Electrical Troubleshooting

VATS is usually a GM (General Motors), system for helping prevent vehicle thefts.

There are several different procedures depending upon the year and make of the vehicle. Search the internet for "VATS Bypass".

I bypassed the vats on a 1996 Buick by accessing the wiring diagram for the VATS. I measured the resistance of the key chip. There are 15 different values it can be. It is simply a resistor. In this case, the chip measured 3000 ohms. I went to Radio shack and got a combination of resistors that would give me 3000 ohms. There are 2 wires that come out of the VAT module and go up the steering column to the key switch. I located the VAT module behind the glove box. I cut those two wires and soldered the resistors in so that the module always saw 3000 ohms. You have to solder the resistor on the module side, not the key side.

When dealing with HVAC systems, notice that there are two systems at work, and two possibilities for failure: the HVAC control system, and the actual HVAC unit. Determine if your problem is in the function of the controller (blank screen, not turning on, relay does not click on when fan or thermostat is set past current temperature), or in the HVAC unit (thermostat relay is clicking but HVAC fan does not turn on, condenser outside fan is not spinning, etc.) If the air conditioner was 'grounded', as I understand it, I would think your HVAC breaker would trip. This should help you figure out your problem.

coil on side of fly wheel broke of can refit but want to rewire bike from start to finish key start to cut of switch to coil whole bike but need to know lay out of wiring of bike and replace all wireing

The likely problem is the setting of your pressure switch. These models differ and you'll likely have to locate the manual for your switch. The manufacturer and model will usually be listed on the pressure switch case. The dump valve will likely do nothing except cause the air to be released in case of overpressure. It won't cause the compressor to shut off when it should.

The general answer is that a continuity tester is used to check for an open circuit.

You shouldn't have to clean the contacts. If they aren't working, it's often due to damage from arcing. However, you can use compressed air, air in a can, or electrical contact cleaner, only after the main bus is de-energized and the breaker is removed from the panel. Contact cleaner is a liquid, and should not be sprayed into a live bus or breaker.

The VOM is in the classification of a multimeter. This means that the one meter can do several functions. VOM stands for Volt Ohm meter. In the voltage portion, the meter has the ability to measure a variety of voltages from fractional usually up to 1000 volts. This is both AC and DC voltages. In the ohms portion, the meter has the ability to find the resistance of components. On most good multi meter there is a continuity setting that will have a sound emitted if what you are testing has continuity through it. That setting is very useful in troubleshooting control panels to locate open and closed contacts in energized and de energised relays.

One possibility is the manufacturer of that particular board. Many have repair/exchange programs. If the board itself is physically broken and/or burned, it is probably not repairable. If components on the board are bad, it may be repaired. One possibility is the manufacturer of that particular board. Many have repair/exchange programs. If the board itself is physically broken and/or burned, it is probably not repairable. If components on the board are bad, it may be repaired.

A high pot test is a test of a wiring system for high potential (voltages). It's also called a megger test. This test verifies the integrity of a wire's insulation, and ensures that high voltages cannot escape through cracks in the insulation smaller than the human eye can see.

rated current is the current at any given conditions......but full load current is the maximum current that the system design can hav.....it w'll b larger than that f the rated current. rated current is the current at any given conditions......but full load current is the maximum current that the system design can hav.....it w'll b larger than that f the rated current.

G Hz is 1000 bigger than MHz now we can write that1GHz = 1000MHz

The heat generated by any particular resistor depends (at least electrically) solely on the power it dissipates. Power dissipation in a resistor is equal to current squared times resistance, and the current through the resistor is equal to the voltage across it divided by the resistance. If we take a 10 ohm resistor ('your resistor') and put it in a series circuit such that there is 10 volts across your resistor, the current through it will be 1 ampere (10/10=1). the power dissipated will be 10 watts (1^2 * 10=10). If we put your resistor in a parallel circuit that also puts 10 volts across it, then the current and power will be the same. Your resistor does not know or care where the voltage came from. From this point of view, once you get down to the voltage across the resistor, it does not matter what type of circuit it is in. On the other hand, for any given power supply voltage, then the type of circuit and the value of external components certainly does affect the terminal voltage and thus the current through as well as the power dissipated by the resistor. In a parallel circuit, the voltage across your resistor remains basically the same no matter what resistance you put in parallel with it (unless you overload the power supply or the power supply has high internal resistance). In this case, the voltage across the resistor is the same as the power supply, current is I=E/R, R being that resistor only, and power is P=I^2 * R. In a series circuit the current through the resistors is I=E/R, R being the total resistance (including the other resistor(s)). The power dissipation in your resistor will then be P=I^2 * R, I being the series current we just calculated, and R being your resistor only. Since the other resistors affect the current, and since the current is the same no matter where you measure in a series circuit, then the voltage across your resistor and thus the power dissipation will be affected. The voltage across your resistor will be E=I*R, I being the series current we just calculated, and R being your resistor only. So, while the calculation for power dissipated in a particular resistor does not change relative to what type of circuit it is in, the calculation to arrive at the voltage across the resistor and/or the current through it (which you will then need to calculate power) does. Keep in mind there are other mechanical parameters that influence the actual case temperature of the resistor. Physical size of the case, composition, and airflow velocity, if any, will alter the case-to-ambient thermal conductivity. Ambient temperature will also be a factor in the final temperature.

