I have read different forums and watched a few youtubes (in addition to my textbook readings) and the explanations seem to fall short. The issue seems to be how we are first taught about a direct relationship between voltage and current (that is, an increase in voltage renders an increase in current if resistance remains the same) and then we're taught about power lines that have high voltage and low current (because other wise we would need thick wires that carry high current [which would run the risk of overheating due to the joule effect or something or another..). So please don't explain to me the infrastructural reasons why high voltage, low current is necessary for power lines. I just need to know how high voltage, low current is even possible. I've only been studying DC so far so maybe AC has rules that would enlighten me...but I thought the E=IR formula was universal.
That's a really good question. I wanted to keep my video simple to really answer your question you need to look at the transformer from the power plant to the power lines. It's a step down transformer and the designer can decide how much to step down the voltage. It's best to step down a little to keep the voltage high. (ac max)
The problem is that V=IR (i.e. Ohm's Law) is *not* universal. The quick explanation for why high voltage, low current is possible is that because you have to conserve energy from the input to the output of the transformer, the power input must be roughly equal to the power output; and because P=IV, stepping-up the voltage causes the current to step-down (and vice versa). The more complex explanation is that formula (V=IR) really only applies to DC circuits. In AC circuits, there is a more general relationship to be considered V=IZ, where Z is called the impedance of the circuit. Impedance depends upon a variety of factors: the resistance(R), the capacitance(C), the inductance(i), and the frequency(f) of the AC circuit. When you take into account all of these factors with a simple set of transformer coils as shown in the video, all the math works out.
In lament terms… if I have you bench pressing weight. Let’s say you can only benchpress 150lbs. Your power only allows you to lift 150lbs. This is power. There is two ways to make the load easier. Take off the weight or juice you up to become stronger to lift easier without resistance. So while you’re lifting 150lbs that is your max amps before you give out, since resistance is too strong. Now while you’re lifting I inject you with Superman DNA and you can lift more weight because you have more power. Now you have more power to lift weight. That 150lbs which is resistance is nothing to you now because you have more power (voltage) so in terms of current, it is much easier for you to lift the weight. Less amperage with more power makes sense. What wouldn’t make sense is having much less power and using less amps. It’s about power. So the more power I have the less current is needed because simply I have the power to lift all that weight.
Trying to give some thoughts on the actual question the viewers wanted to be answered: How is it possible to increase the voltage without increasing the current? Ohm s law states their are directly proportional. I am still trying to figure it out myself, but here s what i gather from this video, which by the way is very good. First, you notice he s using the Power formula, not Ohm s law. Why? My guess is because in this situation we HAVE to take Power into consideration, since it is given from the start. Ohm s law doesn t include power. 2) P=U*I now, since P is a fixed known value, it means that the increase of either the voltage or current comes at the loss of the other. Basic maths. 3) there is a sort of misconception that high voltage means high current. I ve thought that to be true from the 6th to about 12th grade. Of course, we see it is not always the case. What makes the difference (as far as i can tell) is POWER, specifically whether the value is fixed or to te to be calculated. After all, voltage is a POTENTIAL, so whether it means or doesn t mean high current, depends on other values. This is what i got from the video, i don t know whether it s correct or not, feel free to give me your view. Thanks for the vid!
Well as an 11th grader I think you're pretty much right, the power across a transformer has to be constant, and as you said the increase of one would bring about the decrease of the other value.
I was thinking the same and looking for an answer. I think that the V_rms in the P=V^2/R simply doesn't equal the voltage used by the power plant, but instead it is the voltage drop on the resistor V_r, which is equal to V_r=IR. The drop of voltage of a rezistor doesnt equal the voltage provided by the power plant, because the resistor isnt the only power consumer in the grid. The current however is the same as the one provided by the power plant, becouse all the current has to flow through the wire. Thus we can calculate the current easily as shown in the video. I_low = 6667 A and I_high = 17 A. Based on that we can calculate the voltage drop on a resistors. V_r_low = I_low * R = 6667 * 12 = 80 004 and V_r_high = I_high * R = 17 * 12 = 204. Despite providing higher voltage the voltage drop on a resistor is actualy smaller. So in fact you can use the formula P_lost = (V_r)^2/R and you do get the same results.
