ECN Electrical Forum - Discussion Forums for Electricians, Inspectors and Related Professionals

Page 1 of 2 1 2 >
Topic Options
Rate This Topic
#26132 - 05/30/03 06:13 PM ohms law
james S Offline
Member
Registered: 05/13/03
Posts: 107
Loc: West England
When i was an apprentice i asked the following question to diffrent electricians and was told it does not matter you will never need to know!!great answer.same answer with many other questions also asked. Step up the voltage current goes down, step down volts current increases. Georg simon ohm;the current in a circuit is proportional to the voltage (not inversely)?
Top
Work Gear for Electricians and the Trades
#26133 - 05/30/03 06:29 PM Re: ohms law
maintenanceguy Offline
Member
Registered: 12/02/01
Posts: 300
Loc: Southern NJ, USA
Ohm's law isn't the right law to use to prove or disprove transformers. It's like using the law gravity to prove water boils at 212º. Good law, but not really the right place to apply it.

I guess the best law to justify what transformers do is conservation of energy. Watts in = watts out.

But Ohms law does apply to each coil of a transformer, as long as you're taking your measurements on one complete loop, that is through only one coil at a time. Because the two coils (or however many there are) aren't directly attached but only interact by magnetic fields, ohms law doesn't work across a transformer, only through a continuous path such as through one of the coils of a transformer.

And ohms law says I=E/R, so current through a coil goes up as voltage on that coil goes up, and goes down as impedance on that coil goes up. Impedance goes up as the magnetic field (from a load on the secondary coil) fights the current in the coil you're testing. So ohm's law does apply, just not quite in the way you were thinking.

Man I wish I was better at explaining stuff.
Top
#26134 - 05/31/03 04:34 AM Re: ohms law
pauluk Offline
Member
Registered: 08/11/01
Posts: 7520
Loc: Norfolk, England
The current in a circuit is directly proportional to the voltage, but only for a given value of resistance.

You're dealing with what are effectively two separate circuits (primary and secondary), each of which can have vastly different values of resistance.

Look at it this way:
If you keep the resistance of a circuit constant, then increasing the voltage will result in a corresponding increase in current (just as you would expect from Ohm's Law). You have increased both voltage and current, therefore you have also caused an increase in the dissipated power (P = I x E).

Now let's look at the transformer. Maybe an example with numbers will help. Assume you have a xfmr with a 240V secondary feeding a 2400W load. The primary side is fed with 2400V.

The current in the secondary will be:

2400W / 240V = 10A

And from Ohm's Law the resistance must be:

240V / 10A = 24 ohms.

Now let's see what's happening on the primary side. You can't get more power out of a transformer than you put into it. To get 2400W from the secondary, you therefore have to feed 2400W into the primary (*).

If the primary side is running 2400W at 2400V, then the current must be:

2400W / 2400V = 1A.

Then from Ohm's Law, the primary resistance is therefore:

2400V / 1A = 2400 ohms.

In other words, the primary resistance is a hundred times greater than the secondary. That's why even though the primary voltage is ten times greater than the secondary, the primary current is still only one-tenth of the secondary current.

(*) In practice, of course, no transformer is 100% efficient, so the output power will actually be slightly less than the input due to losses. I've also completely ignored power-factor considerations here to keep it simple.
Top
#26135 - 05/31/03 05:51 AM Re: ohms law
james S Offline
Member
Registered: 05/13/03
Posts: 107
Loc: West England
Thanks for your time maintenance guy / pauluk!

[This message has been edited by james S (edited 05-31-2003).]
Top
#26136 - 05/31/03 09:10 PM Re: ohms law
Scott35 Offline

Broom Pusher and
Member
Registered: 10/19/00
Posts: 2707
Loc: Anaheim, CA. USA
One thing to remember in the formula of E ÷ R = I is that "R" is Resistance, not Reactance.

"R" could be the component of Pure Resistance - dispersing True Power (Wattage), or may be part of an Impedance (Z) figure - which would make the figure more dependant on the total Reactance and Resistance.

For simple calcs across pure resistance loads (like Resistance Heating elements or Incandescent lamps when they are at operating temperature), the true power (P) will be a direct representation of E × I, and the level of current flowing (I) will be a result of the impressed Voltage (E) across the fixed Resistance (R).
Changes in either one will result in a Linear type change overall. The total consumed true power (Wattage) changes in a linear fashion - like how an Incandescent lamp dims when a motor starts.

The linear effect of true power works both ways: If Voltage is decreased across a fixed resistance, the load current is also decreased - which results in a decreased consumption of true power.
If the Voltage is increased across a fixed resistance, the load current also increases, which results in an increased true power consumption on the circuit/element.
That's what makes an Incandescent lamp blow out from higher than rated Voltage being impressed across the filament.

On Reactive elements, only the Resistive element (if any) recieves / dissipates the true power - hence this is the only part with a linear characteristic. Any Resistance within the Reactive elements also becomes affected (for instance - winding resistance of reactor coils).

An Induction Motor is effected by changes in voltage because the TOTAL OUTPUT HORSEPOWER is linear (true power). The Motor works similar to a transformer as viewed by the input KVA (Windings are the primary, the squirrel cage rotor is the secondary).
Only the Reactive component of these circuits is non linear.
Raise the input voltage too far and the motor lets smoke out!

For a simple VA formula (E × I = VA for an AC circuit that's not pure resistance), the equation reflects Apparent Power, not True Power only.