A ground is not expected to carry any current. It's only there in case of a fault condition. If you have a current reading through a ground wire there is a fault that needs to be corrected. Many times a lazy electrician who couldn't find a broken neutral connected a receptacle or light fixture to ground to make it work. This is not a proper use of the grounding system, and it is dangerous and should be corrected if encountered. The grounding conductor should have at least the same ampacity of the largest phase conductor connected to the circuits it protects. That way it is capable of carrying the full current of the largest conductor in case of a fault.

To minimize hysteresis loss

You can use an ohmmeter or continuity tester. Connect one lead to ground and the other to each of your sensor wires. When you have a ground on the wire being tested, your meter will indicate continuity. Make sure the sensor wires are de-energized when testing with these methods, this may require you to disconnect the system power and unplug the backup battery. If you already know which wire is grounded, you'll need a transmitting/receiving device such as a circuit tracer or short tracer. This device will allow you to follow along a wire and detect approximately where the problem is.

All resistors have a rating called 'maximum power dissipation', usually referred to as simply 'power'. Typical values are 1/8 watt, 1/4 watt, 1/2 watt, 1 watt, etc. up to many hundreds of watts for specialized resistors. Basically, if you exceed the power rating for a given resistor, it will fail. If you run it right at maximum power, it will work for a while, but will have a fairly short lifespan and will eventually fail (this is true for all electronic components for the most part). If you run it well below its maximum rating, it will pretty much last forever. Take a 10 ohm, 1 watt resistor and connect it to a 12 volt battery (hypothetically, that is, don't try this at home!) The current through the resistor will be: I=E/R I=12/10 I= 1.2 amperes The power dissipation will be: P=I^2 * R P=1.2^2 * 10 P=14.4 watts A 1 watt resistor that is dissipating 14.4 watts will fail in seconds, accompanied by smoke and quite possibly fire! Now connect a 270 ohm, 1 watt resistor to the 12 volt battery. The current: I=12/270 = 0.044 amperes (44 ma.) The power: P=0.044^2 * 270 = 0.52 watts This resistor will certainly get warm, but is operating well within its power rating and will last a long time.

This is not possible. You must replace the diode.

I think by 'cross wiring' you mean reverse polarity. This means the hot wire is connected to the neutral screw and the neutral wire is connected to the hot screw. This shouldn't have any impact on an AC motor, since AC voltage already changes polarity 60 times per second.

3 disposable type (1.5v) batteries would get you to 4.5 volts, and 4 rechargeable (1.2v) type batteries would get you up to 4.8 volts. As far as the size... You can get 4.8 volts from any four 1.5 volt batteries. The batteries will power the load longer, though, if they are bigger. 4 rechargeable 'D' cells and 4 rechargeable 'AAA' cells will both run a 4.8v motor, but the 'D' cells will do it for much longer. Estimate how long the motor is to be used, or how much load will be on the motor, and make your best guess as to the battery used... Normally, you wouldn't use 'AAA' or even 'AA' cells for a motor actually doing something. I would probably go with 'D's.

Are you referring to the original Intel 8051, or one of the many variants? The 8051 has weak internal pullups on the i/o pins, and can source only about 60 ua, but can sink 1.6 ma, still not much when it comes to driving the led in an opto. Some 8051-based dervatives can sink much more current. Atmel's 89C2051 for instance, can sink up to 20 ma per i/o pin. This can directly drive most optos.

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This can not be answered with the information you have given. 60Hz does not relate the the current draw so wire size can not be calculated.

This seems like a question from an electrical course, and is probably best answered by your course materials. It's your test question, not ours, and there won't always be someone to ask for the answer. Earn your diploma.