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Electrical energy transmission and transformer

by sgstudent
Tags: electrical, energy, transformer, transmission
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sgstudent
#1
Mar5-12, 08:27 AM
P: 645
When transferring electricity from a generator to a factory or anything we would use a transformer to step up the voltage right? As P=VI and P is constant in the primary and secondary transformers (ideal case), with an increased V as compared to the primary transformer it will in turn result in a smaller I as compared to the primary transformer.

Then again, the values of V and I are fixed in accordance to the new voltage output by using the formula Vp/Vs=Np/Ns and VpIp=VsIs. in most calculation questions, the R of the cable is given and the values fit nicely to get the total power. However, if i were to change the resistance of the cable what will happen? Since the values of V and I are constant, how will the V=RI values change. This is because the value of the power, voltage and current are fixed to a certain value due to the usage of the coil formula and the conservation of energy formula. So I'm completely clueless about the scenario that i change the values of resistance within the cable.

I'm hoping that you will be able to help me in this. Thanks so much for the help! :)
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Philip Wood
#2
Mar5-12, 11:06 AM
PF Gold
P: 963
This set-up is a well known for bringing out misconceptions! Here are one point that I think you might be confused about...

Is is not fixed by the transformer equations. Is is given by
Is = Vs/(Rlines + Rload). (For example, if the load resistance is infinite, that is the lines don't have anything connected to their far end, then Is = 0.)
Is then fixes Ip via your second transformer equation.
sgstudent
#3
Mar5-12, 04:38 PM
P: 645
Quote Quote by Philip Wood View Post
This set-up is a well known for bringing out misconceptions! Here are one point that I think you might be confused about...

Is is not fixed by the transformer equations. Is is given by
Is = Vs/(Rlines + Rload). (For example, if the load resistance is infinite, that is the lines don't have anything connected to their far end, then Is = 0.)
Is then fixes Ip via your second transformer equation.
Huh, but then it current is dependant on the resistance, if resistance increases current will decrease? But then again power is fixed so as V is constant and P is constant, how will the values add up again? Thanks for theft help!

sgstudent
#4
Mar6-12, 12:43 AM
P: 645
Electrical energy transmission and transformer

Quote Quote by Philip Wood View Post
This set-up is a well known for bringing out misconceptions! Here are one point that I think you might be confused about...

Is is not fixed by the transformer equations. Is is given by
Is = Vs/(Rlines + Rload). (For example, if the load resistance is infinite, that is the lines don't have anything connected to their far end, then Is = 0.)
Is then fixes Ip via your second transformer equation.
So does it mean that the power value will change according to the resistance of the whole load?
Philip Wood
#5
Mar6-12, 02:27 AM
PF Gold
P: 963
Power is not fixed. The only things that are fixed (in the simple transformer treatment you are using) are VP and VS.

The power is indeed dependent on the load resistance.

What will give you confidence is some numerical practice. Try this. A step-up transformer has 2000 turns on its primary, which is connected to an alternating voltage of 1200 V. The secondary has 20000 turns and is connected to a load of 76000 Ω via a pair of long connecting wires, each of resistance 2000 Ω.

Calculate VS, IS, the power input to the transformer, IP, the p.d. across the load, the power used by the load and the power dissipated in the connecting wires.
sgstudent
#6
Mar6-12, 03:04 AM
P: 645
Quote Quote by Philip Wood View Post
Power is not fixed. The only things that are fixed (in the simple transformer treatment you are using) are VP and VS.

The power is indeed dependent on the load resistance.

What will give you confidence is some numerical practice. Try this. A step-up transformer has 2000 turns on its primary, which is connected to an alternating voltage of 1200 V. The secondary has 20000 turns and is connected to a load of 76000 Ω via a pair of long connecting wires, each of resistance 2000 Ω.