@@MrTostek Ah, so if we knew the exact voltage drop over the line itself, perhaps by measuring the voltage at the transmission center and at the distribution center, we *could* use the V^2/R. But that would be more difficult to measure, as you would need a shared reference at two points potentially hundreds of miles apart! Whereas you can measure the current anywhere along the line and have pretty much the same value everywhere. So the reason we use I^2/R is because of the relative convenience of measuring I, not because the math "doesn't work" for the voltage. That seems to make sense. Thanks!
But sir a question then comes to my mind that is how we put electric energy in a circuit. The answer is giving some potential difference. Which is actually a Voltage. (Is there any other way also?) You said that you are putting some energy per second. How do you put it? If ohms law is correct then, current should increase with increased voltage over resistance.
The theory was mostly correct. The only thing that I have issue with is that you used 120v on the secondary side of the power company's transformer. In the US, the secondary side of the transformer is always 240v in single phase. The phase only gets broken up at the branch circuit level in a residential wiring configuration. Every Breaker Panel is 240v, Every disconnect, every 240v fixed appliance.
What if to find current I use 48000/12 then my current through wire will be 4000A and in case of 120V will be 10A ??? Then this is entirely opposite to what you told
You need to divide the power emitted by the power plant P[W] with the voltage U[V] to get the current I[A]=P[W]/U[W]. From what I can read out of what you've written. It looks like you have switched P[W] and U[V] up. So you divide the voltage with the power and not power with voltage. Which is the correct way.
If I understand correctly, you can't do it this way. You are using I=V/R to calculate the current through a resistor, which is usually fine, if the voltage is not high enough. However, if the voltage is high enough, the I-V curve of the resistor is no longer linear. The electrons in the resistor has a limit of how fast they can move which I guess is due to scattering. Beyond a certain point, the current will not increase as fast as the voltage does. On the other hand, P=IV is the total power transmitted through the resistor, but not the power dissipated by the resistor: that would be P=I^2R. The difference between these two are made of the power consumption in your houses.
@@tengwenxiang1692 I think the whole concept of resistance is to measure the energy dissipation rate. So for dissipation, P=I^2V is probably more basic than P=IV. It's just that in the linear regime V=IR which allows us to use the second equation.
this might seem like a dumb quisition, but isnt voltage equal to voltage*current? So why is the voltage's resistance different than the wire's resistance?
Dr.E, i have always had problems with voltages and currents because they're basically interchangeable in equation. The explanation you give is cool and all but i have a really big doubt wandering within me. So hear me out. We're talking equations. Power=Current * Voltage---(1) Voltage = Current * Resistance---(2) From equation (1), Power output of a given poweplant is constant . So, voltage is inversely proportional to current which basically means BIG CURRENT =small voltage and BIG VOLTAGE =small current. Now you use (2) to convert (1) into P=I*(I*R) P=RI^2 But but but, here's the deal, i can also convert this equation in terms of voltage. Which means P=(V^2)/R---(3) Here, the two results clash From (3), Power dissapated or lost = (Voltage^2) /Resistance Here if you use high voltage, you LOSE more power More power is dissapated. How do you explain that?
@@PhysicsNinja i already watched it But it still didn't clarify my doubts. Yes, current does go through and voltage is indeed the potential difference between two points, which is being increased and that increases the power dissapation, doesn't it? Could you please give an indepth explaination?
Yeah, I had the same doubt. Especially in terms of equations since everyone says H is proportional to I^2 but only proportional to V, and hence heat loss is lesser with high voltage. But the problem here is every time you increase voltage, current also essentially increases (V=IR) so what's the point? I don't understand why we don't consider both to have the same heat loss
sorry I got it because the total current depends on both (the wire resistance + the load resistance), calculating the current only by using the wire resistance will give a wrong value.
Sometimes it takes that one video to finally understand a concept. This was it for AC current for me! I didn't ever think about the RMS of current, only for voltage. So stupid of me! Thank you so much for this video!!!
But why if i use the other formula V^2/R instead of I^2*R im getting larger numbers in the high voltage scenario, what am i missing. 48000^2/12 is much bigger than 120^2/12.
