Effect of the supply internal resistance

1V & 9V. :c)

Rich, The battery's internal resistance is in series with its ideal voltage. The internal resistance is always part of the circuit. If the load resistance is much larger than the internal resistance then it won't matter much, if not , its another resistor in your circuit. Draw the circuit being fed with your supply voltage and a resistor in series.

Tom

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Only when there\'s a load on the battery.

| V out |----------------------|------------------- | | | | | | R2 R1 | | | | |--------B-------- I R ------------------| X Y

B= ideal battery with zero internal resistance.

Lets say the battery voltage is 10.1V open circuit.

1 Ohm Internal Resistence (I R). R1 = 90 Ohms, R2 = 10 Ohms.

0.1 A of current flows.

10V between X and Y.

9V across R1. 1V across R2.

0.1V across I R. (V = 1A * 0.1 Ohm = 0.1 V.)

Wait, there is 10V across X and Y, yet if B represents an ideal battery with zero resistance, there should be 10V across I R!

The current is 0.1A a. So, using V= I * R voltage across I R is, 0.1A * 1 Ohm = 1V.

I know that the actual voltage across I R *is* 0.1V

But looking it another way there is supposed to be 10V across X and Y. Which is effectively 10V across I R.

Which

No, there's still 10V across the ideal battery. You can't ignore that when figuring your voltage drops between X and Y.

I think I've figured it.

Youve got to draw a center line down the circuit.

! !

Looking at the above circuit, because B is ideal battery and zero resistance, it looks like R1 and R2 are practically in parallel because I R (Internal resistance), is 1 Ohm.

But you must look at the left hand side of IR and then the right hand side.

Add up the voltageson either side and see what IR sees (Z & Y).

| V out |------------------- | - | | R1 9V | | + I R ------------------| Y

| V out |-------------------- | + | | R2 1V | - | |--------B-------- I R X - + Z 10.1 V

We see that there are two voltages as far as I R is concerned. And they are opposing each other.

On the right, I R sees + 9V

On the left, I R sees 10.1V - 1V = 9.1V.

So, there is *not* 10V across I R. There is 9.1V - 9V = 0.1V

So, there is 0.1V across X and Y. There is 1 Ohm between these points. Yet

*because of battery B* you cannot say that R1 and R2 are practically in parallel. Despite the low impedance of the suply source.

And I think this sort of stuff is going to explain Op Amps.

In fact there is something special about point Vout here. It's like a reference point. Like a ground.

Here is the way I learned to draw circuits: And simplify them in College:

R(int)=1 ohm |--/\\/\\/\\----------- +| | --- | | B | \\ |10v| R1 / 90 ohm |___| \\ | / -| |_____________ V(out) | | | \\ /\\ | / | | R2 \\ 10 ohm | | / | | | \\/ |__________________|____________ 0 volts

V(out) = (output / input) * V(source)

output is I * R2 Input is I *[(int) + R1 + R2]

then V(out) = (10 ohms / 101 ohms) * 10 volts v(out) = 0.9901 volts

Your equivelent source impeadance will be R2 parallel [R1 + R(int)]

R eq = product / sum = 10 ohms * 91 ohms / (101 ohms) = 9.01 ohms

There: at V(out) the voltage will be 0.9901 volts with an equivelent source resistance of 9.01 ohms

Shaun

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Read this:

sa3ro4l4ddcsv1c6mfdd6vv2tgmrpeadh7@4ax.com

JF

That's some wierd email address, and if I click it, Outlook express pops up.

link didn't work

Shaun

I'm slightly wong.

There is a point along the circuit where positive meets negative.

Going one way is like going into negative, going the other way like being into positive. Of course anywhere along the circuit one part can be considered more positive, or negative relative to another.

In the circuit above the center point, which is kind of a reference point is within R1. It's 40.5 Ohms into R1 on one side. Other side is 40.5 Ohms. Adding resistances from the center point either way adds up to 50.5 Ohms, so total is 101 Ohms. True center point, which I suppose is a reference point is total resistance divided by 2.

