Electrical Engineering

Much of the newer overhead cables is made of aluminum with steel reinforcing it for strenght purposes. Aluminimum is much cheap, and much lighter, thus it is cheaper to use vs. copper for high voltage transmission, and copper has a lower resistance which means more electricity can get through the wire/ cable, which means more electricity can get around the world quicker. most people should use copper becasue it has a lower resistance where aluminuim has a bit of a high resistance not to much but a bit so that is why you use copper not alominuim.

Time delay protective relays used in the power industry can be time over current or voltage, or definite time over / under voltage / over current.

Definite time type relays use an over / under voltage / current element that picks up when voltage / current reach a specified level. When this picks up, a timer is started. If current / voltage stays above / below the pickup value for long enough to allow the timer to expire, the relay trip contacts are closed.

Time over current elements use an inverse time - current curve to define speed of operation, so the more current is present, the faster the relay will operate.

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the voltage standing wave ratio is defined (1+p)/(1-p), where p is the the reflection coefficient magnitude. p = 1 for an open circuit, therefore the VSWR will approach infinite.

To be sure, you need to contact the utility. My guess from your description is this is not really deemed high voltage. It is most likely a distribution to sub transmission voltage (2.4kV - 69kV); If it were higher than this, it is very likely that the utility would have purchase agreements or own the property directly under the line for 30-80 feet either side. this is very typical in rural farming areas. The utility often allows farmers to farm right up to structures...but if anything goes wrong (such as a crop duster clips the line), the utility has every right to come in and grind the crops to dust in the process of repairing the line.

Electrical engineering is a large field. Certain jobs can be dangerous; others are not so dangerous. If you know what you're doing and are careful, the risks tend to be minimal.

In my line of work that is an Electrical High Voltage substation that is brand new. It is being built from a green field into an energized fully functional substation yard. There is no existing substation or remodeling of an existing one being done.

The more turns of wire in an electromagnet the stronger the magnetic field.

Nikola Tesla invented the poly-phase induction motor in 1883.

line transformation ratio would be the turns ratio (ie voltage ratio), while the phase transformation ratio is most probably the phase shift introduced by a 3 phase transformer.

Impedance is a vector sum using the formula Z = square root (XL2 + R2); where Z = impedance, XL = inductive reactance, and R = resistance. Therefor the formula for a circuit where XL = 64ohm's and R = 36ohm's is Z = square root(642 + 322); Z = 71.6ohms.

By Ohm's law, resistance is voltage divided by current, so the resistance of a light bulb can be measured by observing the voltage across it simultaneously with observing the current through it.

Interestingly, the hot resistance is significantly different that the cold resistance, so measuring resistance with an ohmmeter will not give a meaningful resistance.

This is because the resistance of a light bulb has a positive temperature coefficient. Take a typical 60 W 120V light bulb, for instance... Its cold resistance is about 16 Ohms. Calculate current and power at 120 V and you get 7.5 A and 900 W. The truth is that at 60 W, the bulb pulls 0.5 A and has a resistance of 240 Ohms.

That's like asking how many meters in a liter.

Ampere and Volt are two DIFFERENT measurements.

Ampere is how much electricity you are using, while volts are how much pressure the electricity is under(Think water).

If you want to figure out how many amperes your appliance is using you could use this formula: P=UxI (Watt=Volt x Ampere) or U=RxI(Volt=Resistance x Ampere).

Sure - an unconnected power supply of any kind - a battery is + at one end and - at the other.

you run the voltage up to 2x the operating voltage + 1000v

this detects exposure to moisture and pinholes in insulation

if it cant pass a hi-pot test chances are it wont pass signals reliably either

The auxiliary contact is to be utilized for interlocking of another relay or contactor or indication signal from another equipment.

What motor is NOT suitable for use as a DC machine?

A capacitor is like a 0.001 ohm resistor when the capacitive reactance is 0.001 ohms.

Capacitive reactance, in ohms, is defined as -1 divided by 2 pi f C, where f is frequency in hertz, C is capacitance in farads, and the -1 means that current leads voltage.

Plugging in 60 Hz, and solving for C, you get 2.65 farads. That is a very large capacitor. At 6000 Hz, you get 26.5 millifards, which is still very large. At 6 MHz, you get 26.5 microfards, but at 6 MHz, you need to consider parasitic inductance.

Both of are power supply devices. But there are many difference between IPS & UPS.

1. IPS means Instant power supply, UPS means Uninterrupted Power system.

2. IPS backup system is so high, But UPS backup system is low

3. IPS use a large number of electronic equipment, But UPS use a small equipment

4. IPS provide large time backup, But UPS provide short time backup

5. For IPS must be use big size battery, But UPS is not.

6. IPS can be operate more than one electronic equipment like Computer, TV, Fan etc at a time, But UPS can be operate only one equipment.

No. What you are describing is a series-parallel circuit, not a parallel circuit.

Rated current is typically defined for all electric devices. Either the device has a rated current, wattage or Volt x amperes (VA). Current can be computed by dividing the wattage or VA rating by 120 volts (For resistive loads). In residential situations the maximum current is governed by the protecting breaker on a circuit. If you all up all the rated currents for devices on a circuit they should be less than or equal to 80% of the protecting breaker current rating.

V=I*R -- Potential (Voltage, V) = Current (Amperage, I) * Resistance (Ohms)


2A*5Ohms = 10V

this happen due to sudden amount of voltage drop in the main feeder due to large current drawn by the heater , so this drop in voltage will let the bulb operate by avltage of a mount (V-Vd) [where V represent the supply voltage and Vd represent the voltage drop in the main feeder of the circuit] , which is less than the voltage before the heater is connected and due to this situation the current passes through the wire of the bulb will be less and therefore the brightness of the bulb becomes dim.after a while the wire of the heater will has high temperature which increase its resistance and due to this the current drawn by the heater will decrease than the current at the first time , therefore the drop in voltage will also decrease , which implise increase in current drawn by the bulb and therefore the dimness decrease .