I made a follow up because the example in the first wasn't the best, hopefully this helps answer your question. ruclips.net/video/C_2dBitJrGM/видео.html
Well it starts at the Power Plant That Makes The High Voltage Lines Then The Lines Go To A “Sub Station” That Makes The Low Voltage Lines That Go To Homes.
If you going to tell it tell it right. Power plant to switch yard to transmission lines to substation to sub- transmission to sub to distribution to service
Yeah numbers were the best choices. I did another case here. Power Transmission: High Voltage vs Low Voltage Comparison ruclips.net/video/C_2dBitJrGM/видео.html
I was shocked once pretty good from the entrie cable coming into the home from the road I was on an 8ft step ladder and was taping around the knuckles where it connects to one another it was just black tape around them instead of plastic housing my hair was wet from sweating an it felt like and sounded like getting hit up side the head with a 2x4 I dont remember the fall or anything just came back to on the ground standing looking up thinking to myself did I just get shocked? I bit 2 holes in my tongue both side from my k9s I still had all my tools in my work pouch an landed on my feet somehow it was really crazy experience
I dont know if I am correct or not, but I think it has something to do with the fact that it is Power being transmitted and not just voltage from the power station. You have to use Power and Voltage to get the current thats being outputted from the power station, instead of just using the voltage and resistance thats found in the wire.
Because the Vrms given is for the power station and the not the wire. If we were given the theoretical voltage dissipated on the wire then we would be able to compute using Ohm's law. [The voltage given was for the power station and the resistance given was for the wires. That is why we had to use Vrms (120v) and the Power from the station(800 kW), both of which come from the same place(device), to compute for the Irms. Which would be the same for the Power station, wire, and transformer (since this is a series configuration). This is also assuming that this is a HVDC (High Voltage Direct Current) transmission system. The computation would get a lot more involved was this an AC transmission system. One thing to also note is for the computation of the turns ratio for the transformer, it also didn't account for the voltage drop between the power station and the transformer. Which may or may not be severe enough to warrant addressing in a real-life situation.
Hi Physics Ninja! I liked the video a lot and I subscribe the channel! Please, can you say me, in high tension powerline, what temperature a overheated cable can reach?
Blah blah blah, because too much energy is lost as heat. Took you 6 minutes of boring math to answer the question. If we want to know the math we will look for a video called "How to do the math..."
What a beautiful explanation as compared to the confusing explanation in textbooks. Excellent explanation.
Thank you!
I have read different forums and watched a few youtubes (in addition to my textbook readings) and the explanations seem to fall short. The issue seems to be how we are first taught about a direct relationship between voltage and current (that is, an increase in voltage renders an increase in current if resistance remains the same) and then we're taught about power lines that have high voltage and low current (because other wise we would need thick wires that carry high current [which would run the risk of overheating due to the joule effect or something or another..). So please don't explain to me the infrastructural reasons why high voltage, low current is necessary for power lines. I just need to know how high voltage, low current is even possible. I've only been studying DC so far so maybe AC has rules that would enlighten me...but I thought the E=IR formula was universal.
That's a really good question. I wanted to keep my video simple to really answer your question you need to look at the transformer from the power plant to the power lines. It's a step down transformer and the designer can decide how much to step down the voltage. It's best to step down a little to keep the voltage high. (ac max)
The problem is that V=IR (i.e. Ohm's Law) is *not* universal. The quick explanation for why high voltage, low current is possible is that because you have to conserve energy from the input to the output of the transformer, the power input must be roughly equal to the power output; and because P=IV, stepping-up the voltage causes the current to step-down (and vice versa). The more complex explanation is that formula (V=IR) really only applies to DC circuits. In AC circuits, there is a more general relationship to be considered V=IZ, where Z is called the impedance of the circuit. Impedance depends upon a variety of factors: the resistance(R), the capacitance(C), the inductance(i), and the frequency(f) of the AC circuit. When you take into account all of these factors with a simple set of transformer coils as shown in the video, all the math works out.
@@SkepticMM so if i touch it before and after the transformer i will feel the same pain if the frequency is the same right ?
