I think I get it now. Thanks to all of you I finally can imagine how transformer works. Although I need to study some examples to be sure that I truly know how it works, I feel that i just understand it. If I have a problem, I'll ask you.

Your community is just great, you're really helpful.

 

Hi,

First thing I wanted to say is I searched the forum for similiar topics and I found some, but none of them made me understand the issue.

So here's what I can't understand:

We use step-up transformer in order to increase voltage on a power line - which voltage do we increase, phase or line voltage?

Increasing voltage gives us smaller current - how? The resistance of the power grid stays the same, so how can we decrease current by increasing voltage? Increased voltage makes the electrons go faster and that makes the greater current. I know, that P=R * I^2, but I don't understand why the power stays the same before stepping-up voltage and after that.

It's like a situation with a 60W bulb. If I apply 20V to it, some amount of current will flow, but the amount of that current is I = U * R, where R is the resistance of that bulb. If I increase the voltage to 40V there will be 2 times the current there was before (because 2*U = 2*I*R). That 60W mark on a bulb shows us only the maximum power it can handle, the power won't always be equal to this 60W, it depends what voltage we apply to it. It might be just 30W or 40W if we apply low voltage.

When I hear about that higher voltage with lower current thing, I think about that bulb described a while ago. If higher voltage gives lower current in power grid, why doesn't it apply to a bulb? Why in the power grid higher voltage gives us smaller current and in the circuit with a buld higher voltage gives us higher current?

Please help me with this issue, because it holds me back and makes it impossible for me to understand more complex things about power transformation.

Greetings:
Stepping up the voltage does not AUTOMATICALLY reduce the current. It allows you to reduce the current while maintaining the same power level.

 

Hi everyone,

I'm also a little confused by the same question that Loreno asked: How does stepping up voltage decrease the current through the same load? I understand that the output current will always be smaller than the input current in a step-up transformer, but looking at the output side only how can power losses be reduced by increasing the voltage on the output side? For example:

My load is 9 ohms and transmission line resistance 1 ohm (the transmission line and load form a series circuit?). I want to transmit 100kW of power to the load.

The only way that I understand this can work is if we apply 1000V to get I=1000/10 = 100A to obtain P = 100*1000 = 100kW.

Now power losses would indeed be lesser if we could deliver 100kW at 100kV and 1A. So let's try using a step-up transformer to step up 1000V to 100kV.

At 100kV, the same load would draw I = 100kV/10 = 10000 A to give P = 100kV * 10000A = 1000 MW! This selection of voltage and current values is producing another power level altogether!

So it seems to me that it's not up to us to decide any suitable value of current that we please (i.e 1A in this case). Because for a fixed load, two different values of voltage and current can not produce the same power.

I know I'm missing something, but I can't figure it out.

 

looking at the output side only how can power losses be reduced by increasing the voltage on the output side?

You can't just look at the output side. Assming an ideal (lossless) transformer, the power available from the transformer output equals the power supplied to the input.

 

You can't just look at the output side. Assming an ideal (lossless) transformer, the power available from the transformer output equals the power supplied to the input.

What I meant to say is that it's the secondary circuit of the transformer that is connected to the load and we are concerned with the power delivered to the load. So increasing voltages there must result in an increase in current also. The fact that this new power is still going to be equal to that of the primary was never in question.

 

My load is 9 ohms and transmission line resistance 1 ohm (the transmission line and load form a series circuit?). I want to transmit 100kW of power to the load.

The only way that I understand this can work is if we apply 1000V to get I=1000/10 = 100A to obtain P = 100*1000 = 100kW.

First, that won't deliver 100 kW to the load. The load is 9Ω, not 10Ω. You will have 100 A flowing through 9Ω yielding 90kW to the load (the remaining 10kW being lost in the transmission line).

Now power losses would indeed be lesser if we could deliver 100kW at 100kV and 1A. So let's try using a step-up transformer to step up 1000V to 100kV.

You can certainly do this, but you then need to step it back down at before the load since you need 1000 V at the load. So step it up and see what your line losses are and how much power reaches the load.

 

I think we're missing something and that's a discussion of the WIRE. The two way resistance is R = pL/A, The p is the Greek letter Rho and it's the resistivity in units of ohm-length, L is the length. For 2 conductors going to a load, we have to multiply L by 2. A is the cross sectional area of the wire. This means, the larger the diameter, the less drop.

