# Internal Resistance and Watts

#### Pikester

Joined Nov 27, 2005
13
1) Every electrial components has some Internal resistance (because of friction, at least, thats what i think...). How would you find out these resistances?

2) Watts is just really confusing for me, I know that Voltage is the "potential energy", and Currents(or amps) is the actual energy that is release from the potential energy(correct me if I'm wrong), so where does that leave Watts? I thought that it was the pressure of the current, but people said that was wrong... So, what is it?

Thx

#### n9xv

Joined Jan 18, 2005
329
On the first question, what specific devices are you referring to? A battery has a value of internal resistance. A signal generator has a value of internal resistance/impedance.

To find the internal resistance/impedance of a voltage source;

Zs = (ZL(Vs - VL))/VL

This is a "Universal Ohm's Law." It applies to anything and everything that constitutes a circuit in the electronics relm.

Zs is the internal resistance/impedance of the source.

ZL is the resistance/impedance of the load.

Vs is the source "unloaded" voltage of the source.

VL is the voltage across the load.

An example is a resistor connected to a battery.

On the second queston, you seem to have a good understanding of the idea of voltage & current. Watts or power is simply the product of voltage and current.

Volts * Amps = Watts

Power is defined as the "time rate of doing work".

The "time" here is normally considered to be 1-second.

The rate is in terms of 6.25 * (10^18) electrons moving past a point in a circuit for a period of 1 second. The definition of an amp.

So, to put it together;

If 1-volt pushes 1-amp of current through a circuit then the resulting power consumed is 1-watt. More voltage and/or current means more watts etc.

#### Pikester

Joined Nov 27, 2005
13
So, Power is the amount of time to fill/pass thru a compontent? (kinda like the "pressure" of the current"?)

#### n9xv

Joined Jan 18, 2005
329
Well, power is the time rate that work is performed. The "time" is considered to be fixed at period of 1-second. The "rate" means X number of electrons moving past a point in a circuit or wire etc durring that time period of 1-second. Power relates the amount of "work" (electrons pushed) performed durring the specified time interval (1-second) with a given voltage. The work performed in any electrical/electronic circuit is the movement of electrons. If the electrons did'nt move then no work could be done.

Let me try a water analogy;

Voltage concept

A big tank holds X-gallons of water with a pressure of X lbs/sq. in.

Current flow concept

A valve (switch) is opened up to allow the water to flow through a pipe.

Now, we know that a voltage or pressure is applied and that current is flowing. But we need to harness the idea of how much "work" is being done by that process.

How much current/water flowing past a given point for a specified unit of time with a given pressure applied would be an indication of the actuall work being performed.

So,

lets say we installed a "flowmeter" that indicated how many gallons of water are actually flowing. This flowmeter is comparable to an ammeter that measures electric current. BTW, in electronics an amp-meter is called an ammeter.

We have a given pressure applied and X number of gallons flowing. Thus, we have establish the idea of voltage & current.

Now, tie it all together.

Power concept

If we said the water was flowing at a rate of X-gallons/second when the pressure was X-lbs/sq. in. in the tank then we have just given meaning to the concept of "work" performed by the flowing water.

This rate of flow per unit of time under a given pressure is a measure of work performed or power.

Are you with me?

#### Pikester

Joined Nov 27, 2005
13
So, using that concept, it's the X amount of water flowing at specific point in the pipe, at X amount of time? Or does the time ALWAYS has to be 1 second? Also, what happens when the the Xlbs/sq. is not consistent?

#### spork891

Joined Jan 17, 2006
3
Watt is defined as Energy per second, so the 1 second is merely implied in the same sense that there is ONE hour in miles per hour. It's a rate.

So if you wanted to know, for instance, the energy dissipated as heat in a resistor in 40 seconds and you knew that the resistor was dissipating 1/4 watts, then all you have to do is multiply the watts by 40 seconds to get the total energy: 10 Joules.