So, in my circuit Vout, is not the center point.

And the voltage across I R is + 5.05V on one side, + 4.95V on the other, so there is 0.1V across I R (internal resistance of battery).

Yet between point X and Y, you would measure 10V, and for a moment it looks there is 10V across I R, without closer analysis.

40.5 Ohms and 49.5 Ohms.

The supply internal resistance will give you a voltage drop, lowering the source voltage available at the battery that depends on the current you voltage divider and load require. V(battery) = V(10 volts) - I * R(internal). I is the total circuit current with your voltage divider and whatever is connected to it. V(10 volts) is your open circuit voltage.

Shaun

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What a surprise!

It\'s a USENET message ID, not an email address...

JF

No, you're considerably wrong, and/or trying very hard to confuse yourself (and the rest of us, too), and succeeding.

No, you are forgetting the voltage source within the battery - it provides 10.1 volts, and there will be 0.1 volts (in the opposite polarity) across the internal resistance, to give 10.0 volts between X and Y.

Let me steal the drawing that John Fields made in another post:

+---------+--E1 | | [1R] [90R] R1 | | +--Ebat +--E2 |+ | [BAT] [10R] R2 | | +---------+ | GND

I have added a "Ground" label so that we have a reference point to measure voltages against. You are confusing things by adding your "center point", rather than selecting a connection point as the zero volt reference.

In electronics, "Ground" might better be called "Common", or "Reference point" - it is the point in the circuit that the designer has decided to call "Zero Volts". In the above circuit, it is the negative terminal of the battery.

Assume that the unloaded battery voltage Ebat is 10.1 volts.

The total circuit resistance including the battery's internal resistance is 101 ohms. This means that the current in the circuit will be 0.1 amp, giving 0.1 volt across the battery's internal resistance, so the voltage across the battery terminals (between GND and E1) is 10.0 volts. There will be 9 volts across R1, and 1 volt across R2.

--
Peter Bennett, VE7CEI  
peterbb4 (at) interchange.ubc.ca  
GPS and NMEA info: http://vancouver-webpages.com/peter
Vancouver Power Squadron: http://vancouver.powersquadron.ca

NO! NO! NO!

You have the battery and R(internal), This is your source. Then you have your voltage divider and you want the voltage across the 10 ohm resistor. See Neds description of how to work out the circuit, it is the proper method that is taught in schools and books. You don't use an imaginary reference point, you're over complicating the circuit, the only reference point to use is battery neg (common).

Shaun

One think I'm shaky about is this concept of reference and ground. Especially when I try to grasp Op Amps.

Of course, I know that there is 0.1V across 1R (the internal resistance) not

10 Volts.

10.1 Volts at Ebat is referenced to ground. We can draw 1R as connected to that point. Of course that one side of 1R is 10.1V above ground is not that significant, if it is above ground is a fact that does not assist with the understanding of how to calculate the voltage across 1R.

What is significant is the voltage across 1R. In fact what is significant is what 1R sees as being it's source.The following circuit shows what that there across the interal resistance of the battery. It's usefullness is in showing more clearly how only 0.1V PD apears across 1R. It ignores any kind of reference point to ground, although in practice you may need to know the voltage across 1R and ground.

+---------+--E2 = 0.1V | | [10R] R2 | | | +---E1 | | [90R] R1 | | [1R] +--Ebat | |+ | [BAT] | | | | | | | +---------+

BAT is suppossed to be a device with zero resistance. So, you can erroniously conclude there is 10V across 1R, thinking one part of 1R is at ground. The voltage stated is a voltage *reference to ground*. One end of 1R is + 10V above ground or the negative terminal that's for sure. But in fact the other side of 1R is *not* connected to ground at all through a zero resistance, so there is not + 10V across R1. This shows how careful one must be when associating a zero resaistance value to some ideal component. I must be careful in doing that.

+-----------+--- 10V | | | [R1] [1R| | | | | | [BAT] [R2] | | | | +-----------+ | GND