110 volts or 12 volts on your pocketbook?

110 volts or 12 volts on your pocketbook

I have tried everything that I can think

of to prove

that alternative energy

should be our future.

Except science.

This time let us try a little science,

very little because I am not a scientist.

Therefore, I am using these two links.

Electrical current is measured in amperes

(Amps),

Amperage is the amperage that a light bulb is designed to operate

at under normal conditions

And

At the voltage, the bulb was designed to operate at

Almost all bulbs

have their power wattage

reported on them,

For example

the bulbs can be 40, 60, 100, etc Watts.

This number is

the POWER

used by the bulb.

The power consumes,

P,

is related to the operating voltage,

V,

and

the current drawn, I, as the following:

Power (P) = Voltage (V) x Current (I)

So to find the current drawn,

simply divide the power by the operating voltage.

In the United States,

most household appliances work at 110 Volts.

Therefore,

the calculation is divided the Wattage by 110.

For example:

A 40- Watts bulb draws 0.36 Amps to operate.

A 60 -Watts bulb draws 0.54 Amps to operate.

An 80 -Watts bulb draws 0.72 Amps to operate.

A 100-Watts bulb draws 0.90 Amps.

https://snapguide.com/guides/calculate-the-electric-current-drawn-by-light-bulb/

A 12-volt light bulb runs off a battery, solar, wind

and/or

Other alternative power (Watts).

I = P/V or I = 100 W/6 V = 16.67 amps

What would happen if you use a 12-volt battery and a 12-volt light bulb to get 100 watts of power?

I = 100 W/12 V = 8.33 amps

In an electrical system, increasing either the current or the voltage will result in higher power.

Let us 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 I = P/V,

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.

Therefore, you can rearrange the equation to solve for

I –

and substitute in the numbers

I = P/V or I = 100 W/6 V = 16.67 amps

What would happen if you use a 12-volt battery

and

a 12-volt light bulb

to get 100 watts of power?

I = 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 = I*R*I, or P = I2*R

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.

To learn more about electricity

and

related topics, check out the links on the next page.

http://science.howstuffworks.com/environmental/energy/question5011.htm