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Current through 6 resistors in series and parallel

  1. Feb 16, 2010 #1
    1. The problem statement, all variables and given/known data
    See attached figure.
    All resistors have the same resistance of 2.00kilo-Ohm. Assume V = 12V. Calculate the current through resistor R6 and resistor R5.

    2. Relevant equations
    V = IR
    I = V/R
    R total (series) = R1 + R2.....etc
    1/R total (parallel) = 1/R1 + 1/R2...etc


    3. The attempt at a solution
    Since the current would travel towards A, it would first go through R5 (in parallel with R1 and R3), and I = V/R so for R5, I = 12/2000 = .006 A
    This is wrong and I don't understand why.
    Then for R6, the total resistance for R1 - R5 is 10,000 ohm, so the voltage would drop 10,000*the current, but I don't know how to find the original current.
    I'm completely lost. I redrew the figure, but still can't solve.
     

    Attached Files:

  2. jcsd
  3. Feb 16, 2010 #2

    vela

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    You can't just plug in any voltage that happens to be in the problem into Ohm's law. The current through R5 is equal to the voltage across R5 divided by R5, so you need to find that voltage first. The only voltage you know right now is the voltage across the battery.

    Start by combining all the resistors into one equivalent resistance.
     
  4. Feb 16, 2010 #3

    rl.bhat

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    Using series and parallel combination find the resistance across AC. Then find the total resistance of the circuit.
     
  5. Feb 16, 2010 #4
    I found the resistance across AC to be 1999 ohm, and the total resistance to be 3999 ohm.

    There is no voltage drop before the current hits R5, correct? So wouldn't the voltage through R5 still be 12 V? This would make the current .006 A, which is wrong, so I'm not sure what I'm doing wrong.

    Also, I don't know how to calculate the voltage drop over AC because I don't know the current?
     
  6. Feb 16, 2010 #5

    rl.bhat

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    I found the resistance across AC to be 1999 ohm
    Check your calculation for this part.
     
  7. Feb 16, 2010 #6
    R1 and R2 in series= R1 + R2 = 4000 ohm
    1/ (R1&2) and R3 in parallel = 1/4000 + 1/2000 = 1333 ohm
    (R1,2,3) in series with R4 = 1333 + 2000 = 3333 ohm
    1/ (R1,2,3,4) in parallel with R5 = 1/3333 + 1/2000 = 1250 ohm

    So total resistance (R1,2,3,4,5) + R6 in series = 1350 ohm
     
  8. Feb 16, 2010 #7
    sorry I meant total resistance is 3250 ohm
     
  9. Feb 16, 2010 #8

    vela

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    The battery is connected across this equivalent resistance, so you know the voltage drop across the resistance. Use Ohm's Law to calculate the current flowing out of the battery and through the circuit.
     
  10. Feb 16, 2010 #9
    Ok, I understand how to get the current for R6 by calculating the voltage drop from A to C and using the leftover voltage to find the current for R6.

    But, I still don't understand why I'm wrong on the current for R5.
    R5 gets all 12 volts, and R5 has a resistance of 2000 ohm, so the current should be .006, but this is wrong and I don't understand why?
     
  11. Feb 16, 2010 #10

    vela

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    R5 doesn't have a 12-volt drop across it. The voltage drop across it is the difference in potential between points A and C.
     
  12. Feb 16, 2010 #11
    Ah, I see. Thank you so much for your help, I really appreciate it!!!
     
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