Hello Folks,
It's a little quiet here at work today
and I though I would use my lunch-hour to
write a little tutoral on electrical circuits
and wiring. A first installment is given
below. So here goes...
The electrical systems in our LBC operate on
direct current or DC. That means that
current flows in one direction only, unlike
our household power. Another way to think of
that is the electrons flow in one direction
only.
We're going to need to define a few terms
here and that can be perhaps done best by
including analogys.
Voltage - Voltage can be thought of as the
PRESSURE in the electrical circuit. This is
a good time to remember that the water in a
hose doesn't go anywhere if the flow is off,
regardless of the pressure.
Current - This relates to the VOLUME or
AMOUNT of current that is or CAN flow in a
circuit. One might think of this as related
to the diameter of the water hose.
Resistance - As the word implies, this is
the impedance to electrical flow. Think of
the knob on the water faucet (feeding the
water hose) as adjusting the resistance to
flow. Cut it off, infinate resistance, and
no current flows. Open it full, zero
resistance, and in theory, ALL of the water
(or current) will flow!!!
The parameters merge by the simple
relationship, E = IR
Of course, the engineers couldn't make it
easy so here's what that means: E is
electromotive force or VOLTAGE. I is
current and R is resistance. The units
needed to balance the equations are E in
volts, I in Amps and R is Ohms.
There is another pair of relationships that
can help. They are P=IE and P=(ExE)/R. P
is power in watts. The other terms are
explained above.
A quick example,
If a car's headlight lamp is rated at
55watts then given a few other knowns, we
can solve the puzzle.
P=IE is rewritten with the unknown on the
left as, I=P/E or as stated in words, the
current in amps is the quotent of power in
watts divided by voltage in volts. In our
example,
I = P / E
I = 55 / 12
I = 4.58 amps
The practical side of this is that one 55
watt headlight bulb draws about four and a
half amps of current at 12 volts. That is
like saying, to make a headlight bulb
produce light, a pressure of 12 volts moves
4.6 amps of volume through the (hot)
resistance of the bulb's filament. So what
is the (hot) resistance?
E = IR is rearranged to,
R = E / I
R = 12 / 4.58
R = 2.62 ohms
For completeness and to show that it works,
please remember that P=(ExE)/R. If we put
that back together as such,
P = (E x E) / R
P = (12 x 12) / 2.62
P = 54.96 watts (which is as close to 55
watts as one ever needs to get!)
So it all works.
Above I put (hot) in parenthesis because
the resistance of a lamp's filament changes
with temperature. As the temperature goes
up, so does the resistance. That is
exactly backwards from the way a
temperature sensor works for an electric
oil or coolent temperature gauge. The
substance in the sensor has an inverse
resistance profile. The resistance goes
DOWN as the sensor heats up. That's a
topic best left for another day.
We also need to discuss fuses versus the
current expected in the circuit and as
related to the gauge and length of the
wire in the harness. We can do that
later as well but the short story is that
the fuse used on a circuit must be large
enough to carry the component's load (and
the sometimes much larger impulse load
experienced when a circuit is first
energized) but not be so large that the
wire will "fuse" or melt before the fuse
fuses! There are lots of guidelines for
wire gauge and length (resistance increases
with length) but I can't put my hands on
them right now.
We can also talk about diodes later too.
They are devices that allow current to flow
through them in one direction only. More
next time if you find this of any value.
Happy Holidays!
Rick
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