Re: alfa-digest V9 #398

[George Graves <gmgraves@xxxxxxxxxxx>]

Not really, no.

Since power (P) = I  X E:

12 X 10 does indeed equal 120 watts

BUT if the voltage drops to 8 volts

8 X 10 = 80 Watts

However, because of Kirchoff's law, the total voltage drops across all 
the elements in the circuit must add up to the input voltage (in this 
case, 12 volts with respect to the battery return). This means that the 
other 4 volts will be dropped by the 'bad connection' which added an 
additional 50% to overall resistance of the circuit. So schematically,  
you would represent the motor and the 'bad connection' as two resistors 
in series across 12 volts. If one were to measure the voltage between 
the first resistor (the 'bad connection') and the second resistor (the 
motor) with respect to the battery return, one would read 8 volts, that 
means that the 'bad connection' dropped the voltage by 4 volts, and 
that the motor is now only seeing 8 volts. If you measure the voltage 
after the second resistor, you will get ZERO volts and that satisfies 
Mr. Kirchoff (a 4 volt drop PLUS and 8 volt drop equals a 12 volt drop, 
and 12 volts MINUS 12 volts = 0).

Now, one thing that I forgot to mention before. We're dealing with an 
electric motor here, so there is a mechanical component to factor in. 
If one puts an electric motor under load, it draws more current than 
does spinning freely. This is because the motor winding is an inductor 
and the slower the motor runs, the higher the inductive load because 
each leg of the coil is energized longer (it's also, on a more basic 
level, a conservation of energy thing). If one has a bad bearing or a 
seized motor shaft, the current that the motor normally draws when 
running under no-load conditions will increase dramatically. I had this 
graphically illustrated to me with a bad GTV-6 fan. The motor on one of 
my two identical fans had a bad bearing. The other (good) fan drew only 
a couple of amps at 12 volts, but the bad fan which ran quite slowly 
due to the bad bushing, got very hot and drew more than 12 amps which 
would and did blow the fuse.

George Graves
'86 GTV-6
now with 3.0 liter 'S' engine
and Power Steering


On Wednesday, April 9, 2003, at 06:36 AM, alfa-digest wrote:


> ----------------------------------------------------------------------
>
> Date: Tue, 8 Apr 2003 17:15:15 +0100
> From: Jonathan Coates <jon@domain.elided>
> Subject: Ohm's law numbers
>
> Originally, well almost - its getting a bit old now - George wrote:-
>
> Date: Mon, 7 Apr 2003 18:57:27 -0700
> From: George Graves <gmgraves@domain.elided>
> Subject: Re: alfa-digest V9 #395
>
> No. Let's make up some figures to show how this works. Let us, for the
> sake of argument, say that the resistance of the motor is 10 Ohms (It
> doesn't matter for our purposes, the relationships will be the same no
> matter what the motor resistance is. We just use 10 Ohms because it
> makes the math easier). since:
>
> I =E/R then
>
> I = 12/10 or 1.2 Amps
>
> Now let's say there is a bad connection in the circuit, and this
> connection is adding 5 Ohms to the 10 Ohms of the motor.
>
> since resistance in series is additive, 10 Ohms (the resistance of the
> motor alone) + 5 Ohms (the resistance added by the poor or corroded
> connection) = 15 Ohms, so:
>
> I= 12/15 or 0.8 Amps
>
> Which, as you can see is LESS current than the motor alone (1.2 Amps).
> This decrease in current will increase the voltage drop across the
> circuit and the motor will, as a result, now be seeing less than the
> full 12 volts, and will likely run more sluggishly than normal, but it
> won't contribute to fuse blowing. A partial short on the other hand
> will have the opposite effect: Let's say that the motor has a short in
> one of the motor windings, and this lowers the resistance of the motor
> from 10 to 7.5 Ohms:
>
> I=12/7.5 or 1.6 Amps
>
> If this car has a 1.5 Amp fuse in this circuit, then this condition
> would likely blow it.
>
> Clear now?
>
> George Graves
> '86 GTV-6
> now with 3.0 liter 'S' engine
> and Power Steering
>
>
> Hmmm
> So the motor needs, lets say 120 watts of power, thats 10 amps at 12 
> volts - looks OK if the fuse is 12 Amp.
> BUT if there is 50% more resistance in the circuit, the motor only 
> sees 8V.
> 120/8 = 15 Amps.
>
> With an additional 50% resistance the amps go up by 50%
>
> Make sense to anybody?
> Jonathan Coates
> ( Grinning )
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