Electrical Troubleshooting

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.

Re open the switch junction box after turning the power back off. If you can locate the wire that bring the voltage to the box make sure that this wire goes to the top of the switch. If it is tied into more black wires this could be receptacle circuits that are on all of the time. If this is the case, from this group add a pigtail from this group to the top of the switch. from the bottom of the switch find the wire that is going to the light. It is probably in the group that is tied together. Once found terminate this wire to the bottom of the switch. Put the breaker back on and try the switch.

It can, but if you're wanting to run a 120v light bulb on DC, you'll need 120v DC to get the rated output. That's a lot of batteries. It's easier, and more sensible, to find a DC rated light bulb, such as an RV bulb.

It depends on the type of fan, whether it is a range hood, bathroom fan, or room ceiling fan. There are several ways they are mounted as explained below. Preferably the power to the fixture in question should be off.

1. If there are thumb screws near the cover (mainly on range hoods and some light only ceiling fixtures), you would remove them. Then you would unscrew the old bulb and screw in the new one. Then you would put the thumb screws back in.

2. This may be just for a light-only fixture, but if the fixture is recessed and the cover is almost completely flush with the ceiling, you would pry it from the ceiling. Then you would replace the bulb. Getting the reataining springs back in may be tricky, and you would need to protect your eyes from debris up in the ceiling.

3. With most bathroom fans that use a square acrylic light cover, you would press into each side and pull it loose. Then you would replace the bulb and snap the cover back on.

4. If you mean a standard ceiling fan, it depends. For most, you would simply unscrew the old bulb and screw a new one in. For others, there are retaining screws holding the globe on, and for a few, the globe unscrews.

5. For a few ceiling fans, you don't replace the bulb, but replace the entire lamp kit. To do that, you remove the 3 screws from the plate above the light kit to drop it down, then disconnect the wiring (cap off the wires from the fan until you are finished), remove the light kit from the plate by loosening the nut. Then follow the instructions with the new light kit and undo the previous steps to reassemble the lower part of the fan with the new light kit.

An open circuit

4 volts = 4,000 mV.

There are many types of motors used in industrial applications. The most common would probably be a DC motor, as the motor's speed is easily controllable and well suited for production processes.

Well, as far as I know, there is no 'gas' A/C. The A/C will be electric, regardless of the type of heat used. Whether electric heat or gas heat would be more economical is really dependent on your electric and gas utility rates and the efficiency of the appliances involved.

http://www.xpedio.carrier.com/idc/groups/public/documents/techlit/fa4a-6pd.pdf