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Homework Help: Simple circuit - Finding Equivalent Resistance

  1. Jan 21, 2009 #1
    1. The problem statement, all variables and given/known data
    Find Vo in the circuit.
    Image of Circuit Diagram Attached
    Variables: Vo, Io
    Given: Current (15 mA)
    Resistors (300[tex]\Omega[/tex], 300[tex]\Omega[/tex], 2k[tex]\Omega[/tex], 10k[tex]\Omega[/tex], 200[tex]\Omega[/tex], 3k[tex]\Omega[/tex]

    2. Relevant equations
    Resistors in Series: Ra+Rb+Rc...+Rn
    Resistors in Parallel: [(1/Ra)+(1/Rb)+(1/Rc)+...+(1/Rn)]^-1
    V_j = I_j*R_j; I_j = R_eq*I

    3. The attempt at a solution
    I've decided my approach in this problem should be:
    (1) Find Equivalent Resistance (R_eq)
    (2) Use the formula I_j=R_eq*I to determine I_o
    (3) Use I_o to find V_o, as V_o= I_o*R_o

    My problem is that I really am unsure how to find the equivalent resistance in this particular problem. What's messing me up is the 2k[tex]\Omega[/tex] and 10k[tex]\Omega[/tex] resistors...and the gap that is between them. I've just never done a circuit like this I guess.

    My attempt to find the equivalent is below:
    In Series, Left Side:300 and 300 ohm = 600 ohm
    In Series, Right Side:200 and 1000 ohm = 1200 ohm
    Simplified Resistor: 600 ohm on Left, 1200 ohm on right, 2k and 10k ohm on top.
    The equivalent can be found by adding up in series: 600, 1200, 2000, and 10000 = 13,800ohms = 13.8kOhm. Is that correct? If not, how do I attack the equivalence problem?
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution

    Attached Files:

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  2. jcsd
  3. Jan 22, 2009 #2
    Hint :-

    Suppress the current source and then analyze the circuit again.

    Hope this is useful. :smile:
  4. Jan 22, 2009 #3


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    Homework Helper
    Gold Member

    It may not correspond to systematic teachings, but the way automatic for me would be to say the 2k and 10k ohm resistors play no part in the problem since no current flows through them. The current merely divides itself flowing through the two branches proportionately to their conductances.

    Work out how it divides itself.

    There is the same potential drop for the two branches. The potential drop in two resistors in series (in either of the branches) is proportional to the resistance, e.g. it will be equal for the two resistors in the left-hand branch. Work out what it is in each case according to V = IR.

    You can work out the total resistance of this circuit from standard formulae you should know and understand, and hence the total potential. Do this as a check, it should come out agreeing with (voltages added up in) previous results. I hope this is not telling you too much according to the rules.
    Last edited: Jan 22, 2009
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