistance due to the windings. 30 to 87a had no resistance because it is straight through a switch, and 30 ro 87 read infinity because it is an open circuit. The control circuit trminals are 86 and 85 and the switch circuit terminals are 30, 87a, and 87. We then calculated how much current would flow through the control circuit if 12v was supplied by using the formula I=V/R which equals 12/75 which equals .16A. The switch circuit that is normally open is from terminal 30 to 37a and the switch circuit that is normally open is from terminal 30 to 87.

We then set up a circuit that involved a switch, fuse, 5 pin relay, and 3 bulbs which come from the switch/consumer circuit of the relay. We performed multiple voltage available tests with the circuit on and off. We found that the voltage changed the most at terminals 85, 87a, and 87. This was due to the valtage drop in the windings and the switch changing position respectively.

-There is no change in the voltage at terminal 86 because this is the input of the control circuit of the relay.

-At terminal 85, the available voltage while the circuit was off was full supply voltage, but when the circuit was switched on, the available voltage was zero, because it was used up through the windings to create the magnetism.

-There is no change in voltage at terminal 30 because this is the main input to the switched circuit and is connected directly to the battery.

-At 87a, there is full voltage because the switch is normally closed, so there will always be full voltage here in the normally closed position. When the circuit was switched on, the voltage became zero because the switch had switched over.

-At 87, there is no voltage available because the switch is normally closed so the switch was not connected to this terminal. When the switch was switched on, the switched moved across to terminal 87 so there was full voltage.

 The final part of relays was to make a circuit to switch between 2 light bulbs. We did this by connecting one light bulb to terminal 87a, and another to 87. So when the switch was changed over, it went from one light bulb to the other.

Diodes

Diodes are important as they only allow current to flow in one direction and are used in places like rectifiers on an alternator.
The first test we did was to measure the resistance of the diode using the 2k setting on the ohm meter. The meter read infinity because the ohm meter does not put out enough voltage to break through the bindery layer of the diode. We then tested the diode using the diode test position on the multimeter. We got a reading of infinity from the cathode to anode position, and .54v from the anode to cathode position which means it it needs .54v to break through the diode.
Next we set up a small circuit using a 12v power supply, a 1kΩ resistor, and a diode. The resistor had a voltage drop of 12.71v, and the diode had a voltage drop of .66v. The diode should not have a big voltage drop because it is not a consumer. The voltage drops add to 13.37v and the available supply voltage was 13.37v. The current flow at the diode measured .013A.
In a series circuit, the voltage drops of each consumer should add to the total supply voltage.
The next test was the same, except we used a bigger resistor, a 9.89kΩ resistor.
Voltage drop across the resistor didnt change a huge amount, but was slightly bigger at 12.81v and .56v across the diode. The current flow however was alot lower at .001A
Because we have a bigger resistor, there was a higher voltage drop and less current.
In the next part of the test we used a light emmiting diode, instead of a normal diode and a 1kΩ resistor. Using the diode tester, we found the LED needs 1.95v to break through it. In doing this we found that the voltage drop of a LED is higher than that of a normal diode.
We found the voltage drop of the resistor was 9.04v and the voltage drop of the LED was 4.3v. The sum of the voltage drops is 13.34v and the available supply voltage was 13.35v. The current flow of the LED was .01A.
In the diode circuit, the diode doesn't use much voltage, therefore the resistor uses more. In the LED circuit, the LED uses mor voltage than the diode, so the resistor doesn't use as much.

Starter Motor bench testing report

Part of  4841 electrical is starting systems. In which we disassembled a starter motor, did a number of tests, and then reassembled it.
The first test we did was a no load test on the starter motor using the bench tester. Voltage read 12.8 volts and current read 38.8A which is normal. After we disassembled the starter, we did a ground circuit test between each commutator segment and the armature shaft. We got readings of infinity which means it is an open circuit, which means it is not grounding to the shaft, which is a pass. Next was a continuity test between each commutator segment. We got readings of 0 ohms on the ohm meter which means there is a circuit, but no resistance, which is a pass. We then tested the commutator diameter and undercut, these were both passes. We then checked the commutator diameter for its circle shape using the dial test indicator, this was in good order.
We then did a test for internal short circuits in the armature using the growler. The hacksaw blade did not vibrate at all so this was a pass.
We then did a visual inspection of the field coil and pole shoes. There were no signs of overheating, burning, physical damage or poling that could cause the starter motor to malfunction, so this was a pass.
We then tested the field coils for continuity. We got a reading of .02 ohms, which means there is a circuit with minimal resistance, which is a pass. Next we tested the field coils for grounding. We got a reading of infinity, which means it is a open circuit and the field coils are not grounding to earth, which is a pass.
We then checked the length of the brushes. All four were 15 to 16mm long which means they all make suitable contact with the commutator, which is a pass. We then tested the solenoid. The first part was the pull in windings. By connecting the power supply to the S and M terminals, the physical action was to pull in the plunger. It did this and drew a current of 20A. We found this to be higher than spec, but after other groups getting the same result, we concluded this to be normal for that type of solenoid. We then tested the hold in windings test. By connecting the power supply to the S terminal and body of the solenoid, it should hold the plunger in. It did this and drew 8A of current, which is in spec, which is a pass. It takes 20A to pull the plunger in and only 8A to hold it in, because it is easier to hold it in that pull it in.
The pinion clutch and bushes were all in good order so these were also passes.
We then reassembled the starter motor, before perfoming another no load test. We found it used 13.2V and 37A which is a pass.

Logic Probe

As part of the electrical course, we were to construct a logic probe. After constructing it, we were to answer a set of questions.

Q1: Why do both the red LED and the green LED go when you connect it to the battery?

A: The current flows through the green LED, through the brass rod, through the red LED and back to negative to complete the circuit.

Q2: Why does the green LED go out when the probe contacts the battery positive?

A: The green LED goes out because when the probe touches the positive, that makes the brass rod positve and the green LED shorts out. This is because the current coming from the positive alligator clip meets the (now) positive brass rod and has no where to earth.

Q3: Why does the red LED get brighter when the probe contacts the battery positive?

A: The red LED gets brighter because all of the current flows through the one LED and it gets all of the available voltage. The current flows from positive, through the probe, through the red LED and back to negative, not through the green LED.