# Why the current goes down when we step-up voltage in the power grid

Discussion in 'General Electronics Chat' started by Loreno, Mar 17, 2015.

1. ### Loreno Thread Starter New Member

Nov 10, 2014
14
0
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.

2. ### MikeML AAC Fanatic!

Oct 2, 2009
5,450
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Suppose you have two 100W light bulbs.

One requires 10V at 10A (P=IE = 10*10 = 100W)
Other requires 100V at 1A (P=IE = 100*1 = 100W)

You are tasked to build a house using either of the above lights. You are paying for the copper wire to wire the house. Which type of lights would you rather use?

Hint, what wire gauge do you need to buy for each choice of bulbs.

Now extend this argument to a 100hp motor (1hp = 746W)
Would you rather run a 100hp motor on 10V? 100V? 480V?

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3. ### ErnieM AAC Fanatic!

Apr 24, 2011
7,338
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What is this mysterious "I = U * R" relationship you offer? Current is inversely proportional to resistance; make one go up the other must go down.

Power thru a transformer is simply power in equals power out plus power lost, or Pin = Pout + Ploss. If you approximate losses as zero (close to reality anyway) you get:

Vin * Iin = Vout * Iout

So yeah, if Vout = 2Vin then Iin = 2Iout.

If you change Iout you also change Iin.

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4. ### alfacliff Well-Known Member

Dec 13, 2013
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the output power of a transformer must stay the same as the input power (less efficiency loss) so to raist the voltage 10 times, the current must go down to one tenth holding p =ExI. turns ration raises or lowers voltage, curent must change to keep power the same.

5. ### Loreno Thread Starter New Member

Nov 10, 2014
14
0
@ErnieM Of course, I made a mistake with that "I = U * R" equation. It's wrong.

I've got to ask a question: Does it mean that the first bulb has R1=1ohm and the second has R2=100ohm? Or maybe that's me who has to connect the wires of that resistance?

For example the second bulb would need a thinner wire than the first one, because the thinner the wire, the bigger the resistance. So for the second bulb I would need to connect it to a 100V source using 100ohm wire, right (I made an assumption that a bulb has 0 ohm)?

Does it mean that the high voltage fragment of power grid (the one that is applied to the higher voltage side of the transformer) has thicker wires than the fragment with lower voltage (voltage that gets stepped-up by transformer)? If it was true, if the thickness of those wires was different with different voltages, it would make sense.

6. ### Kermit2 AAC Fanatic!

Feb 5, 2010
3,672
892
power Must stay the same. ( ideal case where we ignore small losses that create heat)

power = voltage X current

This a very simple algebraic formula.

If power in and power out must remain equal, what happens to current as voltage increases?

7. ### WBahn Moderator

Mar 31, 2012
17,461
4,701
You are confusing a few issues.

If you have a fixed load resistance, say R=10Ω, then if you apply 100V to it you will get 10A through it and have 1000W of power dissipated in it. If you double the voltage to 200V you will get 20A through it and have a power dissipation of 4000W. This much I think you understand.

Now let's power this thing from a 100V AC (rms) supply through a 1:1 isolation transformer. We would expect nothing to change and we would draw 10A from the source and deliver 10A to the load and dissipate 1000W.

But now let's use a 1:2 step up transformer so that we get 200V AC out. When we applied that to the load we would get 20A through the load but the input current to the transformer would be 40A. So 40A drawn from a 100V source gives us 4000W supplied and 20A into a 10Ω load gives us 4000W dissipated. The input to our transformer was 40A @ 100V and the output was 20A @ 200V. So we see that our voltage was stepped up by a factor of two and our current was stepped down by a factor of two.

When doing the step-up and step-down comparisons, you need to compare the input NOW to the output NOW, not the input yesterday to the output a week ago.

8. ### Kermit2 AAC Fanatic!

Feb 5, 2010
3,672
892
Assuming you understand simple algebra, you will now know that for any choosen power level, current gets smaller as voltage gets larger.

power company wants to get maximum power to market. loss in trsnsmission occurs due to wire resistance. This formula is used to figure power loss
current(squared) X resistance = power loss
obviously we want to make the squared value as small as possible. we want as low a current value as we can get.

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9. ### Loreno Thread Starter New Member

Nov 10, 2014
14
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@alfacliff Ok, so the input power of the transformer must be equal to the output power of the transformer. So, let's say we have a transformer and on its input it has 10 turns and on its output it has 100 turns. We apply 10V voltage to its input. On the output side we get 100V.
On the input side the current is, for example, 1A (because the input side has a resistor R=10 ohm connected in series with the input coil). So the power on the input side is P = U* I = 10W.
That means that the same power has to be transformed to the output. The output voltage is 100V, so the current must be 0,1A to get that 10W of power. But what about the resistance on the output side? What if I connect, for example, 1 ohm resistor? Output voltage is 100V, so the current should be I = U/R = 100V/1ohm = 100A! So what current will flow on the output side: 0,1A or 100A?

