# Transformers- When are they useful?

Discussion in 'Homework Help' started by Hauss32, Mar 22, 2014.

1. ### Hauss32 Thread Starter New Member

Mar 21, 2014
7
0
Hello,

First of all, what a great site! I am in college physics, and electronics really grabbed my attention (granted the class is only algebra based, and all of this is a bit overwhelming). I would love to pursue electronics further, but I need solid foundations first. A huge 'Thank You!' to everyone for their responses and support in advance! I saw this during my investigation in trying to figure things out:

So if the ratio were 200:1 instead of 20:1, we would see a secondary 1.2V and 10A. Could this even light the bulb with only 1.2V? Would the wire filament get way too hot with 10A? If power is the same, why is this useful? If transformers are inside of things I plug into the wall to step down voltage for lower voltage-requiring electronics, then won't they carry higher currents? Or is current restricted by resistors in that little plug box I use to plug in? Why can't they just dissipate the voltage with resistors?

I understand these are very elementary questions (and perhaps on the completely wrong thought process), and I understand there are about 100 questions in there. This would help me so much to clarify these concepts, and the videos, worksheets, etc. didn't completely clarify things for me.

A million thanks for your patience and shared knowledge!!!

2. ### MrChips Moderator

Oct 2, 2009
12,633
3,453
One fundamental topic in every course in electricity is Ohm's Law, I = V/R.

Learn Ohm's Law and live, sleep and breathe Ohm's Law and its two corollaries:

V = I x R

and

R = V/I

In the example given above, the 60W load (lamp) takes 5A at 12V.

Calculate the resistance R of the load using the equation R = V/I.

In this example, one assumes that the resistance R of the load remains constant.
With 1.2V applied to the load, what is the current through the load? Use I = V/R.
What is the power consumed by the load? Use P = I x V.

Two reasons:

1) It would not work for a variety of load resistance. They would have to know ahead of time the resistance of the load.

2) The resistors get hot and waste energy. When the transformer is close to ideal, i.e. close to 100% power transfer, there is little waste of energy.

Last edited: Mar 22, 2014
3. ### crutschow Expert

Mar 14, 2008
13,496
3,373
Also note that power equals V * I. From this and Ohm's Law you can derive the equations for the power dissipated in a resistor which are I*I*R (I$^{2}$R) or (V*V)/R (V$^{2}$/ R).

4. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
515
At your level most of physics is about looking for straight line relationships.

That is

Something is proportional to something else

$A \propto B$

We convert this to a useful equation by introducing a constant of proportionality thus

$A = cB$

Where c is the constant of proportionality and is often a new quantity with units in its own right.

This theme runs right through college physics and many of the practical experiments are designed to gather data to show and plot the straight line and evaluate the constant.

So we have

The elongation of a spring is proportional to the applied force : the constant is called Youngs modulus

Extension = EF

Force is proportional to acceleration : the constant of proportionality is called mass

F = ma

Distance is proportional to time : the constant is called speed

d =st

Rate of heat flow is proportional to cross sectional area.

Q(joules per second) = kA

Also for a given cross section

Rate of heat flow is proportional to the temperature difference : The constant is called the thermal resistance.

Q = r(t2-t1)

The last two are interesting because electricity follows this pattern.

The rate of charge flow is measured in coulombs per second and is called the current in amperes.

This is proportional to the voltage difference :

I = Charge (per second) = current = R(V2-V1)

The constant is called the electrical resistance and the equation is called Ohm's law.

Last edited: Mar 22, 2014
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5. ### supermankid Member

May 26, 2013
36
1
I had this feeling few years back. One most important thing you have to understand is.

The current doesnot flow at all when there is no load.

The most important thing to understand is
1. The voltage is transferred according to ratio of the transformer winding

2. But the amount of current on either primary or secondary side is determined by the load or bulb or motor or whatever connected on the secondary winding....

so...the current is not massive at any time...

hope it helps

6. ### amilton542 Active Member

Nov 13, 2010
494
64
Check this out; the peak flux in the core is a function of voltage and frequency, and the impedance seen from the source is by virtue of the secondary, odd huh?

I love transformers .

7. ### Hauss32 Thread Starter New Member

Mar 21, 2014
7
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Thanks everyone, for all of the great responses! I think things are starting to clarify. One question here:

So if we now have 1.2V, then I= 1.2V/2.4Ohms --> I=.5A
Thus, P=.5A x 1.2V --> P=.6W

So this 60W lightbulb is only drawing .5A from the outlet, and thus only 6W from the outlet? 6W probably won't light the bulb, at least not well at all.

Another question then:

Thanks again everyone, I'm learning so much!!!

Last edited by a moderator: Mar 22, 2014
8. ### MrChips Moderator

Oct 2, 2009
12,633
3,453
Correction: 0.6W, not 6W

Correct: The bulb would not light.

9. ### MrChips Moderator

Oct 2, 2009
12,633
3,453

I deleted your question in this regard.

10. ### Hauss32 Thread Starter New Member

Mar 21, 2014
7
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^Apologies. The thought didn't even cross my mind. This is for safety reasons I'm sure. Sorry for the extra work, I've now read the guidelines in full, I promise!

Anyways, this is starting to make more sense. Thanks everyone!

