🚀 go-pugleaf

>	Could someone please explain why it is better to have a circuit with
>high amps and low volts rather than the other way around?  Does increased amps
>lead to increased power in the same proportion as increased voltages leads to
>increased power?  Why are car batteries touted as having more cranking amps, 
>isn't your amperage held constant by:

>			V = I R  V = 12, R = Constant  Power =  V^2/R
 
>	What am I missing here? A brain perhaps?
>	    Thanks in advance, -Mark


The fact that you can ask the question is indication enough that your brain is
functioning quite nicely, thank you; one of my biggest problems as an
electronics instructor has been to get students to *ASK QUESTIONS!!!!* about
anything they don't completely understand.  Now, dragging out my standard
"Voltage, Current, and Power" lecture.....

The basic definition of power is voltage times current, or, in equation form:

                     P = V * I

(The power dissipated in a circuit element is equal to the potential across
that element (a resistor, say) in volts, multiplied by the current through
that element in amperes).  Now, since Ohm's Law gives us the relation between
voltage, current, and resistance (or, in the more general form, impedance),
we can substitute into the above as follows:

             IF we know voltage and resistance, then

                    I = V/R, so  P = (V/R)*V, or P = V^2/R

             IF we know current and resistance, then

                    V = I * R, so P = (I * R) * I, or P = I^2 * R

Either expression is equivalent to the original definition of power, through
Ohm's Law.

Obviously, then, increasing either the current through a resistance, or
the voltage across it, will increase the power which that resistance
dissipates.  But, equally obviously, increasing either or these also increases
the other, by Ohm's Law.  They're really two ways of looking at the same
thing - there is really no distinction between "increasing the voltage across
the resistor" and "increasing the current through a resistor", as Ohm's Law
shows that these quantities are always directly related.  (One classical
analogy is water - electrical current is analogous to gallons per minute,
or some other volume/time expression, while electrical potential (voltage)
is analogous to water pressure.  You can get the same "power" from a stream
of water in either of the following cases - a large pipe (high current) with
relatively low pressure (voltage), or a small pipe (low current) with 
high pressure (voltage).  The size of the pipe is analogous to the resistance.

In theory, then, there is no difference, power-wise, between high voltage/low-
current situations, and the other way around.  However, in practice, it is
much simpler to deal with high currents than to deal with high voltages.  High
currents simply require bigger conductors.  High voltage, on the other
hand, can require sophisticated insulation and isolation schemes to "keep
the electricity where it belongs", as it wants to keep breaking down (ionizing)
such common materials as the surrounding air, and hence "getting away".  In
addition, high voltage can be extremely hazardous.  You've probably heard that
it's current that's dangerous, and that's true; but people are pretty high-
resistance items, and it takes a fairly high voltage to force a dangerous
level of current through dry, unbroken skin.  Voltages on the order of the low
tens of volts are considered "safe", as they are not high enough to cause
appreciable current flow through the body.  You don't feel much if you grab
the terminals of your car battery by accident, but you'll certainly feel the
line current from an AC outlet - and yet the battery is "capable of supplying
more current".  It's just that it can't supply that current through a high
resistance, since it's voltage it too low.

Now, what about the "cranking amps" rating of the battery?  Well, it's true
that all car batteries are 12V (actually, more like 13V) devices; what
distinguishes them is their internal resistance.  A battery can be modelled
as a voltage source (something which produces a constant voltage across its
terminals, regardless of the current through it), in series with a resistance:

                      ____/\/\/\/\/\________  +
                   +  |      Rint
                     ___
                   |     |
                   |12V  |                    Vout
                    -----
                      |
                   -  _______________________  -

Note that with no external resistance connected, the voltage Vout is exactly
the voltage of the source: 12 volts.  However, as soon as any resistance
is connected, and current begins to flow, there is a drop across the internal
resistance, and the voltage Vout must be something less than the source.
If the load resistor is very small, then Rint acts to limit the current which 
the source supplies.  The  worst case is, of course, a zero-ohm load - a short
across the terminals.  In this case, the maximum current (and so the max.
current that the battery can EVER produce) is  12V/Rint.

All batteries have some internal resistance; it's part of the physical
structure of the battery.  Obviously, the lower this resistance, the better,
at least from the standpoint of our desire to have an "ideal" voltage source.

Hope this has helped!


Bob Myers  KC0EW   HP Graphics Tech. Div.|  Opinions expressed here are not
                   Ft. Collins, Colorado |  those of my employer or any other
{the known universe}!hplabs!hpfcla!myers |  sentient life-form on this planet.