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Equivalent Resistance (Prep for Lab on DC circuits)

  1. Feb 9, 2013 #1
    1. The problem statement, all variables and given/known data
    Use the circuit shown to answer the following questions. At time t = 0 the switch in the circuit is closed. Answer the following questions after the switch in the circuit is closed.
    2cwq6hg.gif

    What is the equivalent resistance of the circuit given R1 = 5 , R2 = 5 , R3 = 4 , R4 = 5 , R5 = 1 and R6 = 2 and a 2 V battery?


    2. Relevant equations
    [itex]\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+...[/itex] for parallel
    [itex]R_{eq}=R_1+R_2+...+R_n[/itex] for series



    3. The attempt at a solution

    Alright, this didn't seem too complicated to me but all of my attempts have resulted in incorrect answers. The way I am seeing it is that [itex]R_4[/itex] and [itex]R_6[/itex] are parallel, and that the equivalent resistance of these two resistors are in series with the rest of the resistors. Here's my attempt:

    [itex]\frac{1}{R_{46}}=\frac{1}{5}+\frac{1}{2}=0.7Ω[/itex] Now, the reciprocal of this is ≈1.43Ω which gives the equivalent resistance of resistors 4 and 6.
    It seems like a simple matter here to add up this equivalent resistance to the rest of the resistors in series: 1.43+5+5+4+1=16.43Ω

    Am I incorrect in my parallel/series diagnosis? I have tried several other options, which I won't type up here, but if my diagnosis is wrong I will offer up my other thoughts. Thanks in advance.
     
  2. jcsd
  3. Feb 9, 2013 #2
    No, the equivalent of R_3, R_4 and R_5 is in parallel with R_6.
     
  4. Feb 9, 2013 #3
    Recalculating. Hadn't seen a resistor setup like that. I'm not sure I understand why they are in parallel. Is it because there is no wire between, i.e. they are connected to each other?


    Edit: Ok, that worked like a charm. Thanks.
     
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