Electrical Engineering

light can be altered by using different types of gas and varying the pressure in the bulb. Gases such as neon, argon, krypton, mercury, and xenon allow electric-discharge lamps to be used in a variety of applications

Work is usually measured in joules, which are a unit of energy. Work/time is called power and is usually measured in Watts (joules/second) other common units of power include Horsepower and ft*lbs /sec.

in short, is is power, which is measures in watts.

Electrical angle is half of mechanical angle in unipolar electrical machines. In multipolar electrical machines, the relationship between the mechanical angle and electrical angle is Electrical angle = (P/2) x Mechanical angle where: P = Number of poles.

Resistance (Ohms) = Potential Difference (Volts) / Current (Amps)

So, 12/0.25 = 48 Ohms.

The incoming voltage from the source goes to L1,L2 and L3 terminals. The Load connects to T1, T2, and T3. The coil voltage goes to A1 and A2. The stop ,start circuit is in series with the N.C. contacts on the contactors overload block. Usually numbered 95 and 96.

Sydney, Australia :)

nitrogen

A photovoltaic transducer is simply another name for a solar panel. A transducer changes one energy to another in this case light energy to electrical energy. The two most common type of photovoltaic transducers are silicon and selenium.

Reduce the harmonic current in the electrical supply system.

Coils are rotated past these magnets which causes the electrons in the coil to align in a particular way. This causes current to move in that direction. The power is actually generated as the coil wires pass through the lines of flux from the magnet.

If you would like to see these lines of flux you can! Put a piece of paper over a magnet and sprinkle some iron filings on the paper. Tap the paper a few times and the flux lines will start to appear. A wire passing through these lines generates a small current flow. The faster you can make this action repeat, the more flow. Many wires are coiled together to make this action occur so often that usable power is made.

No. From what you're saying you're taking the input to the 2kV transformer in a microwave and replacing it with a 1MW input. The transformer will burn in milliseconds of power.

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The installation and maintenance of a 1000 KVA transformer can be worked on in a de energized state. To do so live could be extremely dangerous at any voltage.

Friction in the bearings and in the commutator, and windage loss. Electrical resistance of the armature and field windings and the brushes.

The area required would depend primarily on the efficiency of the system and on its location. A photovoltaic system works better in a desert in the southwest than in Seattle, Washington.

Use a wattmeter, as it only reads 'real power' of your load. Use an ammeter and a voltmeter, and the product of the two readings will give you 'apparent power' of your load. Since apparent power is the vector sum of real power and reactive power, use the following equation to find the reactive power of your load: (reactive power)2 = (apparent power)2 - (real power)2

A series DC motor has to have a starting resistor to limit the current flow before the speed builds up.

In any series circuit, there is one and only one path for current flow. All the current flowing in the circuit will flow through all of the devices in that circuit. A break at any point in the circuit will cause current flow to cease. Lastly, it is current that is the same at any point in the circuit where we'd care to measure it.

in order to establish the Q-point of transistor

V_rms = V_peak / √2 so V_rms = 80/√2 = 56.6 V (3sf)

There are two main devices. One is the circuit breaker. It provides a dual function where in it can detect short circuits with its internal magnetic sensor and overloads with its internal bi metallic strip. The other device is a fuse. It operates by melting a fusible link, that breaks the fault current that is in the circuit, when a high current rush through the fuse.

The angle by which a circuit's current leads or lags its supply voltage.

(R) Resistance = (E) Voltage / (I) Current in Amperes

R = 12/3

R = 4 ohms

With the meter on the voltage scale you will read zero volts across the blown fuse. From the bottom terminal of the service switch fuse to ground you will read zero volts and from the top terminal of the service switch fuse to ground you will read 120 volts.

The multimeter will read 0

To answer this question voltage, power factor, efficiency, and run configuration must be known, none of which was given in the question. I will make some assumptions, to shown how the calculation is done.

Given: 50 HP, three phase. Assumed: 480 volts delta, 0.92 PF, 0.95 EF, and WYE-DELTA start-run configuration.

1 HP (electric motor) is 746 watts, so 50 HP is 37.3 KW. Divide by PF 0.92 and by 0.95 EF to get 42.7 KVA. Divide by 480 V and by 3 to get 29.6 A per winding. Multiply by 1.732 (square root of 3) to get 51.3 A per phase.

Now, 51.3A is a lot of current, requiring large windings and, therefore, a large motor. I would consider a higher voltage motor, such as 4160. Rerun the calculations and you get 5.9A per phase. Does this justify a 4160 service? Its marginal, but you begin to see the issues involved in choosing a motor.