Ohm's Law?

[John Bunting]

Who's seen the letter from Kevin Annells on page 53 of the October RCM&E? If you haven't, please have a look. Is it as wrong as It looks to me, or am I misunderstanding something?

[Guest]

it is wrong

 altho his formula is correct he applies it incorrectly when raising the voltage

the following underlying equation is  Ohms Law  ( at least it was when I was at school )

V volts  = I current times R resistance    ie  V= I x R

and from that we get     power P watts =  V  x I

from which Kevin derived  I = P/V    which is also correct but in raising the voltage Kevin should have gone back to V = I x R  where in raising voltage from 6 to 12 , the current must also double as the resistance R remains constant ( assuming pure resistance )  and Ohms law remains in balance

from this we see that with voltage doubled, current also doubles and as a result the power goes up by 2x2 = 4

so Kevin's statement  "that as volts increases, current drops"  is wrong " 

I trust David will pick up on this and correct the letter next month otherwise Ohm will be turning in the proverbial

[birdy]

I think he is "barking up the wrong tree", and is assuming that the aim is to get the same number of watts.

[Paul Adams]

I agree,

I believe that he has swapped resistance with power in the second sentence of the third paragraph, which should read “So, in a DC circuit with the same power, if the voltage (battery) source increases the current used will actually drop”.

It would then make sense. <!--[if gte mso 9]><xml> <w:WordDocument> <w:View>Normal</w:View> <w:Zoom>0</w:Zoom> <wunctuationKerning/> <w:ValidateAgainstSchemas/> <waveIfXMLInvalid>false</waveIfXMLInvalid> <w:IgnoreMixedContent>false</w:IgnoreMixedContent> <w:AlwaysShowPlaceholderText>false</w:AlwaysShowPlaceholderText> <w:Compatibility> <w:BreakWrappedTables/> <wnapToGridInCell/> <w:WrapTextWithPunct/> <w:UseAsianBreakRules/> <wontGrowAutofit/> </w:Compatibility> <w:BrowserLevel>MicrosoftInternetExplorer4</w:BrowserLevel> </w:WordDocument> </xml><![endif]--><!--[if gte mso 9]><xml> <w:LatentStyles DefLockedState="false" LatentStyleCount="156"> </w:LatentStyles> </xml><![endif]--> <!-- /* Style Definitions */ p.MsoNormal, li.MsoNormal, div.MsoNormal {mso-style-parent:""; margin:0cm; margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:12.0pt; font-family:"Times New Roman"; mso-fareast-font-family:"Times New Roman";} p {mso-margin-top-alt:auto; margin-right:0cm; mso-margin-bottom-alt:auto; margin-left:0cm; mso-pagination:widow-orphan; font-size:12.0pt; font-family:"Times New Roman"; mso-fareast-font-family:"Times New Roman";} span.EmailStyle15 {mso-style-typeersonal; mso-style-noshow:yes; mso-ansi-font-size:10.0pt; mso-bidi-font-size:10.0pt; font-family:Arial; mso-ascii-font-family:Arial; mso-hansi-font-family:Arial; mso-bidi-font-family:Arial; color:windowtext;} @page Section1 {size:612.0pt 792.0pt; margin:72.0pt 90.0pt 72.0pt 90.0pt; mso-header-margin:36.0pt; mso-footer-margin:36.0pt; mso-paper-source:0;} div.Section1 {pageection1;} --> <!--[if gte mso 10]> <style> /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-parent:""; mso-padding-alt:0cm 5.4pt 0cm 5.4pt; mso-para-margin:0cm; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:10.0pt; font-family:"Times New Roman"; mso-ansi-language0400; mso-fareast-language0400; mso-bidi-language0400;} </style> <![endif]-->

[Paul Adams]

Must not write messages in word and copy across

Must not write messages in word and copy across

Sorry

Paul

PS Is there anyway you can delete a message once you have posted to correct the above?

[Brian Parker]

Ohms Law and Electrical power…..

The resistance is the constant.

The Watt is the derived unit of power and is equal to the product of the voltage across the resistance and the current flowing through the resistance.

Thus W=V*I

and I=V/R (Ohms Law)

and V=I*R (Ohms Law)

Then W=V*V/R

 and W=I*I*R

So who is correct and who is incorrect?

  

[Guest]

OHM is correct 

[Doug Ireland]

"The Potential Difference across a conductor will remain constant as long as the temperature within the conductor remains constant"

Ergo, higher temperature = higher resistance and therefore higher power consumption. AC circuits are a bit different. For example: an induction motor running at say 50Hz and producing xxxKw at say 4160v will pull 120A. If you reise or lower the voltage a few volts at a time you will notice the current drop. If you continue this process you will come to a point where the current increases again;  this is termed "Minimum Current Point".

Just my 6d worth.

[Brian Parker]

To take it on a stage..

All batteries have an internal resistance(r) and as current also flows through the battery some voltage will be lost in driving the current through the battery itself.

Thus V=I*R +I*r.

Therefore one could argue that. The best battery is the one with the lowest resistance.

 

Also, the maximum transfer of power from the battery to the load/motor/resister is when R=r.

 

[Tom Foreman]

Perhaps what he meant to say, if power remains constant, then as voltage increases the current decreases.

Of course in the real world this usually means reducing the load on motor by fitting a smaller prop, or with an EDF changing the motor to a lower Kv.

Or as I recently found out different battery chemistries behave different under load, i.e. voltage sag per cell e.g. A123 vs LiPo.

Its all good fun

 Tom

[Brian Parker]

It’s all in the application!

 As you say. All good fun…apart from the almost constant rain!

[Former Member]

[This posting has been removed]

[Peter Beeney]

Another 6d worth.

I've always assumed watts to be derived from V times A times p.f.