For power transmission we set some rules. I'll set two:
1) We can't load the wire more than 80% of it's rated capacity. e.g. 14 AWG 15 A
2) The voltage dropped across the wire can't drop more than 3%

When we do that, we can select the size of the conductor.

The "power company" wants to minimize it's losses, so it choses with distance and load what voltage to use at the primary. Residential primaries might be around 10 kV and steeped down to various voltages like 240 and 120 VAC. Lower currents also translate to less weight of copper.

For power transformers:
Vpri*Ipr =Vsec*Isec; Power = vI, this we see that primary power and secondary power are equal. There are some losses.

 

First, that won't deliver 100 kW to the load. The load is 9Ω, not 10Ω. You will have 100 A flowing through 9Ω yielding 90kW to the load (the remaining 10kW being lost in the transmission line).

Ah, true. Totally missed that.

You can certainly do this, but you then need to step it back down at before the load since you need 1000 V at the load. So step it up and see what your line losses are and how much power reaches the load.

I don't need 1000V at the load. I just need to deliver 100kW to the load while keeping current at a minimum. The minimum part is what I don't understand. From what I know of basic circuit theory, more voltage = more current for a fixed load. How we are stepping-up voltage 100 times and reducing current by the same factor, without any changes made to the load resistance, is more than I can understand.

 

If your load resistance is fixed at 9Ω, then to get 100 kW into it you need 949 V across it. Period.

Where is it that you are getting this notion that stepping up the voltage is going to reduce the current IN THE LOAD? As has been stated over and over and over, when you step up the voltage from input to output of a transformer, the current INTO and OUT OF the TRANSFORMER is what is reduced by that same factor. Furthermore, that reduction applies to the current into and out of the transformer; it is NOT a comparison between the current now and the current when there was no transformer.

 

If your load resistance is fixed at 9Ω, then to get 100 kW into it you need 949 V across it. Period.

Where is it that you are getting this notion that stepping up the voltage is going to reduce the current IN THE LOAD? As has been stated over and over and over, when you step up the voltage from input to output of a transformer, the current INTO and OUT OF the TRANSFORMER is what is reduced by that same factor. Furthermore, that reduction applies to the current into and out of the transformer; it is NOT a comparison between the current now and the current when there was no transformer.

Okay. I know I've been making a lot of mistakes. Thanks for clearing those. But I still don't understand how power distribution at higher voltages implies lower current through transmission lines. Maybe that's partly because I don't really know where the transmission line is placed in the system. Is it between two transformers, connected to the secondary of one and to the primary of the other?

The other thing I've been unable to grasp so far is whether it is possible to transmit power at whatever value of current we please? Because the way I understand, we produce some voltage and apply it across some resistance and leave the determination of current to Ohm's law. It confuses me when people talk about transmitting 100kW at 100kV and 1A or 10kV and 10A or 1kV and 100A. How can you decide the value of current by yourself regardless of the resistance?

P.S: I'm sure you must be beating your head against the wall WBahn. I want to thank you and apologize for my slowness.

 

I suggest you read:
Electric Power Transmission
Directly from the WIKI and pay attention to and note the illustrations, especially those illustrating exactly what you are asking. Then after reading the material return with any questions specific to the article. Pay attention to where things like efficiency and cost factors are discussed. All of which have already been explained in this thread. Think about that as you ask:

It confuses me when people talk about transmitting 100kW at 100kV and 1A or 10kV and 10A or 1kV and 100A.

From a cost point which would you rather do? Why?

Ron

 

Thinking about voltage and current is confusing you, no?

Remember this. POWER transmitted will remain the same(in an ideal lossless system)

Power is most often discussed in terms of WATTS.

If you put 10kilowatts of power into the system, you can only get 10kilowatts out. Period.
Volts times Amps will equal watts.
1000 volts and 10 amps will be 10 kilowatts(10,000)
You can use a transformer to change the voltage and amps but you will NEVER get more than 10 kilowatts
If you raise the voltage to 100,000V then you will only have .1 amps in the system. (10 kilowatts)
If you lower the voltage to 10 V then you will have 1000 amps in the system. (10 kilowatts)

Losses in transmission lines can be approximated by knowing the current in the wire and resistance of the wire.
current(squared) times resistance will give you the power loss in transmission which will appear as heat in the system.
These losses also occur in the resistances of the copper wire in the transformers too.