11. ### Hauss32 Thread Starter New Member

Mar 21, 2014
7
0
One final question to secure my understanding:

If a person were to touch a 120V mains outlet, it could deliver quite a shock and be dangerous. But if you use a transformer, there is nothing you could do to make it more dangerous by increasing current? Either you step-up voltage and it decreases the current in the load (person), or you step-down the voltage, thus decreasing the current through the load?

12. ### MrChips Moderator

Oct 2, 2009
12,633
3,453
Once again, go back to applying Ohm's Law.

The human body has resistance. If we assume that the path to ground is a fixed resistance, then the current flowing through the body is directly proportional to the applied voltage.

What is more important is the current through the human body and what part of the body it flows.

10mA can be fatal.

http://en.wikipedia.org/wiki/Electric_shock

13. ### Hauss32 Thread Starter New Member

Mar 21, 2014
7
0
I was trying to apply Ohm's Law. Here is what I was thinking:

Let's say there is a 120V mains outlet, and a person has a resistance of .1MΩ (in the right ball park I think, but not going for validity here). If this person stuck their fingers in the sockets, then I=V/R --> I=120V/100,000Ω=.0012A

If we use a transformer to scale up voltage, say a 1:100 (1°:2° coil turns) then voltage will be 100x higher, and this will be 100x120V=12,000V. We cannot use this value for voltage in Ohm's Law because energy is conserved, so amperage must decrease. Does this mean current through the load (person) will be 1/100 of what it was before? Therefore 1/100th of .0012A is .000012A?

If we use a transformer to scale down voltage, say 100:1 (1°:2° coil turns) then the voltage would be scaled down by a factor of 100. So 120V/100=1.2V. And I=V/R so I=1.2V/100,000Ω=.000012A

I'm having some trouble figuring out when to use Ohm's Law and when you can't for these special cases. Thanks!

14. ### MrChips Moderator

Oct 2, 2009
12,633
3,453
You cannot use conservation of energy rule because you have changed the conditions.

Simply apply Ohm's Law:

If V = 120V, R = 100,000Ω
then I = V/R = 120V/100,000Ω = 0.0012A = 1.2mA

If V = 12000V, R = 100,000Ω
then I = V/R = 12000V/100,000Ω = 0.12A = 120mA

If V = 1.2V, R = 100,000Ω
then I = V/R = 1.2V/100,000Ω = 0.000012A = 12μA

Ohm's Law still rules.

15. ### Hauss32 Thread Starter New Member

Mar 21, 2014
7
0
Thanks you for all of your responses, MrChips!

So if Voltage going into the transformer is 120V, and let's say we use that same step-up transformer as before at 12,000V and 120mA output.
Then P=IV, so P=.120A x 12,000V = 1440W
On the other side of the transformer, there is 120V input, so P=IV, P(input)=P(output), so 1440W=120V x I --> I=1440W/120V= 12A. There is 120V and 12A on the input, and 12,000V and .120A on the output?

My professor made a Tesla Coil, and he used a transformer from a power line, and ran it backwards I guess. He took 120Vac from the outlet and made what was theoretically 1,000,000V (although because of losses a lot less). Air was ionizing and you could see and hear the electricity from the top. He then touched it and said he couldn't feel a thing because current was so low. If the coil were even at 100,000V, I=V/R, I=100,000V/100,000Ω of the body (maybe more with shoes)=1A. Shouldn't he be dead?

16. ### #12 Expert

Nov 30, 2010
16,685
7,324
I'm a bit late to the party, but I did a very simplistic article on Ohm's Law a few years ago. One side effect of that writing is to help make a crude model of how electricity moves. Take a peek.

The problem is that I can't quite get a grip on your level of expertise...yet.

17. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
515
I would not recommend applying the simple form of Ohms law with transformers, without a lot of thought, for two reasons.

Firstly the output voltage of any transformer will fall with load. This means that the voltage calculated by the transformer equation will not be correct. Consequently for a given load the current through that load will be less than expected.
Note increasing load means decreasing resistance.

The regulation of a transformer is defined as (No load Voltage - Voltage at load) / (No load voltage).

this can be expressed as a % if multiplied by 100.

Secondly transformers operate exclusively on alternating current. This introduces phase considerations and more complicated forms of Ohm's law, appropriate to AC circuitry should be employed.

18. ### crutschow Expert

Mar 14, 2008
13,496
3,373
A true Tesla coil steps up the voltage and the frequency to be in the many kHz range. At that frequency the current does not excite the nerves so no pain is felt. You should hold a metal object though when conducting the current through your body since the skin can be burned if the spark enters you skin at a single point.

19. ### amilton542 Active Member

Nov 13, 2010
494
64
Studiot has raised an issue I-myself have always considered in these kinds of analysis problems.

I have a friend who is a full-time electrician, and we meet frequently for a few drinks in a local pub and inevitably, EE forms the basis of discussion for the night. I'm always in doubt how one can get away with treating alternating quantities as a pure real valued problem by means of applying basic Ohm's law in the realm of d.c. when the fact of the matter is it's a.c.

20. ### MrChips Moderator

Oct 2, 2009
12,633
3,453
Sorry, I overlooked the fact that he was using a transformer.
Of course that transformer cannot deliver 1440W.
The problem is not that Ohm's Law does not apply.
The problem lies in the physics of the transformer. The resistance of the windings, core saturation, high temperature and insulation breakdown will fry the transformer long before it gets to delivering 1440W.