Getting the voltage to drop in a battery?

[mot]

Homework Statement

Suppose that you have a 9.0-V battery and wish to apply a voltage of only 3.5 V. Given an unlimited supply of 1.0-Ω resistors, how could you connect them to make a “voltage divider” that produces a 3.5-V output for a 9.0-V input
r=resistance of 1 resistor=1
R=Total resistance
v=3.5V

Homework Equations

V=IR

The Attempt at a Solution

Note: The correct answer is use 18 resistors and measure the voltage across 7.
I really have no idea how to do this...
The current has to be the same across one resistor as it is over the entire circuit.
V=IR
I=V/R=v/r
9V/R=v/1Ω=3.5V/(n*1Ω), where n is the number of resistors
I got to a point of 3.5/9 *R=n, and I plugged in their answers and it works, but so does using 36Ω total and 14 resistors. I just don;t know what to do with no current...

I've tried a lot to no avail, so not much of a point of posting it here.

Help please?
Thanks!

Homework Statement

Homework Equations

The Attempt at a Solution

 

[Simon Bridge]

Technically you also need to know the total resistance of the load - to make sure you deliver 3.5V to the load.
But the question just wants to have an open-circuit voltage of 3.5V - i.e. the volt-drop across some part of the circuit must be 3.5V

You can reason it out:
If you have 1 resistor - the the volt-drop across is it 9V ... too much.
If you used two of them, the volt drop across each one is 4.5V - promising.
Notice that the actual value of the resistors does not matter?

If you used 9 resistors, what is the volt drop across each one?
At this point the solution should present itself.If you prefer to do it algebraically - then consider you have a total of N 1-Ohm resistors in series with voltage V and you want the voltage across n<N of them to have voltage U.

This is the standard voltage divider circuit with two resistors ... one of n-Ohms and the other of (N-n)-Ohms.
U is taken across the n-Ohms resistor, giving relation:

V/U=N/n

Which means that (3.5)N=(9)n which means that 7N=18n which can be written (N/18)=(n/7)

One equation - 2 unknowns: need another equation!
This is where you got up to.

We do know more than that about N and n!
We know that N is minimum and both N and n are integers.
So what is the smallest integer value of N that allows n to be an integer in that relation?

 

[willem2]

So what is the smallest integer value of N that allows n to be an integer in that relation?

Ydon't need integers. It's possible to make resistors of 11/2 and 7/2 ohm with only 12 resistors, or 11/3 and 7/3 with only 11.

 

[Simon Bridge]

Ydon't need integers.

For N and n? (these were defined to be the number of resistors) ... no I suppose you could just saw a resistor in half and seat a new lead! ... you mean I don't need integer resistances ... then: no, this is correct.

It's possible to make resistors of 11/2 and 7/2 ohm with only 12 resistors, or 11/3 and 7/3 with only 11.

There are configurations of resistors other than series - that is correct.
But I did make a mistake though - N does not have to be minimum ;)

It is not what OP was asking about exactly but would make for a more elegant solution than the one provided :)