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Homework Help: Circuit question, watts!

  1. Jun 17, 2012 #1
    1. The problem statement, all variables and given/known data

    physics questions.jpg

    2. Relevant equations

    What is the current leaving the cell.

    3. The attempt at a solution

    Since all the resistors have the same drop in watts, I assumed both "sections" will lose drop 6V. So each side of the the parallel section drops 6V and the single resistor also drops 6V.

    Then I tried to find the resistance in ohms of each one. P=V^2/R... 60^2/60=60.
    So I assumed every resistor is 60ohms. I then went to find the total resistance of the system (60^-1+60^-1)^-1+60. Rt=90.. Then I used V=IR 120/90 to find current. I end up getting 1.3Amps is this correct ?
    Last edited: Jun 17, 2012
  2. jcsd
  3. Jun 17, 2012 #2
    Hi zaddzad! :smile:

    [strike]Your answer is correct[/strike]. Though, it would be much easier to answer this question using the concept of equivalent resistances(instead of P). You can easily find the equivalent resistance due to the parallel connection, and then due to the series, for total resistance in the circuit.

    Edit : I just realized the given terms is watts not ohms...
    Last edited: Jun 17, 2012
  4. Jun 17, 2012 #3
    Hm, don't think I'v ever heard of this concept. I have a circuits test tomorrow, would you mind showing me how it's done ?
  5. Jun 17, 2012 #4
    Basically, when you have a parallel circuit, the equivalent resistance R is given as,

    [tex]\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + ....... + \frac{1}{R_n}[/tex]

    And for a series circuit, the equivalent resistance is,

    [tex]R = R_1 + R_2 + ......... + R_n[/tex]

    Take a look here for a deeper understanding : http://en.wikipedia.org/wiki/Series_and_parallel_circuits
  6. Jun 17, 2012 #5


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    I don't get 1.3 amps.
  7. Jun 17, 2012 #6
    Practicing delta-star circuits currently, makes me think everything is given in ohms!
  8. Jun 17, 2012 #7
    Thats what I did with the P=V^2/R I found the resistance of each resistor. Then when I found the resistance was 60, I did the total resistance of the circuit. For the parallel part I did (60^-1=60^-1)^-1 = 30 resistance there, and for the series I just added 60. The total resistance of the circut is 90 no ?
  9. Jun 17, 2012 #8
    I assumed each section of the circuit dropped 60 of the 120 volts because they are = in power drops. So wouldn't the top series resistor drop 60 volts, and the parallel resistors as a whole drop 60 volts?
  10. Jun 17, 2012 #9
    Then I used P=V^2/R to find their ohm values ?
  11. Jun 17, 2012 #10
    How is the correct answer found, and what is it?
  12. Jun 17, 2012 #11
    The resistance is not 60 ohms for the resistors.
  13. Jun 17, 2012 #12
    What is it then?
  14. Jun 17, 2012 #13
    It's no help having someone tell you your answer is wrong, some guidance would be kindly accepted.
  15. Jun 17, 2012 #14
    Still needing help...
  16. Jun 17, 2012 #15
    help please!
  17. Jun 17, 2012 #16


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    Please give the complete statement of the problem as it was given to you.
  18. Jun 17, 2012 #17
    Let the resistors have resistances R1, R2, and R3. As you had written, the current and voltage drop across a resistor is:
    I = \sqrt{\frac{P}{R}}, \ V = \sqrt{P \, R}

    Resistors 2 and 3 are connected in parallel, therefore the voltage drop across them is the same.
    V_2 = \sqrt{P \, R_2} = V_3 = \sqrt{P \, R_3} \Rightarrow R_2 = R_3
    This means that the current passing through each of them is the same.

    According to 1st Kirchhoff rule, the current through resistor 1, is twice as big. But, the power delivered is the same. Therefore:
    I_1 = 2 I_2, \ P = I^{2}_{1} \, R_1 = I^{2}_{2} \, R_2 \Rightarrow 4 R_1 = R_2

    We have found the relative resistances of the three resistors.

    What is the equivalent resistance of the three resistors in terms of R1?

    What is the current flowing through resistor 1, in terms of V = 120 V, and R1?

    What is the power delivered on resistor 1, in terms of V, and R1? You know that this power is 60 W. Find R1 from here.

    Go one step backwards and find the current through resistor 1. This is the same current flowing out of the source.
  19. Jun 17, 2012 #18


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    Here's a hint.

    The solution is so easy, I can do it in my head!
  20. Jun 18, 2012 #19


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    I think this is why SammyS just requested verification of the problem.

    The graphic doesn't strike me as a typical homework problem.
  21. Jun 18, 2012 #20
    The job of the cell is to supply the power.
    The total output of the cell must be equal to total output of the load.

    For house electric consumption, you're charged by the amount of load total wattage irrespective how you connect your appliances, series or parallel.
    Last edited: Jun 18, 2012
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