# Getting the voltage to drop in a battery?

**1. The problem statement, all variables and given/known data**

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

**2. Relevant equations**

V=IR

**3. 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!

**1. The problem statement, all variables and given/known data**

**2. Relevant equations**

**3. The attempt at a solution**

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