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?

I would use the word load instead of impedance

In order to ensure that my primary source keeps producing 1000V at 100A

We would never do that because V is regulated. i.e. If the required current goes up, the generator would output more current.
A transformer tap changer is another way to regulate current coarsely. The consumer may choose to regulate power going into a heater (i.e. 0 to 2000 W), but the utility would not. Voltage is the right parameter to regulate in this "simple" system.

An inverter based solar system may decide that locally (substation) regulating power factor is more important. The end user may sometimes include a power factor controller in the buildings electrical system design. We have been simplifying the AC distribution system to make it look like DC where voltages can be stepped up/down with transformers. It's OK to do that to gain some understanding. Be aware, that the rules will change. It would be wise to ignore this paragraph for now.

 

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.

This part still confuses me. To me, it suggests that if a transformer is designed for 100kW, it will supply 100kW regardless of the circuit it is being used in.

Suppose we are supplying 1000V and 100 A to the primary of a step-up transformer with the secondary open-circuited. Then we connect a 10 ohm load to the secondary. How can we be sure that after this change, the transformer will still be producing 100kW of power? In the example you gave, the addition of line resistance reduced the total power to 99.99kW as well!

Everything makes sense to me as long as I start with the assumption that the primary is fixed at 100kW. But that is not enough. I want to know HOW it is to be ensured that the primary keep producing 100kW, regardless of what's happening on the other side of the transformer. Only then can we be sure that the current is going down on the secondary, and that it's not going up on both sides such that the secondary current is still, of course, less than the primary by a constant factor. The latter case would result in a positive change in power, and that is not what we want.

I think the root cause of my confusion is that I'm unable to wrap my head around the notion of assuming fixed power.

 

The notion of "fixed power" is wrong. Power in = power out is correct. The loads change and the generators adjust their output.
Voltage is the constant. Power, current and the loads are constantly changing.

A 16 MW generator does not always produce 16 MW. It's the MAXIMUM it can produce. A 2000 W inverter does not ALLWAYS produce 2000 Watts and a 100 kW transformer does not always operate at 100 kW.

EDIT: Fixed typo.

 

The notion of "fixed power" is wrong. Power in = power out is correct. The loads change and the generators adjust their output.
Voltage is the constant. Power, current and the loads are constantly changing.

A 16 MW generator does not always produce 16 MW. It's the MAXIMUM it can produce. A 2000 W inverter does not ALLWAYS produce 2000 Watts and a 100 kW transformer does not always operate at 1000 kW.

This is music to my ears. Thanks for putting it in a comprehensible way, where I failed.

Now, if you guys remember when I said that it confuses me when people talk about 100kV @ 1A or 10kV @ 10A or 1kV @ 100A, the variability of current and thus power is what I meant to emphasize. While I do understand that they all would theoretically equal 100kW of power, but if my objective is reducing the current so my power losses in the transmission line can be reduced, I must make some arrangements to make sure that stepping-up voltage to 100kV doesn't result in an increase in my desired power level, and the only way I know of making that happen is if I introduce series resistances such that they can restrict my current to 1A.
This is equivalent to saying that my primary current is restricted to 100A. The one implies the other. Up to this point, I understand.

When I asked for an alternative method of determining current i.e some way other than Pin=Pout, I wanted to see if the 10 ohms line resistance connecting the secondary of one transformer to the primary of another is sufficient to produce a current of 1 A @ 100kV and won't be tampering with my desired power level.

Starting by assuming power to be pinned down to 100kW isn't helping me. And I can't figure out another way by myself.

You guys have already helped me a lot. I am extremely grateful. I just need a little more help. Please don't get frustrated.

 

OK, try this:

We have a 100V generator (distance is 50 feet)
Then a transformer that kicks it up to 10,000 V
(distance is 100 feet)
Then another transformer kicks the 10,000 V to 100,000 volts
(distance is 100 miles)
Then another transformer drops the 100,000 to 10,000 V
(distance is 500 feet)
Lots of little transformers drop the voltage to 100V for delivery.

