fusible link questions

Sun Feb 7 05:34:30 GMT 1999

Actually, I have considered all of that stuff.  I have done those
derivations many times.  Read again what I wrote about it.

What you are missing is the difference between a source and a straight
resistance.  You can measure the voltage across a resister and the
current through it, multiply them together, and you'll know how much
power is being dissapated, and how much heat is being generated.

Let's do a thought experiment.  Let's say that you have a battery and
attach a one ohm resister to it.  You measure 14 volts across the
resister (and the battery).  Based on ohm's law, you know that 14 A is
flowing through the resister.  Then, use any of the equations (P=I*V,
P=E^2/R, or P=I^2R) to calculate that 196 watts are being dissapated by
the resister.  The resister is obviously getting hot.

How hot is the battery getting?  It has the same 14 amps of current
flowing through it.  It has the same 14 volts of voltage across it.  Why
isn't it dissapating 196 watts?  Because it is a source.

If you had an ideal source, there would be NO power dissapated by the
battery.  Anyone who has tried to start a stubborn engine will tell you,
however that the battery will get warm.  It won't get near as warm as the
starter, however.

A real life battery is represented by a perfect voltage source in series
with a resistance.  This resistance is refered to as the "internal
resistance" of a battery.

The internal resistance of a car battery is some small fraction of an
ohm.

Now, let's get back to our example.  The power dissapated by the resister
is 196 watts.  The power dissapated (not supplied) by the battery is
I^2R, where R is the internal resistance of the battery.  If the internal
resistance of the battery is 0.01 ohm, the battery will dissapate 1.96
watts while it supplies 196 watts to the external load (the resister).

Let's expand our model a little more.  If we have a battery with an
internal resistance of 0.01 that is supplying 14 V at 14 A into an
external load, the source voltage is 14.14V.  If you increase the load
(decrease the resistance), the output voltage will drop because the
increased current will increase the voltage drop across the internal
resistance of the battery.

Did you ever notice that putting a load on a battery causes its output
voltage to drop?  If not, take a look at the voltmeter next time you
start your car.

Now, back to this battery that is supplying 14V at 14A.  If you could
actually directly measure the voltages inside the battery, you would find
a 14.14V source and 0.14V dropped by the internal resistance.  Subtract
them and you get the 14V.

THE POWER DISSAPATED BY THE INTERNAL RESISTANCE OF THE BATTERY IS NOT THE
PRODUCT OF THE OUTPUT VOLTAGE AND CURRENT OF THE BATTERY.  IT IS THE
PRODUCT OF THE OUTPUT CURRENT AND THE VOLTAGE DROP OF THE INTERNAL
RESISTANCE.

Now, let's do something different.  Increase the resistance to 4 ohms and
reduce the current to 7 amps, and increase the voltage supplied by the
battery to 28 V.  You still have 196 watts being dissapated by the load. 
The current has been cut in half, however.

Doing some calculations, you'll find (with the same 0.01 ohm internal
resistance) that the voltage of the ideal source is 28.07 volts, so the
voltage across the internal resistance is 0.07V.  Any way you slice it,
using any of the three power equations discussed, the power dissapated by
the internal resistance is 0.49 Watts.  Compare this to the 1.96 watts in
the earlier example.

THE POWER DISSAPATED BY THE INTERNAL RESISTANCE REPRESENTS THE "LOSS" OF
THE BATTERY.

OK... so we see that doubling the voltage while supplying the same power
will cut the losses by a factor of 4.

So, the alternater will run cooler while providing the same power (at a
fraction of the current).

Now, for a final example, we'll keep the origional rated current of 14
amps and double the voltage to 28 volts.  The output power is now 392
watts, or twice the origional 196 watts.  The power dissapated by the
0.01 ohm internal resistance is 14*14*.01=1.96 watts.  We have twice the
power supplied to the load, but the same internal losses, which means the
same heat generated.  Our efficiency has gone up.

Try that with a tenfold increase in voltage and you'll find an even
bigger increase in efficiency and capacity.

To apply this to an alternater, we need to either look at the average
output of the three stator windings, or an instantaneous output.  It
doesn't matter - it applies equally well.  The ideal source voltage is
the voltage supplied by the moving magnetic field.  The internal
resistance is the resistance of the windings and the diodes.  There is
some heat generated by the field coil and the hysteresis losses in the
iron, so we won't be able to get quite as much increase as calculated. 
We engineers are used to things being less than ideal in practice,
however.

So... I have spent too much time writing this treatise.  If anyone wants
to argue, I would recommend that they have the courtasy of reading the
entire thing first.  If a point is brought up that has been handled in
this message, I'll simply point back to the message.  I have better
things to do with my time than to repeat myself.

Ray Drouillard, BSEE

On Sat, 06 Feb 1999 16:05:51 -0600 John Hess <johnhess at cris.com> writes:
>Actually, you haven't really considered that:
>
>while P=IE (the basic formula)
>E=IR, therefore substituting IR for E:
>P=I*IR or I^2R
>I=E/R, therefore substituting  E/R for I
>P=E*E/R or E^2/R, etc.
>
>To say that one is not related to the other is to refute Omm's Law. 
>Note
>that this is not electronics theory, it is _LAW_!
>
>
>Raymond C Drouillard wrote:
>
>> >requires closer to 2.5 HP. Pushing the same current at higher voltage
>> >quickly oveheats the alternator due to the increase in these I^2 R
>> >losses, which increase at the rate of a power of two with voltage
>> >increase - double the voltage, 4 times the heat with the same 
>current.
>>
>> I beg to differ, but the I^2 R losses are independant of the output
>> voltage of the device
>>
>> P = I^2 R (hense the name)
>>
>> That is why power is transmitted using the "high tension" (High
Voltage)
>> lines.
>>
>> If you are stuffing current through a resister, the power dissapated
by
>> that resister is dependant only on the current through the resister
and
>> the resistance.
>>
>> P = I^2 R
>>
>> Let me explain...
>>
>> Power is defined as voltage * current
>>
>> P = IV
>>
>> That is, the current through the resister * the voltage across the
>> resister.  The voltage across the resister can be calculated using
Ohm's
>> law.
>>
>> V = IR
>>
>> Taking the origional P = IV equation and substituting IR for the V
yields
>>
>> P = I * (IV) = I^2 R
>>
>> What I am proposing uses exactly the same princable that has always
been
>> used by the electric suppliers to get power from the source to the
>> substations, and finally to your home.
>>
>> One thing that neither of us mentioned is that some of the heat is
>> generated by hysteresis losses in the iron core.  That will somewhat
>> reduce the ultimate power that can be supplied by the system, but
it'll
>> still be more than the unmodified system will provide.
>>
>> Ray Drouillard, BSEE

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