electrical energy transmission and transformerby sgstudent Tags: electrical, energy, transformer, transmission 

#1
Mar512, 08:27 AM

P: 636

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! :) 



#2
Mar512, 11:06 AM

P: 861

This setup is a well known for bringing out misconceptions! Here are one point that I think you might be confused about...
I_{s} is not fixed by the transformer equations. I_{s} is given by I_{s} = V_{s}/(R_{lines} + R_{load}). (For example, if the load resistance is infinite, that is the lines don't have anything connected to their far end, then I_{s} = 0.) I_{s} then fixes I_{p} via your second transformer equation. 



#3
Mar512, 04:38 PM

P: 636





#4
Mar612, 12:43 AM

P: 636

electrical energy transmission and transformer 



#5
Mar612, 02:27 AM

P: 861

Power is not fixed. The only things that are fixed (in the simple transformer treatment you are using) are V_{P} and V_{S}.
The power is indeed dependent on the load resistance. What will give you confidence is some numerical practice. Try this. A stepup 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 V_{S}, I_{S}, the power input to the transformer, I_{P}, the p.d. across the load, the power used by the load and the power dissipated in the connecting wires. 



#6
Mar612, 03:04 AM

P: 636





#7
Mar612, 05:08 AM

P: 861

(1) For 'primary stepup transformer', you mean 'primary coil of the step up transformer'. (2) There are two currents involved: I_{p} in the primary circuit and I_{s} 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. 



#8
Mar612, 06:36 AM

P: 636

(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. :) 



#9
Mar612, 07:26 AM

P: 861

Good! I agree with almost everything in your last posting. The only thing that sounds a bit suspect is




#10
Mar612, 04:37 PM

P: 636





#11
Mar612, 04:48 PM

P: 861

At O level, the type of question you're likely to meet is similar to the one I gave you earlier!




#12
Mar612, 05:02 PM

P: 636





#13
Mar612, 05:27 PM

P: 861

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!]




#14
May3112, 12:10 AM

P: 636





#15
May3112, 12:59 PM

P: 861

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. 



#16
Jun112, 04:54 AM

P: 636

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!




#17
Jun112, 05:25 AM

HW Helper
P: 4,715

Power = V² ÷ 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. 



#18
Jun112, 05:40 AM

P: 861

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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