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A power line with a resistance of 2 ohms has a current of 80 A in it what is The power dissipated in the line is?
Wiki User
∙ 13y ago
P=I^2R. 80^2=6400*2=12,800 Watts. That is for D.C. For A.C. you would use the R.M.S. or average wattage.
Wiki User
∙ 13y ago
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How do you calculate copper losses?
copper losses are power losses due to flow of current in the wires or resistances,if the resistance is R, current is I then copper losses are I2R. for a 3-phase system; copper losses are same but for a single line, total losses are 3I2R.
I want to convert 12 volts DC to 29.5 volts DC by using a 400 watt power inverter will it work?
In short, no. There are ways to make the voltage rise, but inverting the power will not help you. If you don't care about the amount of Amps you have, the simplest way to raise the voltage would be to add resistance to the line. Volts and amps are linked by Ohm's law, which states that V=IR, where V is volts, I is current (in amps), and R is resistance in ohms. Therefore you can create a higher voltage at the cost of amperage by adding resistance to your line.
Would a multimeter connected in parallel as part of the circuit measure current or power consumption or voltage or wattage?
A typical multimeter measures voltage and resistance "in parallel" or current in-line with the circuit. It all depends on the model. 'Multi' means many different modes. Some have more than others. Some multimeters can measure current, but require you to alter the lead configuration and plug into different sockets. If you have the meter set for Current measurement and you put it in parallel instead of series, it will cause a dead short and could damage the meter. (They usually have a fuse inside for protection). Knowing any two of the three parameters you can calculate the third by Ohm's Law: Voltage = Current x Resistance. Knowing Current and Voltage you can calculate power as Volts x Amps = Watts for resistance loads.
Electric power is transferred over large distances at very high voltages Explain how the high voltage reduces power losses in transmission lines?
The conductor used to transmit power has a specific resistance at the given power frequency. Power transmitted is equivalent to I^2 * R, so as the current increases, the power loss from the conductor also increases. To maximize efficiency, power is converted to very high voltages, which decreases the current, which minimizes the power lost in transmission.AnswerThe primary reason for using high voltage transmission lines is because, for a given load, the higher the voltage, the lower the load current. Low voltages would require cables of enormous cross-sectional area making transmission impossible, whereas high voltages allow the use of manageable-sized conductors. A secondary reason is the corresponding reduction in line losses -as described above.
What is a Line Commutating Choke?
Line commutating chokes are used to smooth voltage peaks or bridge commutating dips. In effect, this reduces the effects of harmonics on the inverter and power supply. A line commutating choke is necessary to reduce current peaks if line impedance is <1%
The cord of an electric oven does not glow while the heating element glows.Why?
The cord is manufactured to have as low a resistance as possible, while the heating element is intentionally manufactured with a carefully controlled resistance. The current through the whole loop ... cord plus heater ... is determined by the resistance of the whole loop. The magnitude of the current 'I' is (E/R) ... E = the utility line voltage, R = resistance of the cord+heater. But the power dissipated by each individual resistance in the loop is proportional to the resistance of that section. P = I2R. So the heating element dissipates more power than the low-resistance line-cord does.
What is the amount of power loss on a transmission line that is transmitting 10 MW at 500 kV with 1000 ohms?
Line current = 10MW / 500kV = 20A Assuming the 1000 ohms is the resistance of the entire transmission line, end to end. Power loss = line current ^ 2 * line resistance = 20A ^ 2 * 1000 ohms = 400 KW
Calculate the Power in watts dissipated by a resistor of 1000 ohms with 200 volts across it 2 watts 20 watts 40 watts 400 watts?
Ohms law states that V = I * R I = V/R R = V/I P = I*V Where V = Voltage, I=Current, R = Resistance and P = Power or Watt Watt is the amount of electricity flowing through a line which is (Voltage times Current in (Amperage) = Power or watts) To find the power dissipated by a resistor of 1000 ohms, we first find the current I. The voltage is given as 200volts. Therefore I = V/R = 200/1000 = 0.2Amps We said Power or Watt = I*V Therefore the Power or watts dissipated by a resistor of 1000 ohms will be P=I*V = 200*0.2 = 40 Watts
How is heat energy produced in electric iron?
A resistor placed across the power line: I squared R (current x current x resistance) = heat in watts.
Which mode of transmission would result in less energy loss high current and low voltage or low current and high voltage Explain your reason?
low current high voltage power dissipation in power line = I2R the resistance of the power line is hard to reduce, especially when it is a long transmission line. but reducing the current through the line reduces losses as the square, a dramatic savings. reducing voltage would have no effect and would dramatically increase losses due to increase in current to try to deliver same power.
Why you use 33kv for power transmission?
Power loss during power transmission is predominantly due to the current drawn, which heats the wires. Power dissipated is current x voltage and voltage is current x resistance, so in any wire with fixed resistance the power loss is proportional to the square of the current flowing. The higher the voltage you use to transmit the power down the distribution network the lower is the current, hence power is distributed at the highest voltage that is practical, then the voltage is dropped to the domestic level at the destination sub-station or by a transformer on the power line pole. Major distribution systems operate at over 100 kV, with regional ones at lower voltage.Additional Answer33 kV (not 'kv'!) is the primary distribution voltage, not a transmission voltage, used in the UK. Other distribution voltages are 66 kV (not very common) and 11 kV. Transmission systems operate at 400 kV and 275 kV.
How much current will be taken from a 220V line if the resistance is 55 ohms?
Current = Voltage / resistance (more properly impedance) so the current will be 220/55 or 4 amps.
How do you measure current without an amp meter?
Depends on the current. Put a resistor in-line with the current, then measure the voltage across the resistor. V=RI. So, divide the measured voltage by resistor value. Be careful with the size of the resistor, as Power dissipated in a resistor is R*I^2 or V^2/2. So, a 1-Amp current into a 1 Ohm resistor will result in a 1Watt power dissipated in the resistor. If it's too small, it'll burn. Also, notice that if you do that, you haven't measured the current in the original circuit. You've measured the current when an extra resistor is installed in the original circuit, and that's different.
How much power can a transmission line carry example how much power can a 69kv line carry compared to a 390kv line what is the formula for power capacity of a hvac line?
For a line of given cross section and material, the power capacity will depend on the current carried, since resistance heating is proportional to (current)2 . For a given power, current is inversely proportional to voltage. Thus raising the voltage from 69 to 390 kv would reduce the current by a factor 69/390 = 0.177 , for the same power transmission, and reduce the heating losses by 0.1772 = 0.031. So you can see why high voltage for long distance lines is essential. Obviously the limiting current on a power line has to be set by economic and practical considerations, but if this is predetermined and set, the limiting power will be that which produces that limiting current, and power = voltage x current. The actual limiting curent will depend on the line cross section, material, and length. The power that a line of a certain voltage can carry is calculated by using the following formula: (2.55×(KV)2 /1000) MW.
What is line voltage and line current in star connection define?
In star the voltage from line to neutral is 1/sqrt(3) times the nominal voltage, while the load current equals the line current. In delta the voltage between lines is the nominal voltage, while the load current is 1/sqrt(3) times the line current (for a balanced load). So a delta load needs 3 times the resistance compared to a star load of the same power.
The electric bulb in your household circuits are connected in parallel what happens to the line current and line voltage as you turn on more light bulbs?
The line current increases when more bulbs are switched on in parallel, since more parallel current paths results in lower effective resistance. The line voltage should not change in response to any normal use of electric power in a single house.
Why is energy lost while transmitting low voltages over a transmission line?
Because there is resistance in the line. Pushing a current through a resistance generates heat, which is wasted energy.