We started with a low voltage because it's safer and coincidentally, it's our delivery voltage.
We assign a power to our generator and we calculate the wire sizes so that there is no more than a +-10% drop in voltage. (Utility specs)

In principle, we would "juggle" the transmission line voltages and use "standard values" and pick wires of "standard sizes" to achive the goals.

Grid input
At the power stations, the power is produced at a relatively low voltage between about 2.3 kV and 30 kV, depending on the size of the unit. The generator terminal voltage is then stepped up by the power station transformer to a higher voltage (115 kV to 765 kV AC, varying by the transmission system and by country) for transmission over long distances.

Losses
Transmitting electricity at high voltage reduces the fraction of energy lost to resistance, which varies depending on the specific conductors, the current flowing, and the length of the transmission line. For example, a 100 mile 765 kV line carrying 1000 MW of power can have losses of 1.1% to 0.5%. A 345 kV line carrying the same load across the same distance has losses of 4.2%.[9] For a given amount of power, a higher voltage reduces the current and thus the resistive losses in the conductor. For example, raising the voltage by a factor of 10 reduces the current by a corresponding factor of 10 and therefore the I2R losses by a factor of 100, provided the same sized conductors are used in both cases. Even if the conductor size (cross-sectional area) is reduced 10-fold to match the lower current, the I2R losses are still reduced 10-fold. Long-distance transmission is typically done with overhead lines at voltages of 115 to 1,200 kV. At extremely high voltages, more than 2,000 kV exists between conductor and ground, corona discharge losses are so large that they can offset the lower resistive losses in the line conductors. Measures to reduce corona losses include conductors having larger diameters; often hollow to save weight,[10] or bundles of two or more conductors.

Wikipedia: https://en.wikipedia.org/wiki/Electric_power_transmission

 

This part still confuses me. To me, it suggests that if a transformer is designed for 100kW, it will supply 100kW regardless of the circuit it is being used in.

Does this make any sense? If a scale is designed for 400 lb does that mean that it will register 400 lb regardless of how heavy the person is that steps on it? Or does it mean that it will respond with the weight of whoever is on it at the moment but that if anyone heavier than 400 lb steps on it that it might fail in some way? If a resistor is rated for 10W, does this mean that it will always dissipate 10W of heat regardless of the circuit it is being used in? Or does it mean it will dissipate whatever power the circuit results being delivered to that resistor and if that exceeds 10W then bad things might happen?

Does it make sense from a physical reality standpoint. If you power this transformer from an inverter running from your car that is capable of delivering 1kW of power, just how is it that this transformer is going to manage to deliver 100kW of power to any load?

Suppose we are supplying 1000V and 100 A to the primary of a step-up transformer with the secondary open-circuited.

You keep insisting that you can choose both how much voltage and how much current you supply to a device and we keep telling you that you are wrong!

You get to choose one, the device chooses the other in response. If you apply 1000 V to the primary of a transformer that has an open-circuited secondary, you will draw very little current (no current in an ideal transformer) because the transformer reflects the infinite load impedance on the secondary so that the circuit driving the primary sees an infinite impedance multiplied by the square of the turns ratio.

Everything makes sense to me as long as I start with the assumption that the primary is fixed at 100kW.

Then you clearly don't understand much of anything at all about this because that assumption is patently wrong.

Only then can we be sure that the current is going down on the secondary, and that it's not going up on both sides such that the secondary current is still, of course, less than the primary by a constant factor.

But, as we keep telling you, it IS scaled by a constant factor and, in the case of a step-up transformer, it IS a constant factor less than the primary. That's what a transformer DOES. It scales the voltage up by the turns ratio and the current down by the turns ratio. These ratio applies to the current in the input RIGHT NOW and the current in the output RIGHT NOW. You keep insisting that it applies to the input current last month when the transformer was installed in a oil drilling rig and the current right now while it is sitting on the shelf at the surplus transformer store.

 

Does this make any sense? If a scale is designed for 400 lb does that mean that it will register 400 lb regardless of how heavy the person is that steps on it? Or does it mean that it will respond with the weight of whoever is on it at the moment but that if anyone heavier than 400 lb steps on it that it might fail in some way? If a resistor is rated for 10W, does this mean that it will always dissipate 10W of heat regardless of the circuit it is being used in? Or does it mean it will dissipate whatever power the circuit results being delivered to that resistor and if that exceeds 10W then bad things might happen?

