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The three most basic
units in electricity are voltage (V), current (I) and
resistance (r). Voltage is measured in volts, current is
measured in amps and resistance is measured in ohms.
A neat analogy to help understand these terms is a system of plumbing
pipes. The voltage is equivalent to the water pressure, the current is
equivalent to the flow rate, and the resistance is like the pipe size.
There is a basic equation in electrical engineering that states how the
three terms relate. It says that the current is equal to the voltage divided
by the resistance.
I = V/r
Let's see how this relation applies to the plumbing system. Let's say you
have a tank of pressurized water connected to a hose that you are using to
water the garden.
What happens if you increase the pressure in the tank? You probably can
guess that this makes more water come out of the hose. The same is true of
an electrical system: Increasing the voltage will make more current flow.
Let's say you increase the diameter of the hose and all of the fittings
to the tank. You probably guessed that this also makes more water come out
of the hose. This is like decreasing the resistance in an electrical system,
which increases the current flow.
Electrical power is measured in watts. In an electrical system
power (P) is equal to the voltage multiplied by the current.
P = VI
The water analogy still applies. Take a hose and point it at a waterwheel
like the ones that were used to turn grinding stones in watermills. You can
increase the power generated by the waterwheel in two ways. If you increase
the pressure of the water coming out of the hose, it hits the waterwheel
with a lot more force and the wheel turns faster, generating more power. If
you increase the flow rate, the waterwheel turns faster because of the
weight of the extra water hitting it.
In an electrical system, increasing either the current or the voltage
will result in higher power. Let's say you have a system with a 6-volt light
bulb hooked up to a 6-volt battery. The power output of the light bulb is
100 watts. Using the equation above, we can calculate how much current in
amps would be required to get 100 watts out of this 6-volt bulb.
You know that P = 100 W, and V = 6 V. So you can rearrange
the equation to solve for I and substitute in the numbers.
I = P/V = 100 W / 6 V = 16.66 amps
What would happen if you use a 12-volt battery and a 12-volt light bulb
to get 100 watts of power?
100 W / 12 V = 8.33 amps
So this system produces the same power, but with half the current. There
is an advantage that comes from using less current to make the same amount
of power. The resistance in electrical wires consumes power, and the power
consumed increases as the current going through the wires increases. You can
see how this happens by doing a little rearranging of the two equations.
What you need is an equation for power in terms of resistance and current.
Let's rearrange the first equation:
I = V / R can be restated as V = I R
Now you can substitute the equation for V into the other equation:
P = V I substituting for V we get P = IR I, or P = I2R
What this equation tells you is that the power consumed by the wires
increases if the resistance of the wires increases (for instance, if the
wires get smaller or are made of a less conductive material). But it
increases dramatically if the current going through the wires increases. So
using a higher voltage to reduce the current can make electrical systems
more efficient. The efficiency of electric motors also improves at higher
voltages.
This improvement in efficiency is what is driving the automobile industry
to adopt a higher voltage standard. Carmakers are moving toward a 42-volt
electrical system from the current 12-volt electrical systems. The
electrical demand on cars has been steadily increasing since the first cars
were made. The first cars didn't even have electrical headlights; they used
oil lanterns. Today cars have thousands of electrical circuits, and future
cars will demand even more power. The change to 42 volts will help cars meet
the greater electrical demand placed on them without having to increase the
size of wires and generators to handle the greater current. |