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The transmission voltage of a power station is 5000v and the transmission power is 500kW. The difference between the readings of electric energy meters installed at the beginning and end of the transmission line is 4800kwh in one day and night The power loss on the transmission line, the resistance and efficiency of the transmission line, and how much of the transmission voltage should be increased if the power loss on the transmission line is to be reduced to 2% of the transmission power
5 times root 5 times 10 to the second power
A small hydropower station, output power 20kW, output voltage 250V, transmission line total resistance is 0.5 ohm. How much power loss?
Notice! I divide r by P = u squared and the answer is 125KW!
Using P = I square r to get 3200W is correct!
Why? Isn't P = UI = I trivial r = u trivial divided by R?
Why are the answers different?
What do you mean when I do it for the first time?
What is the lost power used for? The answer is that it is consumed in the resistance of the transmission line, so first calculate the current in the line, and then calculate the consumed power according to I ^ 2R
P = UI, so I = 25000 / 250 = 80A
P loss = I ^ 2R = 80 * 80 * 0.5 = 3200W
The voltage of 250V is loaded on the wire resistance and the resistance in the user's home. Now we only know that the wire resistance cannot be calculated by P = u ^ 2 / R
In a hydropower station, the total resistance of 2.5 ohm transmission line is used to transmit power to users 500 km away, and its output power is 3 × 106 kW
Voltage transmission, transmission line transmission current size is?
Why output power / transmission voltage? Isn't the intermediate resistance consuming part of the voltage
Because the voltage consumed on the resistance is smaller than the output voltage, it can be ignored. If you don't believe it, calculate it
According to I = P / u = 6000a, resistance consumption voltage is 15000V, 15kV / 500kV = 3%, which can be completely ignored
When the resistance wire of the electric oven passes through 5 a current, it produces 6.6 * 10 fourth power heat per minute. Calculate its electric power and resistance when the resistance wire works
P=W/t=6.6x10^4J/60s=1100w
The solution of P = I ^ 2R is r = 44 Ω
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