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Welcome back to Linear Circuits, I'm Dr. Harris.

The topic of this lesson is power, and by the of this lesson, you should be able

to define power, as well as understand how power relates to voltage and current.

Power is extremely important when you're designing electronic circuits and

devices because of power dissipation,

power requirements, so let's look at power and see what it means.

This lesson builds upon two things.

First is the concept of electric current, which is the quantity of charge that

passes through a given area in a specified time, and remember that current is dQ dt.

This lesson also builds upon voltage, which is the energy gained or

lost by a coulomb of charge, and remember that voltage can be written as dw dQ,

or change in energy per coulomb of charge.

So building upon those two concepts, what exactly is power?

Well remember that a charged particle q flows over time to produce a current

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i, dq/dt, and a voltage is produced by the energy that's either lost or

gained by that moving charge, or dw/dq.

The power is the rate at which that charges energy(W), changes over time.

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So, power is the rate of a charges energy changing over time, or dw/dt.

So if you look at our definitions for current, voltage and power, you see

that power can also be expressed as the product of the current and the voltage.

If you use the chain rule, you see that dw/dt is

the same as dw/dq times dq/dt, or v times i.

So power is the product of voltage and current.

We're going to use the variable p for power and the unit watts, which has

an uppercase W, not to be confused with the lowercase w that we use for energy.

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So I've giving that definition that power is the product of voltage and current.

Let's take a short quiz and calculate the power for

a known voltage and a known current.

Pause the video, solve these problems, and then we'll solve them together.

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We know that power is the product of the current and the voltage.

Now in this case the current is going downhill from plus to minus,

and as a mnemonic device I always use the sign that I get to first.

So a positive 5 amps times 2 volts, or

in this case, the power is a positive 10 watts.

Now because power is just flowing downhill, meaning the current

is flowing downhill, it's going, it's not taking any extra energy for

this current to flow from plus to minus, so this power is actually consumed.

Positive power is consumed.

Now let's consider p2.

We know that power is current times voltage, but in this case,

we have four amps of current going from minus to plus.

So it's taking some energy.

The current is going, it's opposing the direction that is really wants to flow.

It's flowing uphill, and again, I take the sign that I get to first,

which is a minus.

So I'll say a (-4A)(3V), or

the power is a -12W, and we required some

energy to get this current to flow from minus to plus.

Which means that, in this case, power is actually generated.

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So by our convention, if the current is flowing downhill from plus to minus,

and you have a positive voltage, positive current,

then we have a positive power, positive power is consumed.

In the other case, power is flowing from minus to plus, so

we end up with a negative power.

Remember that negative power is generated.

Another interesting fact about power, is that the sum of power generated and

consumed in a system is zero, because power is the rate

of change of energy and energy is always conserved,

the sum of power generated and consumed in a system is zero.

We know that energy is conserved, we're neither creating or destroying energy,

and since power is simply the rate of change in that energy,

the sum of power also has to be zero.

So take a look at this system.

You have an unknown power with three other sources

that are generating and consuming 3 watts, -7 watts, and 6 watts.

What is Px?

Think about it.

We have plus 3, plus 6, which is a plus 9, and

a -7 which means that PX must be -2,

because -2 watts and -7 watts totals -9 watts, and

plus 3 watts plus 6 watts equals plus 9 watts.

So, 3 and 6 is 9, 2 and 7 is 9.

PX must be a -2.

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We know that the sum of the powers, Px is unknown power

plus 4 watts plus 8 watts minus 6 watts equals 0.

So, 4 and 8 is 12 minus 6 is plus 6,

that means that, Px has to be a -6 watts.

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Again, we have some unknown power plus 2

watts minus 3 watts plus 4 watts equals 0

to a -3 is minus 1 plus 4 is a positive 3.

So Px has equal a -3 watts.

In this case,

we're looking for the current which is power over the known voltage.

So we have a -3 watts divided by 2 volts so our current is -1.5 amps.

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The key concepts that we covered in this lesson are first,

that power is the rate at which energy changes over time, or dw/dt.

It can also be expressed as the product which is most often the form of V and I.

Most of the time you use power as the product of a voltage and the current, and

finally, the sum of power generated and consumed in a closed system is zero.

So if you know all of the power generated or consumed of several of the elements in

a system, you can use those to find the power of an unknown element, and

you can also use that power to find an unknown voltage, or an unknown current.

Thank you.