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Series and parallel resistors

  1. Mar 5, 2008 #1


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    In the circuit shown, find the current in each resistor and the values of the potential at points A, B, and C.

    First I computed the combined resistance of the parallel resistors and got 4.54 kOhms. Then I added this to 10.5 to get 15.04 kOhms total resistance.

    So from Ohm's law, the total current is I=V/R = 16.5 / 15040 = 1.0971 x 10-3 A.

    I get potentials at A, B, and C of 16.5, 11.5, and 0 volts.

    I think this is correct so far.

    To find the current through each resistor:

    For the 10.5 kOhm resistor, since it is in series, it makes sense to me that the current through this resistor must be equal to the total current: 1.0971 x 10-3 A. And that the current running through the parallel pair must also be 1.0971 x 10-3 A. But the voltage across this resistor is 16.5-11.5, or 5v. Using Ohm's law, I=V/R, I get 5/10500 = 0.000476 A

    Which way is correct???
    1. The problem statement, all variables and given/known data

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    3. The attempt at a solution
  2. jcsd
  3. Mar 5, 2008 #2
    The currents through the two resistors in parrallel will be different. you have to work out the current from the voltage drop across each of the resistance's, which is 5V for both according to your calculations, and the resistance. This will give you the current in each element. It will add up to the total loop current.

    Your using the wrong voltage. 5V isn't dropped across the 10.5kOhm, 11.5V is. As it's 11.5V at one side and 0V at the other.
  4. Mar 5, 2008 #3


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    Thanks! That's where I messed up. 11.5/10.5E3 = the same as the total loop current.

    Thanks for the explanation.
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