-Jason

#### n9xv

Joined Jan 18, 2005
329
Originally posted by Pikester@Jan 18 2006, 12:18 AM
So, using that concept, it's the X amount of water flowing at specific point in the pipe, at X amount of time? Or does the time ALWAYS has to be 1 second? Also, what happens when the the Xlbs/sq. is not consistent?
[post=13204]Quoted post[/post]​
Well, as spork91 pointed out, the "time" is implied to be a constant or fixed at 1-second. You could count the number of electrons flowing past a point in 3600-seconds (1-hour) but that in itself is not meaningfull. If the "pressure" or voltage is not consistent then the flow of "current" must be changing due to the changing pressure thats pushing it. This simply means the "power" consumption is varying accordingly.

With spork891's example, you could look at it in terms of "joules" of enregy.

X number of watts over a period of X amount of time = X amount of joules of energy.

#### Pikester

Joined Nov 27, 2005
13
Ok, so Power = the number of electrons that passed a speific point in a ciruit in 1 sec? Assuming the voltage stays the same.

#### n9xv

Joined Jan 18, 2005
329
You've basically got it. But the voltage can be varying which means a varing current which ultimately means that power is varying. With that scenario, you get into the concept of "average" power. Averaging all those variations. But to make clear your point, yes, X-amps flowing past a point for 1-second with X-volts applied = X-watts of power. Again its the aspect of time that ties together the idea of "measurable" work being done.

#### Pikester

Joined Nov 27, 2005
13
Cool, got it down now.

#### chesart1

Joined Jan 23, 2006
269
Originally posted by n9xv@Jan 19 2006, 08:45 AM
You've basically got it. But the voltage can be varying which means a varing current which ultimately means that power is varying. With that scenario, you get into the concept of "average" power. Averaging all those variations. But to make clear your point, yes, X-amps flowing past a point for 1-second with X-volts applied = X-watts of power. Again its the aspect of time that ties together the idea of "measurable" work being done.
[post=13250]Quoted post[/post]​
One point seems to have been lost here. Power is the product of volts and amps.
One watt of power is produced when one volt of electricity results in one amp of current.

If you multiply the power used in watts by the time in hours, your result is watt-hours. That is the basis of the electrical meter that the electric utiltiy company uses to bill you for the power you use.

A watt is defined as one joule per second.

Your statement should read: X-amps flowing past a point for 1-second with one-volt applied = X-watts of power.

John

#### n9xv

Joined Jan 18, 2005
329
Originally posted by chesart1@Jan 27 2006, 12:25 AM
One point seems to have been lost here. Power is the product of volts and amps.
One watt of power is produced when one volt of electricity results in one amp of current.

If you multiply the power used in watts by the time in hours, your result is watt-hours. That is the basis of the electrical meter that the electric utiltiy company uses to bill you for the power you use.

A watt is defined as one joule per second.

Your statement should read: X-amps flowing past a point for 1-second with one-volt applied = X-watts of power.

John
[post=13551]Quoted post[/post]​

I read my previous post over and over again. I thought thats exactly what it said.
I did'nt throw in the power X time = watt-hours, but the other statement reads the same as yours.

#### chesart1

Joined Jan 23, 2006
269
Hi,

Watts are not equal to (amperes per second) x volts.

Watts = amperes x volts.

Amperes = coulombs per second.

Therefore if you wanted to inlcude time in the equation for power, you would write:

watts = (coulombs per second) x volts.

John

#### n9xv

Joined Jan 18, 2005
329
Originally posted by chesart1@Jan 31 2006, 04:48 PM
Hi,

Watts are not equal to (amperes per second) x volts.

Watts = amperes x volts.

Amperes = coulombs per second.

Therefore if you wanted to inlcude time in the equation for power, you would write:

watts = (coulombs per second) x volts.

John
[post=13702]Quoted post[/post]​
Look at your above post again.

Watts = Amps X Volts

and,

Amps = Coulombs per Second

so,

Watts = (Coulombs per Second) X Volts

or simply,

Watts = Amps X Volts