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 :)

 

[HalfLostAndSearching]

I would approach this problem by first understanding the fundamental principles at play. In this case, the concept of a voltage divider is key. A voltage divider is a circuit that divides a voltage into smaller, proportional voltages. It consists of two or more resistors connected in series, with the output voltage taken across one of the resistors.

In this scenario, we are given a 9.0-V battery and we want to create a voltage divider that produces a 3.5-V output. This means that the output voltage should be 3.5/9 times the input voltage. Using Ohm's law (V=IR), we can determine the total resistance (R) needed to achieve this output voltage.

V = IR
3.5 = I * R
R = 3.5/I

Now, we are given that the individual resistors have a resistance of 1.0 Ω. This means that we can calculate the number of resistors (n) needed to achieve the total resistance (R) calculated above.

R = n * r
3.5/I = n * 1.0
n = 3.5/I

Finally, we need to determine the value of current (I) that will give us the desired output voltage of 3.5 V. This can be done by using Ohm's law again, with the total resistance (R) and the input voltage (V).

V = IR
9 = I * R
I = 9/R

Now, we can plug in this value for I into the equation for n, giving us the number of resistors needed to achieve the desired output voltage.

n = 3.5/(9/R)
n = 3.5R/9

Plugging in the value for R calculated earlier, we get:

n = 3.5 * (3.5/I)/9
n = 3.5 * 3.5/9 * 1/I

This means that we need 3.5 * 3.5/9 * 1/I resistors to achieve the desired output voltage of 3.5 V. However, we are given that the individual resistors have a resistance of 1.0 Ω, which means that the total resistance (R) is equal to the number of resistors (n).

Therefore, to achieve the desired output voltage of 3.5