Calculate VS, IS, the power input to the transformer, IP, the p.d. across the load, the power used by the load and the power dissipated in the connecting wires.
Oh ok! I somewhat understand. However, if the power in the primary step up transformer is reduced, then the current will have to drop in that circuit too. This is because the P is now a constant in this situation. So by V=RI, if my I decreases, my R will increase? Like at first V=10V, R=5ohm and I=2A. So my P=20W. Then when my power drops to like 10W, and V is still constant then my I will reduce. So now I is reduced to 1A. Then since V=RI then the resistance should increase compared to before?
Philip Wood
#7
Mar6-12, 05:08 AM
PF Gold
P: 963
Quote Quote by sgstudent View Post
Oh ok! I somewhat understand. However, if the power in the primary step up transformer is reduced, then the current will have to drop in that circuit too. This is because the P is now a constant in this situation. So by V=RI, if my I decreases, my R will increase? Like at first V=10V, R=5ohm and I=2A. So my P=20W. Then when my power drops to like 10W, and V is still constant then my I will reduce. So now I is reduced to 1A. Then since V=RI then the resistance should increase compared to before?
I'm afraid I find this very confusing...
(1) For 'primary step-up transformer', you mean 'primary coil of the step up transformer'.
(2) There are two currents involved: Ip in the primary circuit and Is in the secondary circuit. You must always say which you're talking about.
(3) It's no use quoting V = IR unless you say what you're applying it to (load or whatever).
(4) "So by V = RI, if my I decreases my R will increase?" The currents don't affect the resistances (assuming the wires don't heat up).
(5) "This is because the P is now a constant in this situation." Power in general isn't a constant for a transformer; it depends on the load resistance. Unless by constant P you mean that the input power VpIp and the output power VsIs are equal.

I strongly suggest that you tackle the problem I gave now. It'll teach you a lot more than this sort of interchange of words. If you need hints, ask.
sgstudent
#8
Mar6-12, 06:36 AM
P: 645
Quote Quote by Philip Wood View Post
I'm afraid I find this very confusing...
(1) For 'primary step-up transformer', you mean 'primary coil of the step up transformer'.
(2) There are two currents involved: Ip in the primary circuit and Is in the secondary circuit. You must always say which you're talking about.
(3) It's no use quoting V = IR unless you say what you're applying it to (load or whatever).
(4) "So by V = RI, if my I decreases my R will increase?" The currents don't affect the resistances (assuming the wires don't heat up).
(5) "This is because the P is now a constant in this situation." Power in general isn't a constant for a transformer; it depends on the load resistance. Unless by constant P you mean that the input power VpIp and the output power VsIs are equal.

I strongly suggest that you tackle the problem I gave now. It'll teach you a lot more than this sort of interchange of words. If you need hints, ask.
oh sorry for being too vague. (1) Before i attach a load, my VpIp=VsIs=20W for example, Vp=10V and Is=2A. (2) So when i increase the resistance of the cables it will result in a lower power for both the primary and secondary coils. So, for example my VpIp=VsIs=10W, as my Vp will still be 10V, so my Ip will drop to 1A right? So before the increase the resistance (1), my Vp=RpIp=5ohmX2A. then when i increase my resistance of the cables/load (2), the voltage in the primary coil still remains the same. however, due to the decreased power, the current has to drop to get the 10W, 10W=10VX1A. But then the primary coil's resistance does not change so when using the formula Vp=RpIp, since V and R is constant, then how can the current change?

(talking about the primary coil now)Because now i have the added resistance in the secondary coil's part. So there is a lower power in both the coils. so by P=VI, since V is constant so I will drop. but since V and R is constant, then how can the current drop unless the resistance drops?

thanks for the help. I'm doing the calculations now. really appreciate the help given here. :)
Philip Wood
#9
Mar6-12, 07:26 AM
PF Gold
P: 963
Good! I agree with almost everything in your last posting. The only thing that sounds a bit suspect is

Quote Quote by sgstudent View Post
Then again when I use V=RI for the primary transformer it would result in an increased resistance too? Since I is reduced.
The transformer equations you're using assume that the resistances of the primary and secondary coils are both zero. So when you apply V = RI to the primary coil, in the form R = V/I, you are not, literally, finding the resistance of the primary coil. What you are finding is a sort of 'effective resistance' or 'transferred resistance'. This is quite a sophisticated idea, and, at this stage, I'd keep off it. Just don't go there, that is don't apply V = RI to the primary coil.
sgstudent
#10
Mar6-12, 04:37 PM
P: 645
Quote Quote by Philip Wood View Post
Good! I agree with almost everything in your last posting. The only thing that sounds a bit suspect is