Even though, generally speaking, a higher voltage results in a lower current draw, this only applies in the way the load device has been designed, not increasing or reducing nominal voltage to a given device without compenasting / connecting it accordingly.

Raising the nominal voltage on XL / XC or Z components anymore than 10% will not lower the current draw, but will swing things into a new power level and result in a higher current draw!
Same applies in the opposite direction!

The device being driven must be able to have connection alternatives available, in order to drive it at a higher voltage - otherwise the true power which is drawn from the supply will be greatly affected.

That old idiom of "The Higher The Voltage, The Less the Amps" is another one in the line of "Voodoo Electrical Statements", and has been repeatedly miss quoted over and over again.
The complete statement is more like:
<OL TYPE=1>

[*] The KVA draw of an AC circuit is a component of the Voltage times the Line Amperes. To reduce a high load current and still maintain a given KVA value, devices may be designed to utilize a higher voltage via connections of the internal components.


[*] To reduce a large input current flow on pure Resistance loads connected to AC or DC, the device can be driven with a higher voltage as long as the device can be connected to allow a fixed wattage to be consumed. This would be from increasing the fixed Resistance in proportion to the impressed system voltage.
</OL>

Get the idea?

Scott

p.s. if I sound harsh, please take no offense! I am only trying to make a few points about the whole true power formula, since the entire point of all this stuff is to transfer the correct level of true power (wattage) from the generating source, to the load device... then transpose it into heat energy (either instantly or eventually).

Funny thing about all the stuff we connect to power supplies - we help make the surrounding air get heated, and we pay money to heat the air / enviroment around a given load!
I could burn leaves and do the same thing! Only need to buy a book of matches!!!
(trying to make humorous joke references to energy transfers, but now it sounds stupid!)

Scott

p.s. had to edit a typing boo-boo!
S.E.T.

[This message has been edited by Scott35 (edited 06-01-2003).]
_________________________
Scott " 35 " Thompson
Just Say NO To Green Eggs And Ham!
Top
#26137 - 06/01/03 04:47 AM Re: ohms law
pauluk Offline
Member
Registered: 08/11/01
Posts: 7520
Loc: Norfolk, England
James,

The considerations that Scott has mentioned above are just some of the things that make AC power calculations so much more complex than DC.

That pesky low power factor due to high levels of reactance can really make a huge different between the apparent power (VA) and the true power (W).
Top
#26138 - 06/16/03 02:24 PM Re: ohms law
james S Offline
Member
Registered: 05/13/03
Posts: 107
Loc: West England
Thanks pauluk and scott35 its great to have somewhere to go for answers
Top
#26139 - 07/04/03 11:48 AM Re: ohms law
james S Offline
Member
Registered: 05/13/03
Posts: 107
Loc: West England
Raising the nominal voltage on XL / XC or Z components anymore than 10% will not lower the current draw, but will swing things into a new power level and result in a higher current draw!
Same applies in the opposite direction!
---------------------------------------------
would it be a correct statement to say that if the motors nominal voltage was to increase by more than 10%, THEN it will start to smoke?



[This message has been edited by james S (edited 07-04-2003).]

[This message has been edited by james S (edited 07-04-2003).]

[This message has been edited by james S (edited 07-06-2003).]
Top
#26140 - 07/06/03 11:48 PM Re: ohms law
Scott35 Offline

Broom Pusher and
Member
Registered: 10/19/00
Posts: 2707
Loc: Anaheim, CA. USA
James S:

Quote:

would it be a correct statement to say that if the motors
nominal voltage was to increase by more than 10%, THEN it
will start to smoke?


Greatly exceed the 10% tolerance (+/-) and drive it at it's max output for a continuous time, should produce smoke.

The key idea here is to create excessive heat dissipation within the Motor's windings, and keep it going for a long period of time.
That results in smoke loss from the windings. Excessive smoke loos leads to flame loss, which leads to Motor loss

Scott35
_________________________
Scott " 35 " Thompson
Just Say NO To Green Eggs And Ham!
Top
#26141 - 07/08/03 05:44 AM Re: ohms law
james S Offline
Member
Registered: 05/13/03
Posts: 107
Loc: West England
What i was actualy getting at was within your original reply to my question you mentioned that raising the nominal voltage by anymore than 10% in a XL/XC OR Z Components would increase the current if the voltage was increased and the same in opposite direction. which is why i was asking if it would be correct to say that after 10% rise in nominal voltage THEN the current would start to rise WITH the voltage within the motor windings eventually causing so much of an increase that the current carrying capacity of the windings within our motor would be exceeded and begin to smoke!

It just confused me when you said voltage becomes proportional with the current after a 10% increase in nominal Voltage within XL/XC AND Z COMPONENTS. ( in a nut shell)


ps just trying to improve my understanding of ac theory
Top
Page 1 of 2 1 2 >

Member Spotlight
Member Since: 02/23/13
Posts: 204
New in the Gallery:
SE cable question
Featured:

2017 NEC and Related
2017 NEC
Now Available!

Shout Box


Who's Online
0 registered (), 53 Guests and 8 Spiders online.
Key: Admin, Global Mod, Mod
 
New in the Gallery:
SE cable question
 
Top Posters (30 Days)
Admin 49
HotLine1 43
gfretwell 18
Ruben Rocha 12
Trumpy 9
 
Newest Members
Freecrowder, clee512, Jdscott2005, FAIZAN, Regitest2

ECN Electrical Forums - sponsored by Electrical Contractor Network - Electrical and Code Related Discussion for Electrical Contractors, Electricians, Inspectors, Instructors, Engineers and other related Professionals