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Unknown resistance

  1. Oct 17, 2006 #1
    This problem seems pretty straightforward.. and yet I haven't made any significant progress at all.

    "A parallel combination of an 7 resistor and an unknown resistor R is connected in series with a 12 resistor and a battery. This circuit is then disassembled and the three resistors are then connected in series with each other and the same battery. In both arrangements, the current through the 7 resistor is the same. What is the unknown resistance R?"

    So I drew the circuit out, and I concluded that for the last statement to be true, the equivalent resistances must be the same (since every resistor in a series circuit feels the same current, by virtue of being in series.) So I wrote

    [(1/7) + (1/R)]^-1 + 12 = 7 + 12 + R

    Unfortunately this equation is woefully uninformative, as when it is simplified, it yields 1 = 1. Could I have a hint or two from the masters?

    Thanks,

    Stephen
     
  2. jcsd
  3. Oct 17, 2006 #2

    OlderDan

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    You have to achieve the same potential difference accross the 7 ohm resistor in both cases. Get an equation that expresses this voltage in terms of the unknown R for each circuit, and set them equal.
     
  4. Oct 18, 2006 #3
    Why would the potential drop be the same for the 7 ohm resistor in both resistors? In series, wouldn't the voltage drop depend on where the 7 ohm resistor is connected?
     
  5. Oct 18, 2006 #4

    SGT

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    The voltage drop on a resistor is RI. Since the problem states that the current is the same in both arrangements, the voltage drps must be equal, as OlderDan said.
     
  6. Oct 18, 2006 #5
    So, here's what I have..

    for the first circuit,

    V = I1(7Ω) = I2(R); I1 + I2 = Itot

    for the second circuit,

    V = (7Ω + R)*Itot

    so we have I2(R) = (7Ω + R)*Itot; R = Itot (7Ω) / I2. The thing is, Itot is defined in terms of the resistance and the voltage drop.. so I keep getting an identity (i.e., 1=1). So I really don't know what to do.. could you guys help me out?
     
  7. Oct 18, 2006 #6

    SGT

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    No, for the second circuit V = 7I1, just as for the first. The difference is that or the second circuit the current I1 passes also through the series connection of the two other resistors.
    Since the battery is the same, the voltage drop on the two known resistors must be equal to the voltage drop in the 12 ohm resistor for the first circuit.
     
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