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Re: [vox-tech] ohms law
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Re: [vox-tech] ohms law



On Fri 02 Feb 07, 12:34 PM, Jimbo <evesautomotive@charter.net> said:
> Greetings:
> This is a technical question that concerns dc voltage so hopefully someone 
> that has knowledge in this area can help me with this.
> 
> I know that computers use low dc voltage of 12, 5 and 3.3 volts so 
> hopefully this will fit the mail list criteria.
> 
> I am a mechanic by trade.  I am good at diagnosing electrical and 
> drivability.  I have seen a few times that high resistance in the negative 
> leg of a circuit can take out components like computers, modules and even 
> not-so-complicated devices like bulbs and switches.  What I don't 
> understand is why.  Ohm's law states that E=IXR.  If this is the case then 
> if resistance is high it will decrease amperage.  I would tend to think 
> that just the opposite would happen...component would just lose power and 
> not fry.
> 
> Please enlighten me,
> 
> Jimbo 




Hi Jimbo,

This is the teacher in me.  Ohm's Law doesn't state E=I*R.  That's the way
to get into big trouble when you're trying to calculate things.

Ohm's Law states that (for a linear device) the *change* in potential is
equal to the current times resistance:

   \Delta V = I R

Your voltmeter doesn't actually read potential (measured in volts).  It
actually reads a potential difference (also measured in volts) between two
points in a circuit.

There's actually no such thing as "a" potential at "a" place.  The potential
itself can be anything you like.  For example, you can declare to the world
that your battery is 99999999999.5 volts, and you'd be correct.  The only
provisio is that the potential difference between the two terminals is 1.5
volts, so your negative terminal would be at 99999999998.0 volts.



Consider a battery hooked up to a home stereo and speakers:



                         R=4 Ohms           R_s=2 Ohms
E = 12v -----------------****---------------****--------------0


Here, *** represents a resistor, 0 is ground.   R is the internal resistance
of your home stereo system (which we'll pretend runs on DC from a battery)
and R_s is the resistance of your speaker.  Suppose your speaker is rated
at 8 Watts, but will blow if it receives 10 Watts or more.

We'll apply Ohms law between the battery and ground to obtain the total
current through the system:

         Delta V
   I =   -------
         R_total

where "Delta V" is the change of potential between the battery and ground
(which we'll take to be at 0 volts) and R_total is the total resistance of
the circuit.  So:

         Delta V      E         12 v
   I =   ------- = --------- = ------ =  2 Amps
         R_total    R + R_s    6 Ohms

Now, the question is, how much power is your speaker getting?  Let's first
calculate the potential drop across your speaker:

   \Delta V_s  =  I * R_s  =  2 Amps * 2 Ohms  =  4 volts

and now the power delivered to your speaker (note the Delta V_s means "the
potential drop across only the speaker):

   P_s  =  I * Delta V_s =  2 Amps * 4 volts  =  8 Watts



Imagine that your speaker's internal resistance increases for some reason.
Perhaps the wiring gets frayed.  Perhaps the contacts get dirty (dirt and
corrosion increase resistance.  I've had to replace the cables to the
alternator on my motorcycle because dirt in the contacts between the
alternator and regulator raised the resistance between the tangs of the
contact, increasing the heat, and melting the contact.  I have replace those
cables about every 3 years).  Even heat increases resistance (usually).  For
whatever reason, your speaker is now 8 Ohms.



                         R=4 Ohms           R_s=8 Ohms
E = 12v -----------------****---------------****--------------0


In this scenario, the battery is under slightly more load.  In real life, E
will be slightly less than 12 volts, but just ever so slightly (if we added,
say, 100000 Ohms, then the real characteristics of the battery would come
into play, but for these numbers, and a typical battery, E would still
essentially be 12 volts.

Let's redo the calculation.  Here's the amount of current flowing through
your stereo and speaker:


         Delta V      E         12 v
   I =   ------- = --------- = ------- =  1 Amp
         R_total    R + R_s    12 Ohms


Sure enough, the current decreased, as you said it would.  By half, even!
But look at the potential drop across your speaker:

   \Delta V_s  =  I * R_s  =  1 Amp * 10 Ohms  = 10 volts

This is the heart of the matter.  The current might have dropped, but your
speaker now has a larger potential difference across its terminals (its
"voltage increased").  By quite a bit -- by a factor of 150%!

And take a look at the power delivered to your speaker:

   P_s  =  I*Delta V_s  =  1 Amps * 10 volts  =  10 Watts

Oops!  The speaker has now blown.  I hope it had a fuse!    :-)



The main issue here is not so much current as it is power delivery.  It is
very possible that the current goes down and the power delivered goes up (as
it did here).  You can even find out the resistance that would maximize the
power delivery to your speaker, but this uses calculus, so I'll end here.

Hope there are no math mistakes in here for the world to see in the
archives.   ;-)

Pete

-- 
How VBA rounds a number depends on the number's internal representation.
You cannot always predict how it will round when the rounding digit is 5.
If you want a rounding function that rounds according to predictable rules,
you should write your own.
              -- MSDN, on Microsoft VBA's "stochastic" rounding function

Peter Jay Salzman, email: p@dirac.org web: http://www.dirac.org/p    
PGP Fingerprint: B9F1 6CF3 47C4 7CD8 D33E  70A9 A3B9 1945 67EA 951D
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