 

Transmission lines are everywhere and at various voltages. In a typical residential neighborhood there might be a ~10 Kv transmission line. The top wire on the pole usually uninsulated.

Every 4 or so houses has a transformer connected to that transmission line, The house could have a 200 A service. That means the house could draw 48 KW. Round numbers 4 houses 100 kW.
We have: 240*200*4=10,000*I; So we CAN deliver 100 kW using only 19 Amps at 10,000 Volts. The 4 is for 4 houses.

We can have multiple step-down transformers (10,000 V : 240 V) serving 4 houses each from the same 10 kV transmission line.

 

Transmission lines are everywhere and at various voltages. In a typical residential neighborhood there might be a ~10 Kv transmission line. The top wire on the pole usually uninsulated.

Every 4 or so houses has a transformer connected to that transmission line, The house could have a 200 A service. That means the house could draw 48 KW. Round numbers 4 houses 100 kW.
We have: 240*200*4=10,000*I; So we CAN deliver 100 kW using only 19 Amps at 10,000 Volts. The 4 is for 4 houses.

We can have multiple step-down transformers (10,000 V : 240 V) serving 4 houses each from the same 10 kV transmission line.

Not to mention the savings in fuses!

Ron

 

Okay. I know I've been making a lot of mistakes. Thanks for clearing those. But I still don't understand how power distribution at higher voltages implies lower current through transmission lines. Maybe that's partly because I don't really know where the transmission line is placed in the system. Is it between two transformers, connected to the secondary of one and to the primary of the other?

The other thing I've been unable to grasp so far is whether it is possible to transmit power at whatever value of current we please? Because the way I understand, we produce some voltage and apply it across some resistance and leave the determination of current to Ohm's law. It confuses me when people talk about transmitting 100kW at 100kV and 1A or 10kV and 10A or 1kV and 100A. How can you decide the value of current by yourself regardless of the resistance?

P.S: I'm sure you must be beating your head against the wall WBahn. I want to thank you and apologize for my slowness.

Do a search and read about power transmission.

It's been stated a few time, including very recently by me, that you have a transformer at both ends.

Let's use your example, but change the values to make things a bit easier to use.

I have a 10Ω load that I want to get 100kW of power into, meaning that I need to supply 1000V and 100A to it. So let's assume that I go get a 1000V power supply (which will assume is sinusoidal at a reasonable frequency so that we can use transformers without any problems) capable of delivering that current. Now I connect that supply to my load but it is connected through a transmission line that also has a total of 10Ω of resistance. What will the power be that gets to the load?

Well, I have a total of 20Ω of resistance so the supply will only output 50A of current and the voltage that makes it to the load will only be 500V, so the power delivered to it will only be 25 kW with another 25 kW of power wasted in the transmission line.

Let's see if we can do better without changing the supply

Now let's put a 100:1 step-up transformer at the source and a 100:1 step-down transformer at the load. We could do the full-blown analysis, but what we will see is that we have rendered the effect of that 10Ω transmission line all but negligible. So let's start with that and then work backwards to see it's effect.

Ignoring the transmission line resistance, we would have the following:

At the source: 1000 V and 100 A. (100 kW)
In the (lossless) transmission line: 100 kV and 1 A (0 kW)
At the load: 1000 V and 100 A (100 kW)

This shows how much we have rendered the 10Ω transmission line meaningless. With just 1 A flowing in it we only drop 10 V and that 10 V is not out of 1000 V, but out of 100kV. So we have this is what we are looking at on our second cut:

At the source: 1000 V and 99.99 A (99.99 kW)
At the source end of the transmission line: 100 kV
Current in transmission line: 0.9999 A (0.01 kW)
At the load end of the transmission line: 99.99 kV
At the load: 999.9 V and 99.99 A (99.98 kW)

Now you can see that we are delivering 99.99% of the power to the load.

You can also see that the current and voltage ratios for each transformer are as expected, namely that if one goes up by a factor of 100, the other goes down by that same factor of 100.

 

Do a search and read about power transmission.

It's been stated a few time, including very recently by me, that you have a transformer at both ends.

Let's use your example, but change the values to make things a bit easier to use.