In lament terms… if I have you bench pressing weight. Let’s say you can only benchpress 150lbs. Your power only allows you to lift 150lbs. This is power. There is two ways to make the load easier. Take off the weight or juice you up to become stronger to lift easier without resistance. So while you’re lifting 150lbs that is your max amps before you give out, since resistance is too strong. Now while you’re lifting I inject you with Superman DNA and you can lift more weight because you have more power. Now you have more power to lift weight. That 150lbs which is resistance is nothing to you now because you have more power (voltage) so in terms of current, it is much easier for you to lift the weight.
Less amperage with more power makes sense. What wouldn’t make sense is having much less power and using less amps. It’s about power. So the more power I have the less current is needed because simply I have the power to lift all that weight.
Trying to give some thoughts on the actual question the viewers wanted to be answered:
How is it possible to increase the voltage without increasing the current?
Ohm s law states their are directly proportional.
I am still trying to figure it out myself, but here s what i gather from this video, which by the way is very good.
First, you notice he s using the Power formula, not Ohm s law. Why? My guess is because in this situation we HAVE to take Power into consideration, since it is given from the start. Ohm s law doesn t include power.
2) P=U*I
now, since P is a fixed known value, it means that the increase of either the voltage or current comes at the loss of the other. Basic maths.
3) there is a sort of misconception that high voltage means high current. I ve thought that to be true from the 6th to about 12th grade.
Of course, we see it is not always the case. What makes the difference (as far as i can tell) is POWER, specifically whether the value is fixed or to te to be calculated.
After all, voltage is a POTENTIAL, so whether it means or doesn t mean high current, depends on other values.
This is what i got from the video, i don t know whether it s correct or not, feel free to give me your view.
Thanks for the vid!
THIS this is what I'm stuck on
Well as an 11th grader I think you're pretty much right, the power across a transformer has to be constant, and as you said the increase of one would bring about the decrease of the other value.
Can you tell me why he doesn't use the formula of power dissipated = v²/R
Sir please explain why we can't use P=V^2/R to compare this instead of P=I^2*R
THIS! I'm going all over the web trying to find this out. My teacher just says it doesn't work but can't/won't explain *why*!
I was thinking the same and looking for an answer. I think that the V_rms in the P=V^2/R simply doesn't equal the voltage used by the power plant, but instead it is the voltage drop on the resistor V_r, which is equal to V_r=IR. The drop of voltage of a rezistor doesnt equal the voltage provided by the power plant, because the resistor isnt the only power consumer in the grid. The current however is the same as the one provided by the power plant, becouse all the current has to flow through the wire. Thus we can calculate the current easily as shown in the video. I_low = 6667 A and I_high = 17 A. Based on that we can calculate the voltage drop on a resistors. V_r_low = I_low * R = 6667 * 12 = 80 004 and V_r_high = I_high * R = 17 * 12 = 204. Despite providing higher voltage the voltage drop on a resistor is actualy smaller. So in fact you can use the formula P_lost = (V_r)^2/R and you do get the same results.
@@MrTostek Ah, so if we knew the exact voltage drop over the line itself, perhaps by measuring the voltage at the transmission center and at the distribution center, we *could* use the V^2/R. But that would be more difficult to measure, as you would need a shared reference at two points potentially hundreds of miles apart! Whereas you can measure the current anywhere along the line and have pretty much the same value everywhere. So the reason we use I^2/R is because of the relative convenience of measuring I, not because the math "doesn't work" for the voltage. That seems to make sense. Thanks!
Bro we can’t use V^2/R because the V is potential difference not the voltage measured with respect to ground. Hope it helps.
Thanks a lot sir for explaining this concept in easiest way,it really help me a lot in learning.😊
800kw dissipated as heat would completely melt the wire, it's not just that it's inefficient, right?
But sir a question then comes to my mind that is how we put electric energy in a circuit.
The answer is giving some potential difference. Which is actually a Voltage. (Is there any other way also?) You said that you are putting some energy per second. How do you put it?
If ohms law is correct then, current should increase with increased voltage over resistance.
If there was no transformer at the end and of the circuut and it was just wires, where does the rest of the power get used in the high voltage case?
The theory was mostly correct. The only thing that I have issue with is that you used 120v on the secondary side of the power company's transformer. In the US, the secondary side of the transformer is always 240v in single phase. The phase only gets broken up at the branch circuit level in a residential wiring configuration. Every Breaker Panel is 240v, Every disconnect, every 240v fixed appliance.