Where p. f. = Power Factor.

I would think that generally within a DC circuit the p.f. is very close to one, and thus is insignificant.  However, as soon as we talk about electric motors we start to introduce an element of inductance and capacitance. Now we're talking VI's, (volt amps) as opposed to watts.

Does this unbalance the applecart at all?

Just thought I'd mention it.  

Pete.

[Former Member]

[This posting has been removed]

[empeabee]

Wow what a bag of Apples Oranges and Plums are being compared here!

It seems to hinge on his stated misconception that Power == (is equivalent  to) Current.

This is OK if you stick with the same Battery Voltage 'cos when power consumption increases so does Current & Current is what most people look at when testing.

So :-

I = Current (not power)(Amps)

V = Voltage (Volts),

R = Resitance (pure with no stray electromagnetic happenings)(Ohms)

& P = Power (Watts)

I does equal V divided by R

P does equal V times I

so if R is constant (as stated) then if V goes up I goes up and P goes up.

If you replace Resistance with Impedance, then Ohms law works fine with AC calculations but you must keep a weather eye on the frequency of the AC.

But still the I (current) from the battery is still a good measure of what is going on.

Mike

[Steve Hargreaves - Moderator]

Yes Kevin is wrong......Ohms law states that  voltage =  current x resistance. If resistance is constant & voltage increases then current MUST increase to keep the equation true.

I think Kevin has got his Watts mixed up with his Amps. What he says about current reducing when voltage is increased is true if you keep the power constant hence you can get the same power with less current if you raise the voltage.

The acid test is to replace a 3S battery with a 4S one, change nothing else & watch what happens to the current......I suspect you will release the "magic smoke"......maybe Kevin could demonstrate his theory for us using the above example on one of his models.....

[Steve Hargreaves - Moderator]

For all you equation fetishists out there this might be interesting........

Equations

[Ron van Sommeren]

alert!!

To muddle things a bit more, Increasing the voltage in a BLDC system will give a squared increase in power and a squared increase in motor-current

 Vriendelijke groeten Ron
int. electric fly-in, Nijmegen, Netherlands
NL RC flying forum

[empeabee]

Steve Hargreaves

I Like the Engineers answer to the Astrologists chart

 Ron

Trust you

[Ron van Sommeren]

Sorry guys, it should read 'cubed increase in power' instead of 'squared increase in power'.

 Mike, ask your wife what 'truste' means

[Noel Branson]

I just wonder how Kevin survived this long...

[Noel Branson]

I still am an electronics engineer and was quite horrified when I read Kevin's letter. I was going to reply and set the record straight but having followed this thread I see that it's not necessary. I just hope someone at the magazine gets on to it before we have little clouds of smoke all over the country and a rush on ESC's!

[Peter Eve]

I think it's a great pity that a letter like this wasn't checked before publication.  There are not too many people out there who understand even the basics of Ohm's Law and such confident "corrective" statements only serve to confuse as this thread shows.  In itself the letter is right having assumed that the power is constant.  The article in the mag was also correct in the statement it made because it assumes that the resistance doesn't vary.  In the actual case, the article mentioned a receiver which ,may have an internal voltage regulator (the manufacturers could help here) in which case it would depend on how that corrected the voltage down to working level before we could say what will happen to the current drawn from the battery.

This is just another reason why I stick to I/C!!  How about a manufacturer comment to clear this up? 

 There was also the adamant letter about trimming and alignment.  Who do we believe on that?

 Let's not turn letters into slanging matche, please.

Pete 

[Frank Skilbeck]

What worried me was that the letter was written by an electrical engineer, I hope he's not designing my ESC's

[John Cole]

Kevin Annells assumes a constant-power situation when he contradicts Tim Mackey. What TM says is absolutely correct in describing a simple case: when a voltage (potential difference) is applied across a resistive circuit, increasing the voltage (PD) will increase the current (in direct proportion to the voltage (PD) increase).  Or as Ohm put it: V = I *R,  or Potential Difference (volts) = current (amperes) times resistance (Ohms!).

I think KA is probably wrong in his assumptions, because if you change the voltage fed to a radio system it does NOT necessarily take constant power when the servos move.  It's not an electrical issue.

In electrical systems, power is quoted in Watts.  Power is the product of current and voltage (= V * I). Power reflects the rate of using energy and doing Work. The energy or Work capacity of a battery is the product of its voltage and its current-capacity (the latter is how long it will deliver a certain current and is measured in ampere-hours). So the energy (or Work) capacity of a battery will increase if you increase its voltage OR if you increase its current capacity.  If you do both you get a double effect.

A radio system is not simply a resistive load, and the peak consumption is when the servos are moving (and they move against a modest force-load, using electrical energy and doing Work).  The Work they do is the product of the force-load and the distance moved.  The power they generate (equals the power they use, divided by their electrical efficiency) is the RATE at which they are doing Work (Work per unit time).  When you increase the voltage, the servo moves faster but it gets to its target position quicker (so it's working for a shorter time).  It's using more Watts but for a shorter period.  So if the forces are not speed-related its energy consumption does not necessarily increase.  Whether they are or not is not a question of Electrics but of Mechanics and Aerodynamics!  If the quicker servo movement increases the physical load on the servo (as it would if it were e.g. moving a paddle through the air, as air resistance increases with the square of the speed) then it will need a bigger battery as the Work done would also square, requiring an increase in both battery voltage and Ampere-hour capacity.  Working in a vacuum against just speed-independent friction forces it would not: the work done in a constant length flight might stay the about same so a smaller Ampere-hour capacity might suffice, as the increased volts increases the energy or Work capacity at constant Ampere-hours.

In the real world: put a big one in. Model helicopters don't fly well in a vacuum!