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Getting the voltage to drop in a battery?

  1. Feb 10, 2014 #1

    mot

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    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
     
  2. jcsd
  3. Feb 10, 2014 #2

    Simon Bridge

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    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?
     
    Last edited: Feb 10, 2014
  4. Feb 11, 2014 #3
    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.
     
  5. Feb 11, 2014 #4

    Simon Bridge

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    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.

    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 :)
     
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