I think he was only trying to give a surface level concept
Is the current not dictated by the supply voltage divided by the line resistance as in Ohms Law
What if to find current I use 48000/12 then my current through wire will be 4000A and in case of 120V will be 10A ???
Then this is entirely opposite to what you told
You need to divide the power emitted by the power plant P[W] with the voltage U[V] to get the current I[A]=P[W]/U[W]. From what I can read out of what you've written. It looks like you have switched P[W] and U[V] up. So you divide the voltage with the power and not power with voltage. Which is the correct way.
If I understand correctly, you can't do it this way.
You are using I=V/R to calculate the current through a resistor, which is usually fine, if the voltage is not high enough. However, if the voltage is high enough, the I-V curve of the resistor is no longer linear. The electrons in the resistor has a limit of how fast they can move which I guess is due to scattering. Beyond a certain point, the current will not increase as fast as the voltage does.
On the other hand, P=IV is the total power transmitted through the resistor, but not the power dissipated by the resistor: that would be P=I^2R. The difference between these two are made of the power consumption in your houses.
@@frankartanis1290 Hello May I ask that why is power dissipated would be P=I2R?why cant use p=VI to look for power loss
@@tengwenxiang1692 I think the whole concept of resistance is to measure the energy dissipation rate. So for dissipation, P=I^2V is probably more basic than P=IV. It's just that in the linear regime V=IR which allows us to use the second equation.
solid explanation! thank u for helping me visualise this :)
this might seem like a dumb quisition, but isnt voltage equal to voltage*current? So why is the voltage's resistance different than the wire's resistance?
Power is equal to voltage * current. Voltage is resistance* current (ohms law)
Dr.E, i have always had problems with voltages and currents because they're basically interchangeable in equation.
The explanation you give is cool and all but i have a really big doubt wandering within me.
So hear me out. We're talking equations.
Power=Current * Voltage---(1)
Voltage = Current * Resistance---(2)
From equation (1), Power output of a given poweplant is constant . So, voltage is inversely proportional to current which basically means BIG CURRENT =small voltage and BIG VOLTAGE =small current.
Now you use (2) to convert (1) into
P=I*(I*R)
P=RI^2
But but but, here's the deal, i can also convert this equation in terms of voltage.
Which means
P=(V^2)/R---(3)
Here, the two results clash
From (3),
Power dissapated or lost = (Voltage^2) /Resistance
Here if you use high voltage, you LOSE more power
More power is dissapated.
How do you explain that?
I did another video on this that may help. Power Transmission: High Voltage vs Low Voltage Comparison
ruclips.net/video/C_2dBitJrGM/видео.html
Remember that current goes through but V voltage is the difference of potential between 2 points.
@@PhysicsNinja i already watched it
But it still didn't clarify my doubts.
Yes, current does go through and voltage is indeed the potential difference between two points, which is being increased and that increases the power dissapation, doesn't it?
Could you please give an indepth explaination?
Yeah, I had the same doubt. Especially in terms of equations since everyone says H is proportional to I^2 but only proportional to V, and hence heat loss is lesser with high voltage. But the problem here is every time you increase voltage, current also essentially increases (V=IR) so what's the point? I don't understand why we don't consider both to have the same heat loss
Great explanation, thanks for the video.
Hii.... I am in 12th grade..... ur way explanation is fantastic.... Keep going.... Love from India ❤️
Very simple but impressive explanation
Excellent Video
Great explanation. Thank you so much!
Glad it was helpful!
But why can't we use the ohm's law in this case?
Thanks for the amazing explanation BTW
Power Transmission: High Voltage vs Low Voltage Comparison
ruclips.net/video/C_2dBitJrGM/видео.html
See the follow up video
how to oscillate AMP to charge battery fast . will resistance lower at oscillated AMP
why did not divide 120v/12 ohms for the first case which will give a current of 10 A?
sorry I got it because the total current depends on both (the wire resistance + the load resistance), calculating the current only by using the wire resistance will give a wrong value.
Lmfao what are you an AC man 😂😂 simple ass boy
You can't decrease current by increasing voltage . Do a experiment itself with 12V and 220V.
