Electrical Engineering

Voltage is a measure of electric 'push'. When there's nothing to push (no circuit) it has no meaning, but that isn't to say it isn't there.

Just the same as if you push a car. Take the car away and your push goes away. (or does it?)

A short circuit which offers very low, practically zero, fault resistance is called "dead short circuit". further more it bypasses the entire load current through itself.

Aim: To Calibrate LPF wattmeter by Phantom Loading. Apparatus: 1) Voltmeter - 300V Digital --> 1 2) Ammeter - 10A Digital --> 1 3) Phase shifting Transformer - 415V/15A --> 1 4) Auto Transformer - 240V/0-270V,10A --> 1 5) Power Factor meter - 230V,10A-Dynamometer --> 1 6) Wattmeter(LPF) - 230V,10A-Dynamometer --> 1 7) Connecting wires Procedure: 1) Keep connections as per circuit diagram. 2) Keep Auto Transformer in minimum position and On toggle Switch. 3) Vary the transformer at Power coil till voltmeter reads rated value. 4) Vary phase shifting transformer to set power factore at 0.2 lag. 5) Now slowly vary the auto transformer at Current coil in Steps. 6) Note the values of wattmeter at each step of current. 7) Repeat the above step till Ammeter reads rated current. 8) Calculate percentage of error. Plot the graph between current and percentage of error, keep current on X-axis and % error at Y-axis. Calculations: Wa = VxIxCosQ watts. Wm = watts. % error = [(Wa-Wm)*100]/Wm. Precautions: 1) Avoid loose connections. 2) Avoid parallax errors. *3) Keep wattmeter and P.F meter in parallel. Result:

Check that you have proper fuel pressure or if you have a clogged converter

unitless

Watts is volts times amperes. Volts is joules per coulomb. The two units are not directly related without some other parameter, such as resistance or current. Please restate the question.

opening of circuit..

It's probably only a 1.2V LED. It just depends if the batteries are connected in series or parallel.

If E=I x R then it would require 1232 volts to push 16 amps through a 77 ohm resistance.

Usually rating of CTs within 50- 80 percent. Of the

maximum load current of the consumer does no change

a tubelight consists of filaments at each of its respective ends. the function of the starter is to develop ionic discharge inside the tube to make the current flow. therefore once the tubelight is switched on and it glows. the function of the startea ceases and thus it can be removed. on removing a starter from the glowing tubelight the tubelight will continue to glow unless switched off. but once switched off the tubelight will require the starter to glow again on being switched on.

The angle between the expected and actual secondary current is known as phase error.

The conclusion phase of the decision making process results in one of two things. Either the problem is resolved or the failure leads to a return to a previous phase.

A car battery normally supplies 12.5 volts at full charge. If the light bulb you are trying to light is a 12 volt bulb then make a connection with the +(positive) battery terminal to the small lead button at end of the bulb, and the - (negative) terminal to the metal side of the bulb. If the battery is charged, the light should work. If the bulb is a higher voltage type you must connect more batteries in series until the proper voltage is reached. WARNING...With more batteries connected in series, the voltage could become dangerous. Do not use more than two batteries in series unless you have an experienced electrician to apply safety techniques.

If your battery is 12 volts, an auto lamp up to 24 volts will light up. the 24 volt bulb will be half as bright. Over the years, I have replaced dashboard lamps with 14 volt bulbs because they will last longer because they are rated for a higher voltage but fed with a lower voltage.

No, Electricity from Chemical reaction may be DC.e.g Battery

Yes, these fall in a class called metamaterials.

You have to reverse the oil flow through the motor. This is usually done with a two way valve.

You don't mention what device, but the answer is probably NO. 50hz and 60hz signals will act differently in reactive components such as transformers, coils, capacitors, etc.

Paulev is correct that it would act differently on inductive/reactive devices, especially transformers. The device to be discussed here is the transformer as it is most present in all electronic equipment. Based on theories, a 50Hz transformer will perfectly work on a 60Hz Power system but not vice versa. Assuming the transformer's specs are, P=1kVA, L=1H, R~0 ohm. At 50Hz the impedance will be Z=sqrt(Inductive Reactance=2 x Pi x 50 x L), thus the ampacity (Ip=sqrt(P/Z)) of the magnetic wire at the primary is 1.79A.

Using the same steps at 60Hz. The current drawn by the the transformer will be 1.63A which is within the ampacity of the magnetic wire. therefore a 50Hz appliance can be used on a 60Hz supply without any problem.

yes, it is a passive component.

I x R = E

20 x 6.4 = 128 Volts

That is not the case. Scroll down to related links and look at "Measurement of input impedance and output impedance".

Yes you can, depending on the level of education required for a given position, experience, expertise and abilities, prior performance record, etc.

On a race motor, to convert to alcohol, the only real thing we have to do is put all stainless steel in the carb. The brass will not hold up to the alcohol. If I recall the flame travel, or rate of combustion is slower on alcohol, so, we advanced the timing, I think, and richened, I think, the air fuel mixture. Another neat footnote, whereas you had to keep your car cool, now, you will have to keep it warm. Alcohol burns much cooler than gas, thus lowering your operating temp, leading to sluggish acceleration, and part failure.