Voltage Divider Design

Discussion in 'Homework Help' started by yan500, Jul 17, 2011.

  1. yan500

    Thread Starter Member

    Jul 12, 2011
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    Hi all,

    Hope everyone's having a great weekend. I have a problem that I'm trying to solve and it's tripping me up. http://www.allaboutcircuits.com/worksheets/e_divide.html

    It is Question 35 on the worksheet located at the link above. Since the problem states that you can assume there is a bleed current of 1.5 mA, I calculated the total current in the circuit to be 13.5 mA. I believe that this is what's throwing me off. Can someone tell me what the total current in the circuit is and how they calculated it?

    Thanks!
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    To get I_total you simply need to add the load current plus the bleeding current.
     
  3. yan500

    Thread Starter Member

    Jul 12, 2011
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    Right, but for R3 + Load 1 you get 4.5 mA, which gives you the right answer for R3. But shouldn't the total of R2 + Load 2 give you 4.5 mA as well because of the loop rule?
     
  4. ian_gregg

    New Member

    Jan 19, 2011
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    There is 1.5mA through R3, 4.5mA through R2 and 16.5mA through R1, you know the voltages, V=IR and away you go.
     
  5. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    I don't understand you problem
    R2 = (9V - 4.8V)/4.5mA = 933.3
     
  6. yan500

    Thread Starter Member

    Jul 12, 2011
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    Ah, I see. Thank you. But how did you know to subtract 9 from 4.8 to get the voltage going through R2?
     
  7. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Because I know Ohm’s law and KVL/KCL all too well.
    Simply Load 2 need 9V and Load1 needs 4.8V.
    So the voltage drop across R2 must be equal 9V - 4.8V
     
  8. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Maybe this will help you withe KVL
    And remember that if elements are connect in parallel they will have the same voltage across them.
    [​IMG]
    VR4=V3+V2 or VR4=VB-V1, VR5=VB1, V1+V2=VR2
     
  9. yan500

    Thread Starter Member

    Jul 12, 2011
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    Thanks a lot guys. I need to practice my KVL/KCL, I'm so bad at them...
     
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