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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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