I have a 10Ω load that I want to get 100kW of power into, meaning that I need to supply 1000V and 100A to it. So let's assume that I go get a 1000V power supply (which will assume is sinusoidal at a reasonable frequency so that we can use transformers without any problems) capable of delivering that current. Now I connect that supply to my load but it is connected through a transmission line that also has a total of 10Ω of resistance. What will the power be that gets to the load?

Well, I have a total of 20Ω of resistance so the supply will only output 50A of current and the voltage that makes it to the load will only be 500V, so the power delivered to it will only be 25 kW with another 25 kW of power wasted in the transmission line.

Let's see if we can do better without changing the supply

Now let's put a 100:1 step-up transformer at the source and a 100:1 step-down transformer at the load. We could do the full-blown analysis, but what we will see is that we have rendered the effect of that 10Ω transmission line all but negligible. So let's start with that and then work backwards to see it's effect.

Ignoring the transmission line resistance, we would have the following:

At the source: 1000 V and 100 A. (100 kW)
In the (lossless) transmission line: 100 kV and 1 A (0 kW)
At the load: 1000 V and 100 A (100 kW)

This shows how much we have rendered the 10Ω transmission line meaningless. With just 1 A flowing in it we only drop 10 V and that 10 V is not out of 1000 V, but out of 100kV. So we have this is what we are looking at on our second cut:

At the source: 1000 V and 99.99 A (99.99 kW)
At the source end of the transmission line: 100 kV
Current in transmission line: 0.9999 A (0.01 kW)
At the load end of the transmission line: 99.99 kV
At the load: 999.9 V and 99.99 A (99.98 kW)

Now you can see that we are delivering 99.99% of the power to the load.

You can also see that the current and voltage ratios for each transformer are as expected, namely that if one goes up by a factor of 100, the other goes down by that same factor of 100.

Thanks again for so detailed a response! This post has been quite illuminating. (as have most of your others).

Still I would like to ask, is there any way to find the current flowing in the transmission line without using Pout = Pin? I assume we would have to know the primary coil inductance of the step-down transformer for that?

 

is there any way to find the current flowing in the transmission line without using Pout = Pin?

Measure it. Amp meters exist. If that is inconvenient, you can use a "current transformer" to measure a fixed percentage of the current in a power line.

 

Measure it. Amp meters exist. If that is inconvenient, you can use a "current transformer" to measure a fixed percentage of the current in a power line.

I asked that because I want to see what's happening in the primary side of the step-down transformer in terms of Ohm's law. Assuming 100kV and 1 A and transmission line resistance 10 Ohms, the primary coil impedance must be 99.99k ohms right?

In order to ensure that my primary source keeps producing 1000V at 100A, we would have to use transformers that offer just the right amount of resistance so as to not change the power from our desired value. For example if the primary coil impedance in the example above is lower than 99.99k - say at 800k ohms - then both my primary and secondary currents would go up a little and the power would shift from 100kW to a somewhat larger value.

Is my reasoning correct?

 

The primary of the step-down transformer in my example had 99.99 kV across it (remember the 10 V lost in the transmission line) and 0.9999 A of current. So it's impedance looks like 100 kΩ. This is not a coincidence. A transformer is also known as an impedance transformer. The impedance on one side appears at the other side to be multiplied by the square of the turns ratio. The turns ratio in this case is 100, so the square is 10,000 and that multiplies the 10Ω to yield 100 kΩ. The same thing is happening at the first transformer except in the opposite direction. So the 10Ω transmission line looks like 1mΩ and that 100 kΩ load looks like 10Ω. So instead of seeing 20Ω, the power supply sees 10.001 Ω.

Notice that this has nothing to do with the physical impedance of the transformer windings. It is how the transformer "reflects" impedances from one side to the other, which is according to the square of the turns ratio. This is for an ideal transformer, which has zero winding resistance and infinite inductance. Obviously we can't make such critters. But we don't have to go too far in order to make real transformers that are close enough to ideal for most purposes.

 

Impedance of transformers is another field of study...or can of worms. The impedance of a transformer doesn't stay the same, it's a reflection of the load. If your load changed to use half as much current, the transformer impedance would seem to double. As long as you have been talking about steady amounts of power, you can imagine that your transformers are going to operate where your simple model expects them. When you study frequency changes, transformers get even more complex in their behavior.