10. ### WBahn Moderator

Mar 31, 2012
17,461
4,701
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.

11. ### Veracohr Well-Known Member

Jan 3, 2011
540
75
You're neglecting the effect of the transformer on impedance (resistance). The impedance seen by the source is the load impedance times the square of the turns ratio:

Z'l = Zl*(Np/Ns)^2

Say you have a 10V source, a 10Ω load, and a 1:1 transformer. The source sees a load of 10Ω*(1/1)^2 = 10Ω, so the current delivered by the source is 10V/10Ω = 1A, and the power delivered by the source is 10V*1A = 10W. The transformer secondary is also 10V because the turns ratio is 1:1, so the current delivered to the load is 10V/10Ω = 1A, and the power delivered to the load is 10V*1A = 10W.

Next consider you have a 1:2 transformer. The source now sees a resistance of 10Ω*(1/2)^2 = 2.5Ω, so the current through the primary is 10V/2.5 = 4A and the power delivered by the source is 10V*4A = 40W. The secondary voltage is 20V, the current through the load is 20V/10Ω = 2A, and the power delivered to it is 20V*2A = 40W. The power delivered by the source and the power delivered to the load are the same, the voltage delivered to the load is doubled, and the current delivered to the load is halved.

12. ### mossman Member

Aug 26, 2010
131
3
Let me dumb it down a bit (no offense intended)...To step up voltage using a transformer, the secondary coil must have more windings than the primary coil so that the magnetic flux generated by the primary "cuts" through more wire on the secondary. The more wire you have the higher the resistance. Higher resistance at a given voltage results in less current. V = IR, Voltage is constant, resistance is constant, therefore current (I) has to go down to satisfy ohm's law. Therefore, stepping up voltage results in a proportional step down in current. A transformer is a passive device and can't magically create power, hence the term "power in equals power out" (lossless transformer).

Last edited: Mar 17, 2015
13. ### MikeML AAC Fanatic!

Oct 2, 2009
5,450
1,066
Everybody stop! The OP doesn't understand series circuits or Ohms Law, so all of the discussion about transformers is lost on the OP.

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14. ### Loreno Thread Starter New Member

Nov 10, 2014
14
0
@MikeML I do understand circuits and Ohms Law, I can assure you, I just could't understand how it happens that the same power that is delivered to the transformer is being given to the output.

I really like how @WBahn and @Veracohr describe things happening in the transformer. Could you please tell me what would happen in the situation presented in the attachment? What values would the currents have?

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15. ### alfacliff Well-Known Member

Dec 13, 2013
2,449
428
the same powe on both sides, it is not how much power is being deliverfed, its about how much power is being pulled from the output. then you use the turns ration to figure the input power. or how much power on the input to figure how much power is being pulled from the secondsry output. sources of opwer do not deliver more than is used. the power used is determined by the load.

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16. ### Loreno Thread Starter New Member

Nov 10, 2014
14
0
You guys are truly awesome, with every post of yours I understand more and more. I feel like I'm really close to understanding it fully, but could someone please tell me what would happen in an example I posted in an attachment in my previous post? I think that it would make me understand it completely.

//EDIT
BUT, is my drawing correct? I don't really know if that resistor R1=5 ohm doesn't ruin it. Doesn't it take the whole V1 = 100V? I mean, is there any voltage on the input coil? I think there is none, so that drawing is wrong. Probably I should have given some resistance to the coil, so that it had a voltage drop on it. Am I right?

17. ### uwed Member

Mar 16, 2015
64
17
Example:
I1=19.4A, I2=1.94A

18. ### Loreno Thread Starter New Member

Nov 10, 2014
14
0
I tried doing my example form the attachment without the R1 resistance - so there is no resistance on the input side and there is resistance R2 = 15 ohm on the output side. I got 666W on both input and output, which is right. The formula Z'l = Zl*(Np/Ns)^2 given by @Veracohr helped a lot! I didn't know how the output resistance affects the input circuit. But what if we add a resistance to the input?

@uwed
I got the same input current, but the output is different. My calculations gave me 66,6A. I think that I need to use a formula similiar to Z'l = Zl*(Np/Ns)^2, but this time for the output impedance. Could you please show me your calculations?

19. ### uwed Member

Mar 16, 2015
64
17
R1=5ohm ruins it totally, takes nearly all the voltage.

20. ### uwed Member

Mar 16, 2015
64
17
I1=100/(5+15/10^2)=19.4
I2=I1/10=1.94