Confusing

Discussion in 'Homework Help' started by obliviga, Mar 9, 2013.

  1. obliviga

    Thread Starter New Member

    Mar 9, 2013
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    Hello all. I am a beginner with circuits and I am having trouble with this homework problem. We are asked "How many milliwatts of power is delivered by the battery to the entire circuit?" and to find the voltage drop across R6. How do I go about doing this? It is confusing me because the voltage source is in the middle of the network. Thank you for your help!
    Also, all of the resistor values are 2000 Ohms and the voltage source is 5 Volts...

    I attached an image of the circuit in this post. Thanks again.
     
  2. WBahn

    Moderator

    Mar 31, 2012
    17,757
    4,800
    Would it be less confusing if the battery were on the left side of the circuit?

    If so, then extend the top and bottom wires out to the left and then slide the battery along the wires over the left hand resistors and out to the left. As long as the components remain connected to the same wires, you can move them where ever you want if it makes things more familiar to you.

    So make you best attempt and show your work to give us something to go on.
     
  3. obliviga

    Thread Starter New Member

    Mar 9, 2013
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    Okay, so I simplified the circuit down to the voltage source and a 1.665 kΩ resistor. So the total current is 3 mA. So is the current going through R6 1.5 mA?
     
  4. WBahn

    Moderator

    Mar 31, 2012
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    4,800
    Yes, it is. But why do you think it is?
     
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  5. obliviga

    Thread Starter New Member

    Mar 9, 2013
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    Because since the resistance is equal on both sides of the voltage source, the current will split in half.
     
  6. WBahn

    Moderator

    Mar 31, 2012
    17,757
    4,800
    Correct.

    I recommend that, if you use that reasoning on anything you turn in (whether it be to an instructor or to a boss) you put a note such as:

    The total current is 3.0mA. Due to symmetry, R6 will have 1.5mA.
     
  7. obliviga

    Thread Starter New Member

    Mar 9, 2013
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    0
    Good idea! Thank you so much for your help.
     
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