How the energy(power) is put in cicuit?
Sometimes it takes that one video to finally understand a concept. This was it for AC current for me! I didn't ever think about the RMS of current, only for voltage. So stupid of me!
Thank you so much for this video!!!
But why if i use the other formula V^2/R instead of I^2*R im getting larger numbers in the high voltage scenario, what am i missing. 48000^2/12 is much bigger than 120^2/12.
I made a follow up because the example in the first wasn't the best, hopefully this helps answer your question. ruclips.net/video/C_2dBitJrGM/видео.html
What about I = V/R
Need more people so spread this video more
Well it starts at the Power Plant That Makes The High Voltage Lines Then The Lines Go To A “Sub Station” That Makes The Low Voltage Lines That Go To Homes.
The explanation was great . Thanks brother for this helpful video. It helped me the a lot. Keep going. Take 💕💕
If you going to tell it tell it right.
Power plant to switch yard to transmission lines to substation to sub- transmission to sub to distribution to service
That sounds like an engineering explanation. I’m a physicist and just trying to explain some of the basics.
How power dissipation can be more than generated power? At max it can be 800KW transmission loss. There is something wrong with calculation.
Yeah numbers were the best choices. I did another case here. Power Transmission: High Voltage vs Low Voltage Comparison
ruclips.net/video/C_2dBitJrGM/видео.html
Thanks for the answer now i got a clear explanation
I was shocked once pretty good from the entrie cable coming into the home from the road I was on an 8ft step ladder and was taping around the knuckles where it connects to one another it was just black tape around them instead of plastic housing my hair was wet from sweating an it felt like and sounded like getting hit up side the head with a 2x4 I dont remember the fall or anything just came back to on the ground standing looking up thinking to myself did I just get shocked? I bit 2 holes in my tongue both side from my k9s I still had all my tools in my work pouch an landed on my feet somehow it was really crazy experience
Ouch
So when voltage increases the power lost decreases ??
correct
Something doesn’t add up when your calcs shows power dissipated in the wire is greater than the power output of the power plant!
I agree, there a too many assumptions built into this but it was meant just as an illustrating case of High voltage vs Low voltage.
high voltage lines use aluminium.
Why 120volts wher i live we use 240volts AC ?
Marv vlogs Because he is explaining about the US system that uses 120 Volts AC
Vapervop US system is 240 Volts RMS in the secondary coil of the single phase step down transformer.
The best explanation on the entire internet... no kidding
Thanks
Thanks Sant Kumar Hooda
When calculating the current for low voltage why I cannot use the Ohm's Law to get the current of 10 Amps?
I dont know if I am correct or not, but I think it has something to do with the fact that it is Power being transmitted and not just voltage from the power station. You have to use Power and Voltage to get the current thats being outputted from the power station, instead of just using the voltage and resistance thats found in the wire.
Thats current per second....total current transmitted will depend on total power to be transmitted
Because the Vrms given is for the power station and the not the wire. If we were given the theoretical voltage dissipated on the wire then we would be able to compute using Ohm's law. [The voltage given was for the power station and the resistance given was for the wires. That is why we had to use Vrms (120v) and the Power from the station(800 kW), both of which come from the same place(device), to compute for the Irms. Which would be the same for the Power station, wire, and transformer (since this is a series configuration). This is also assuming that this is a HVDC (High Voltage Direct Current) transmission system. The computation would get a lot more involved was this an AC transmission system. One thing to also note is for the computation of the turns ratio for the transformer, it also didn't account for the voltage drop between the power station and the transformer. Which may or may not be severe enough to warrant addressing in a real-life situation.
Thank you
Hi Physics Ninja! I liked the video a lot and I subscribe the channel!
Please, can you say me, in high tension powerline, what temperature a overheated cable can reach?
I think that depends on the material in the cable!
Low voice boss
thanks
Why don't we just make everything run on high voltage? It seems to be much more efficent.
If Homes Go On High Voltage They Will Explode Because Of All The Volts
ruclips.net/video/QVB2Yfott74/видео.html
See this..
❤
A wholesome 👍
Blah blah blah, because too much energy is lost as heat. Took you 6 minutes of boring math to answer the question. If we want to know the math we will look for a video called "How to do the math..."