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Homework Help: Kirchoff's Law - Wire in parallel with light bulb

  1. May 31, 2015 #1
    1. The problem statement, all variables and given/known data
    If you measure the resistance of the three lightbulbs (independently) and get the following values. What is the equivalent total resistance of the circuit, and it’s uncertainty.
    [tex] R_1 = (50 \pm 7) \Omega, \ \ R_2 = (720 \pm 30) \Omega, \ \ R_3 = 140 \pm 20) \Omega [/tex]
    Now you connect a bare wire (with negligible internal resistance) between to the left and right of L2 respectively in the circuit diagram. What will happen to the brightness of each bulb (i.e. will it increase, decrease, or remain the same)? Justify your results by finding the power of each light bulb in terms of ∆Vbat.

    2. Relevant equations
    [tex] \frac{1}{R_{eq}} = \frac{1}{R_{L2}} + \frac{1}{0} [/tex]

    3. The attempt at a solution
    [tex] \frac{1}{R_{eq}} = \frac{1}{720 \ \Omega} + \frac{1}{0} [/tex]
    [tex] R_{eq} = 0 \Omega [/tex]

    Knowing that the resistance in the top half of the parallel component of this circuit is now 0, does that imply that the current, I1, into the parallel component is now equivalent to I2, the current into L2 and the wire? I.E. is there any power going through L3 after the wire is connected?

    Thank you in advance!
  2. jcsd
  3. May 31, 2015 #2


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    Well, what do you think? What is the voltage across L2 (and L3)?
  4. May 31, 2015 #3
    I think that all (minus a negligible amount) the current will flow through the wire, making the Voltage drop across L2 and L3 zero.
  5. May 31, 2015 #4


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    Science Advisor

    Yes. Therefore, the voltage across L1 is...
  6. May 31, 2015 #5
    It should be equal to the voltage of the battery. Thanks for your help.
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