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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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