Does it make sense from a physical reality standpoint. If you power this transformer from an inverter running from your car that is capable of delivering 1kW of power, just how is it that this transformer is going to manage to deliver 100kW of power to any load?

Of course it makes no sense. That's why I said it is "confusing me" because that's not how I know things to work but I was misinterpreting your post and thought that you were implying exactly that. There, I was mistaken.

You keep insisting that you can choose both how much voltage and how much current you supply to a device and we keep telling you that you are wrong!

You get to choose one, the device chooses the other in response.

Yes that's exactly the mess I had created for myself without even realizing it. I'm looking for an explanation where current isn't directly determined by us and at the same time making the assumption that we have 1000V @ 100A. I'm embarrassed.

P.S: You never explicitly told me I was wrong for insisting on that.

These ratio applies to the current in the input RIGHT NOW and the current in the output RIGHT NOW. You keep insisting that it applies to the input current last month when the transformer was installed in a oil drilling rig and the current right now while it is sitting on the shelf at the surplus transformer store.

Actually, I've been trying to say the same thing. That when you have power at 100kW at one instant, you can't be sure it will stay that way after you make changes to either the primary or the secondary side. However, I was contradicting my own claim by always insisting on the numbers 1000V and 100A. I only see that now.

Now if you have any patience at all left for me (I know I wouldn't if I were in your place), will it be more appropriate if we take an example where we only know the transmission line resistance, and the total power that we need to deliver (let's take 100kW again ). Our objective is minimizing the losses in the line.
What should the primary and secondary circuits (the voltages and resistance) be like to meet this goal?

I'm grateful to you more than I can express.

 

Now if you have any patience at all left for me (I know I wouldn't if I were in your place), will it be more appropriate if we take an example where we only know the transmission line resistance, and the total power that we need to deliver (let's take 100kW again ). Our objective is minimizing the losses in the line.
What should the primary and secondary circuits (the voltages and resistance) be like to meet this goal?

Fill in the blanks in my post #65.

the think is we could use 100 kW and call that the maximum capacity of the plant)
We could say we want no more than a 10% loss at delivery.
Now you would have to select wire sizes to satisfy that.

I had about 4 segments, so each can drop 10/4% in power.
You can then find the max R and then the wire size that satisfies that.

 

Fill in the blanks in my post #65.

the think is we could use 100 kW and call that the maximum capacity of the plant)
We could say we want no more than a 10% loss at delivery.
Now you would have to select wire sizes to satisfy that.

I had about 4 segments, so each can drop 10/4% in power.
You can then find the max R and then the wire size that satisfies that.

Thanks a ton KeepItSimpleStupid!

 

However, I was contradicting my own claim by always insisting on the numbers 1000V and 100A. I only see that now.

Sometimes we have to beat our head against a false wall repeatedly before the tidbit that is keeping us from seeing things the right way gets broken loose.

Now if you have any patience at all left for me (I know I wouldn't if I were in your place), will it be more appropriate if we take an example where we only know the transmission line resistance, and the total power that we need to deliver (let's take 100kW again ). Our objective is minimizing the losses in the line.
What should the primary and secondary circuits (the voltages and resistance) be like to meet this goal?

The approach you take depends on what you have control over and what you have to live with. But in general you start with the load end (i.e., the goal you are trying to achieve) and work backwards. Post #55 give you a skeleton for doing that.

You can pretty much never "minimize" the losses in the line because no matter what you do you can always, at least on paper, do something to make them even smaller. The goal is to reduce them to an acceptable level.

Let's work a simple example (that would likely never apply in practice, so take it as a paper example to get practice with the concepts).

You want to deliver 100 kW to a 10Ω load losing no more than 10% of the generated power in the 10Ω transmission line. The generator outputs 100V and can deliver as much as 2 kA. You have control over the transmission line voltage and the transformer ratios at each end. The transmission line is only used by this load (so no current is flowing through it to other loads).

