Why is Electrical Power Transmitted at High Voltage AC and Three Phases
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- เผยแพร่เมื่อ 5 ส.ค. 2023
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High voltage is more efficient. AC allows transformers to step the voltage up and down. Three phases is 30% more efficient than single phase.
Video about transformers: • Transformers
This is a great explanation of electricity power transmission, and a crime that it has only 5600 views... and dancing cat videos get 30M views. We are doomed as a society.😵💫
About 10 minutes ago out of the blue I thought to myself...man I miss Bob's videos and channel. I decided to look up Bob and low and behold I see his channel. That made my day. Glad to see your back Bob. You're the best in the industry for teaching electronics !!
The break even point for high voltage DC lines depends also on the available inverter technology. It’s easy to rectify AC since quite some time, but the opposite transition is not that easy for 500kV.
I was super confused on the residential 2 phase thing until you got to the split phase part. That was always my understanding. I actually didn’t know houses used to get 2 line phases at the house. That’s wild.
Sir, please start series on Sinusoidal Oscillator like Hartley, Collpits oscillator. They are very difficult to understand and no one has explained it in very intuitive manner anywhere in either book or on net. Your explanation is understandable and realistic. So kindly explain the working of Sinusoidal Oscillators using BJT, FET etc
Where I live, outlets are either 10A or 20A, and while the layout is the same, in a V shape, left if Neutral, bottom is Ground and Right is Hot, the difference between 10A outlets and 20A outlets are the size of the plugs, 20A plugs and outlets are 0.4mm larger than they 10A siblings, so 20A plugs dont fit on 10A outlets, but 10A appliances plug on 20A outlets
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Brilliant! The best explanation of this stuff I've ever heard. 🎉
Quick question about 15 & 20 Amp plugs....they seem a bit backwards in a way. Plugging a 20A load into a 15A socket at worst will blow the breaker. However, a 15A load in a 20A circuit will FAIL to blow the circuit when sometimes it should.
So in the interest of safety, the breaker should err on the "low" side, no?
The breaker is sized to protect the wires and the outlet, not the appliance. A smaller load is fine. A fuse inside the appliance will protect it, if necessary.
@@gt5713 Thanks so much for that! I get it now! 👍
When the voltage of a wire increases, isn't it against Ohm's law for the current to decrease? You mentioned that Ohm's law is also used in AC circuits. So, isn't there a contradiction in this matter? Does the voltage undergo an imaginary change, why should the current decrease when we increase the voltage? If voltage increases, current should increase as well.And yeah , you should talk about Fenno-Skan transmission line on the next video.
I think your looking at it the wrong way. Remember, P over I x E. You want to deliver 1,000,000 Megawatts. If your voltage is 500,000 then that is 1,000,000 Watts divided by 500,000 Volts = 2 Amps.
@@mikesradiorepair if current decrease when voltage increase , impedance must increase with voltage .Then what thing causes to increase the impedance. transformator ? Long wire ressistance?
@@kirchofffontaine6845 That's because you are not applying that voltage JUST across the wire! Consider this to be like a voltage divider or potentiometer where the wire is one part of 2 series resistors and the load is the other part. You can't use the ohm's law unless you know the load resistance as well. How can you determine the current otherwise? Your series current is Vs/Total resistance, according to ohms law. In the case with higher voltage, the load resistance also increases to keep the product power constant, our constraint to provide constant 1MW. And now since that load resistance increases, the overall series current will decrease according to ohm's law itself!
The formula for ohm's law is V/I = R = constant. The formula for power or load is V * I = P = constant (here = 1MW). Notice how the direct proportionality changes to inverse proportionality? So now you will quickly realize that if you increase the voltage and want to deliver the same load of 1MW, your product V*I = 1MW needs to reduce the current. It's like he explained: 5V * 1A = 5W and 50V * 0.1 amps = 5 watts. To draw 1 A from 5V, the overall equivalent resistance of the network comes to be 5 ohms. Out of this, the wire is 0.1 ohm so the load will be 5 - 0.1 = 4.9 ohms. Similarly for the 50 V case the equivalent resistance needs to be 500 ohms to draw 0.1 amps. This leads to load resistance of 499.9 ohms. Or you can flip the logic by using various voltages and various load resistor values, but the power dissipated in the load resistance needs to be constant at whatever you choose. You will find that you end up with an inverse relation between applied voltage and current, unlike the ohm's law. But in no way is ohm's law being violated! It's just that the load resistance is changing its value in the background to decrease the current derived via ohm's law. Why? To maintain our constant power contraint!
@@Swanicorn yeah that explains the situation.Thanks
I'am confused about the formula, by OHMS law 500kV divided by 0.1 ohms would equal 5million AMPS?
The 0.1 ohms is just the resistance of the wire... The load on the downstream side where the power is used is essentially more ohms. Also the step up process after the generation reduces current.
As another comparison, imagine shorting a 9v battery with a 0.1 ohm wire. The math says 90 amps should be going through the wire, however the battery itself can't do that much and essentially has its own internal resistance.
A curiosity from Scandinavia: Wikipedia(Fenno-Skan).
I have the same question as kirchofffontaine6845. I don't understand how 500kV in a circuit with 0.1 Ohms total resistance will deliver only 2 amps.
I think your looking at it the wrong way. Remember, P over I x E. You want to deliver 1,000,000 Megawatts. If your voltage is 500,000 then that is 1,000,000 Watts divided by 500,000 Volts = 2 Amps.
I answered them already. The thing is 0.1 ohms is not the total resistance, you are ignoring the load resistance. You are delivering power of 1MW to a load right? Is that load going to be zero resistance? Try thinking about the load as the appliances in your home, like maybe an incandescent lightbulb of 60W will make it easy to understand. Your household supply is 120V (assuming America), so 60W bulb should have a resistance of V^2 / R = 60W which gives an R = 240 ohms. This 240 ohms is your load resistance which is to be added with the wire resistance to determine the series current.
@@Swanicorn The transmission line voltage is stepped up at the generation plant and is stepped down a couple times before it gets to your home. Your home or the load your talking about is on the secondary side of the step down transformer outside your house.
That must be a mistake in the beginning. 400 KM copper wire even at a 7 cm diameter has to greater than 0.1 ohm.
Yes it would be around 29.5 cm in diameter to have a resistance of 0.1 ohm
At only 2Amps you can easily afford 100Ohms for power loss on 400km. In fact it would be more like 1..2kA to transport GWs over such a line. But even then a resistance loss of a few kW is nothing compared to the capacitive loos over that distance.
That would be a very expensive and very heavy transmission line.@@Nicholaechs
All that was due to Tesla