How to get (v1+v2)/n for odd n with minimum resistors

[LongApple]

For example, the simplest case n=3 I am even stuck on much less the general problem we can start with that and evidently it may be able to be soled with only two or three resistors if you are clever

Also we are not allowed to use op amps only resistors and we are given two voltage sources.

My first plan was to use two resistors to get the average of the voltages and then connect that node with the average to more resistors- however, we can't use op amps and therefore I can't isolate that average of the voltages. I imagine we need to come up with the 1/3 or the 1/5 using a more clever and indirect way

 

[meBigGuy]

You are assuming V1 and V2 are equal. Why not just put V1 and V2 in series with a 2 ohm/1 ohm divider?

 

[Baluncore]

Seems to be one of those problems where you have k voltage inputs.
And you want an output voltage = Sum( V1 ... Vk) / n.
This general problem turns out to have a very simple solution. But it can take some people a long time to reach it.

Since this has now transmuted into a homework problem I won't give you the answer.
The first thing to realize is that the input voltages are all unknown, but must all be treated equally.

Start by selecting an arbitrary scale value of resistance, say r = 1k ohm.
Connect each of the k inputs to the output through a resistor having a value of k * r.
Does that meet the above requirement of treating all equally ?

Now, to get an output voltage that is 1/n of the average input, you will add one resistor to ground.
How might you work out the value of that resistor ?

 

[NascentOxygen]

Why not just put V1 and V2 in series with a 2 ohm/1 ohm divider?

Good thinking! Unfortunately, the problem statement doesn't assure us there is not a common earth.

 

[jim hardy]

V1 + V2 =N*SomethingOdd ? ... Well, anything_even -1 is odd.

 

[Baluncore]

Without an op-amp, the student, who now appears to have withdrawn, cannot use the negative resistance of an amplifier.

It is also odd that n must be an integer.

The reason why odd and integer were specified may suggest that all resistors needed be composed of series / parallel combinations from the same batch of resistors having the same value.

 

[jim hardy]

Voltage divider has been suggested

R1/(R1+R2) = 1/N ?

but i think in pictures , and since this is homework forum won't post my circuit

Spoiler: said i wouldn't

said i wouldn't

 

[berkeman]

the problem statement doesn't assure us there is not a common earth

Maybe, but it's very common in lab supplies and in simulations and with batteries, etc., to stack voltage sources in series...

 

[jim hardy]

Odd might be a red herring... ?

 

[NascentOxygen]

Maybe, but it's very common in lab supplies and in simulations and with batteries, etc., to stack voltage sources in series...

Sure, either solution should be fine. The solution with the fewer limitations should attract more marks.

 

[Baluncore]

we can start with that and evidently it may be able to be soled with only two or three resistors if you are clever

Now, let's turn that advice around. "... and evidently it may be able to be solved with only three resistors, or two if you are clever"
That really only changes the punctuation and implied brackets.

If the voltages could be summed by stacking then it would never need more than two resistors. (You are clever).

If the voltages are all ground referenced inputs, then for odd n, three resistors are necessary in all but the one case of n = 1. That cannot be solved without stacking.

 

[jim hardy]

If the voltages could be summed by stacking then it would never need more than two resistors. (You are clever).

and it'd work for any N .ge. 1, even or odd, integer or non-integer.