We've already established that we need 1000 V and 100 A at the load. The 10% spec means that we can dump no more than 10 kW in the 10Ω transmission line, which means that the current in it can't be any more than 31.62 A. That means we need a step down transformer at the load with a turns ratio of (100 A)/(31.62 A) which is 3.162:1. That would give us a transmission line voltage, at the load, of 3162 V. At the source end the voltage needs to be 316 V higher in order to allow for the voltage drop across the transmission line. That means it needs to be 3478 V. So we need a step up transformer with a turns ratio of (3478 V)/(100 V) which is 34.78:1. The current drawn from the generator would be 31.62A*34.78 = 1100 A. From this is clear that the power supplied by the generator is 10% more than the power consumed by the load, thus meeting spec.

Note that I've been giving the ratios as something:1 and spec'ing whether it is a step-up or step-down (voltage) transformer.

 

Sometimes we have to beat our head against a false wall repeatedly before the tidbit that is keeping us from seeing things the right way gets broken loose.



The approach you take depends on what you have control over and what you have to live with. But in general you start with the load end (i.e., the goal you are trying to achieve) and work backwards. Post #55 give you a skeleton for doing that.

You can pretty much never "minimize" the losses in the line because no matter what you do you can always, at least on paper, do something to make them even smaller. The goal is to reduce them to an acceptable level.

Let's work a simple example (that would likely never apply in practice, so take it as a paper example to get practice with the concepts).

You want to deliver 100 kW to a 10Ω load losing no more than 10% of the generated power in the 10Ω transmission line. The generator outputs 100V and can deliver as much as 2 kA. You have control over the transmission line voltage and the transformer ratios at each end. The transmission line is only used by this load (so no current is flowing through it to other loads).

We've already established that we need 1000 V and 100 A at the load. The 10% spec means that we can dump no more than 10 kW in the 10Ω transmission line, which means that the current in it can't be any more than 31.62 A. That means we need a step down transformer at the load with a turns ratio of (100 A)/(31.62 A) which is 3.162:1. That would give us a transmission line voltage, at the load, of 3162 V. At the source end the voltage needs to be 316 V higher in order to allow for the voltage drop across the transmission line. That means it needs to be 3478 V. So we need a step up transformer with a turns ratio of (3478 V)/(100 V) which is 34.78:1. The current drawn from the generator would be 31.62A*34.78 = 1100 A. From this is clear that the power supplied by the generator is 10% more than the power consumed by the load, thus meeting spec.

Note that I've been giving the ratios as something:1 and spec'ing whether it is a step-up or step-down (voltage) transformer.

Fantastic. Thank you so very much, WBahn! I truly appreciate it!

 

You are still trying to take yesterday's measurements and apply them to today's circuit!

When you apply 10V to the input you will get 100V at the output.

Stop.

That's all you know.

The fact that your load drew 1A yesterday when it had 10V across it is irrelevant to what the transformer is going to draw today when the load has 100V across it!

Today, with 100V across it, it is going to draw 10A and the transformer is going to draw 100A.

If you want to still draw just 1A from the source then you need to increase the resistance of the load so that it only draws 0.1A from the 100V output of the transformer. So you need to replace the load with a 1000Ω resistor. If you do that, then the transformer will draw 10W (10V·1A) from the source and deliver 10W (0.1A·100V) to the load.

What about the voltage drop in your output resistor ? you won't get the 100V on the load but 100V-Vr

 

What about the voltage drop in your output resistor ? you won't get the 100V on the load but 100V-Vr

Another newie necro post.
This thread is over 8 years old, so you are a little late to the party.

But welcome to All About Circuits.
I think you'll find it very interesting if you are into electronics.

 

What about the voltage drop in your output resistor ? you won't get the 100V on the load but 100V-Vr

You are waking up a thread that died a natural death nearly a decade ago. But at least you are continuing the discussion and not hijacking it for some other purpose.

By "Vr" are you talking about the voltage across the load? IIRC, the load in this discussion was connected directly to the output of the transformer, so Vr = 100 V. In the real world you will get some droop because of the resistance of the windings in the transformer, but that is often neglected in calculations that aren't pushing the transformers design very hard.

 

Another newie necro post.
This thread is over 8 years old, so you are a little late to the party.

But welcome to All About Circuits.
I think you'll find it very interesting if you are into electronics.

HAHA Thank you !