The transformer equations you're using assume that the resistances of the primary and secondary coils are both zero. So when you apply V = RI to the primary coil, in the form R = V/I, you are not, literally, finding the resistance of the primary coil. What you are finding is a sort of 'effective resistance' or 'transferred resistance'. This is quite a sophisticated idea, and, at this stage, I'd keep off it. Just don't go there, that is don't apply V = RI to the primary coil.
Oh ok,so actually changing the resistance will affect many variables? But I thought that even if I don't increase the resistance, then P=VI, the V is equal to resistance times current? But now the change in resistance is affected by this complicated concept? Anyways at O levels will there be such questions whereby they change the resistance? Thanks for the help!
Philip Wood
#11
Mar6-12, 04:48 PM
PF Gold
P: 963
At O level, the type of question you're likely to meet is similar to the one I gave you earlier!
sgstudent
#12
Mar6-12, 05:02 PM
P: 645
Quote Quote by Philip Wood View Post
At O level, the type of question you're likely to meet is similar to the one I gave you earlier!
Oh, then it won't be so complicated as in this case. BTW is the name of the transferred resistance called impadence?
Philip Wood
#13
Mar6-12, 05:27 PM
PF Gold
P: 963
Well, it is a sort of impedance, but it's mainly resistive, if the transformer has a resistive load connected to its secondary! [A fuller discussion of these matters would, I think, be of no help to you at the moment!]
sgstudent
#14
May31-12, 12:10 AM
P: 645
Quote Quote by Philip Wood View Post
Well, it is a sort of impedance, but it's mainly resistive, if the transformer has a resistive load connected to its secondary! [A fuller discussion of these matters would, I think, be of no help to you at the moment!]
hi Philip Wood, I was also thinking if I have a step up transformer which steps up the voltage to 1000V and it has a total effective resistance of 10ohms (including wires and all at the secondary transformer circuit), then the power developed will be P=VI so its 100000W right? But now if I use the same circuit just that I step it down to 250V, then won't my power be lower now (6250W)? So when doing questions how can they assume the total power developed when using the different values of voltage in the secondary circuit (includes power losses in wires)? Won't the power be different in other step up/down voltages?
Philip Wood
#15
May31-12, 12:59 PM
PF Gold
P: 963
Hallo sgs. I'm afraid I find the details of your post quite hard to follow. I can't help but feel that you need to tackle some specific calculations on transformers. They will give a better basis for discussion on this forum, though you may well find that just doing them makes things come clear. You could start with the exercise I recommended for you to do in post 5 on this thread.

I'd keep off 'effective resistance'. It's quite an advanced concept which is likely to confuse you at this stage. Sorry if this seems patronising, but I really think you should concentrate on using the basic equations you mentioned in the second paragraph of your first post.
sgstudent
#16
Jun1-12, 04:54 AM
P: 645
Hi sorry for being vague Mr Wood, I meant that if I have the same resistance in my secondary circuit, and when I change the number of turns of the coil (hence changing the voltage output), then will my power also change? I'm able to do the basic stuff just that I'm not sure about these stuff... thanks!
NascentOxygen
#17
Jun1-12, 05:25 AM
HW Helper
Thanks
P: 5,475
Quote Quote by sgstudent View Post
I meant that if I have the same resistance in my secondary circuit, and when I change the number of turns of the coil (hence changing the voltage output), then will my power also change?
Power will increase if you apply a higher voltage to a fixed resistance, that's a certainty.

Power = V R

Whatever power is drawn by the load on the secondary side, is the power that is delivered to the transformer on its primary side.
Philip Wood
#18
Jun1-12, 05:40 AM
PF Gold
P: 963
Ah, this is now clear. If you increase the secondary voltage output, then the power taken by the secondary load will increase, since [itex]P_{load} = \frac{V_{sec} ^2}{R_{load}}.[/itex]

Assuming 100% transformer efficiency, [itex]V_pI_p=V_sI_s[/itex]. So the power input to the primary is also given (approximately) by [itex]P = \frac{V_{sec} ^2}{R_{load}}[/itex].

This is basic stuff!

Challenge you to give me the answers to the exercise I recommended in post 5.


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