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Re: volts and amps



Everything John Denker says (below) is true, but it seems to me we ought to
say it in simpler terms. The misconception John is trying to clear up is
common enough that we ought to make the clarification simple enough to be
understood by persons not wanting to use words like impedance, Norton,
Thevenin, etc.

(1) A device using electricity as its energy source requires a certain
amount of power (energy per second).

(2) Although electrical power can be calculated as volts times amps, such
that 12 volts time 2 amps = 24 watts, most devices are designed for a
specific voltage. A device designed for 12 volts and 2 amps (24 watts)
typically will not operate properly from a power source that could deliver 6
volts and 4 amps (24 watts). (This has to do with the resistance of the
device.)

(3) Power sources (both batteries and plug-in supplies) deliver a particular
voltage and have a maximum amount of power they can deliver. (This has to
do with the resistance of the power supply which often depends on the
physical size of some of the critical components.)

(4) Therefore, to decide if a power source will work with a particular
device, first make sure the voltage is correct. Once that has been
established, then make sure the power supply is capable of delivering *at
least* as much power as required. If a device requires 12 volts and 24
watts (2 amps), it would be fine to use a 12-volt 24-watt supply. But it
would also be okay to use a 12 volt supply capable of 50 watts, or 100
watts, or 1000 watts. As long as the device is operating properly, and the
power supply is delivering the proper voltage, the device will not consume
more power (draw more amps) than it is designed to draw even if the power
supply were capable of providing more power. (This has to do with the
resistance of the device.)

Ignoring inefficiencies (like heating), it does not make any difference if a
kindergarten child lifts a book from the floor to the table, or if a
football linebacker lifts the book from the floor to the table. We might
want to reserve a greater task for the linebacker, but if he happens to be
available, asking him to pick up the book is not a mistake. On the other
hand, the linebacker could lift a box of encyclopedias from the floor to the
table whereas we ought not ask the kindergarten student to do that.

(5) Hence, the example John gave... A 12-volt flashlight light bulb
requiring low power (perhaps one watt) will run okay on D cells or on a car
battery. The car battery is "overkill" (capable of more), but it works just
fine. However, a 12-volt car headlight requiring high power (perhaps 50
watts) will operate okay on the car battery, but not the D cells. The D
cells cannot supply enough power (cannot provide enough current).


Michael D. Edmiston, Ph.D. Phone/voice-mail: 419-358-3270
Professor of Chemistry & Physics FAX: 419-358-3323
Chairman, Science Department E-Mail edmiston@bluffton.edu
Bluffton College
280 West College Avenue
Bluffton, OH 45817

John Denker wrote:

That statement is not correct, thereby illustrating the problem with the
previous statement. In fact, a flashlight battery rated for 12V will work
just fine on a car battery.

The converse would have made a better example: If you take a car headlight
and attach it to a 12V supply consisting of 8 C-cells in series, it isn't
going to work.

As Ludwik pointed out, a power supply can be characterized by its
a) open-circuit voltage, and
b) impedance.

For linear devices, this is equivalent to characterizing it by
a') open-circuit voltage, and
b') short-circuit current.

As they say, two points determine a line..... The word "Thevenin
equivalent" and "Norton equivalent" should be mentioned somewhere near
here....

It is standard good practice to use a first-order model to describe real
voltage sources such as batteries. That is, the battery is modelled as an
_ideal_ constant-voltage voltage source in series with an ideal impedance.

.
.
.
V0 R0 . V1
[ideal voltage source]--------[impedance]---X----------|
| . |
| . |
| . [load] R1
| . |
| . |
| . ground |
|---------------------------------X----------|
.
.
.

where everything to the left of the dotted line represents the
battery. The X symbols mark the terminals of the battery, which is a
two-terminal device.

For homework, consider the open-circuit voltage (V0) to be fixed, and
consider the load (R1) to be fixed.
A) Calculate the battery-terminal voltage (V1) as a function of battery
impedance (R0).
B) Calculate the power delivered to the load as a function of battery
impedance.

Explain why, even though this is properly called a "linear circuit", each
of the foregoing functions is nonlinear.

Explain why, as Ludwik said, if the impedance is "low enough" making it
lower doesn't hurt anything.