1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Power dissipated from resistance

  1. Oct 22, 2012 #1
    1. The problem statement, all variables and given/known data
    This was a diagram, so I will describe it as accurately as possible.

    12 Volts go through one 2 Ohm resistor which then proceeds through a set of resistors in parallel. One is 20 Ohms, the other two are 10 Ohms.

    How much power is being dissipated by one of the 10 Ohm resistors?

    Unknown: Current.

    2. Relevant equations
    V = IR
    P = V * A
    1/R(Parallel-Equivelant) = Sum(1/Rn)

    3. The attempt at a solution

    Can I calculate the Current after the first resistor using V = I * R ? If so, I know how to calculate the total resistance by using 1/R = Sum of (1/Rn). Would the resistance of one resistor just be 1/R = 1/10 --> R = 10?

    THanks for any help
    Last edited: Oct 22, 2012
  2. jcsd
  3. Oct 22, 2012 #2


    User Avatar
    Gold Member

    Yes. But only if you know the values of I & R.
    First you will have to determine the equivalent resistance of the 3 large resistances in parallel. Once you have that, combine it with the 2 ohm resistor, and you can figure out total circuit resistance. This will yield your total circuit amperage from your equation.

    Btw, is this your circuit?
  4. Oct 22, 2012 #3
    Yes, that is my circuit. When I first viewed it was completely illegible, so I just deleted it. Could have been a rendering error so thank you for correcting that for me.

    You say that I must know the values of I and R. Well by V = IR I should be able to calculate this: 12V = I*(2 Ohms), I = 6A. Is this not the current directly after the first resistor?

    Conceptually this doesn't make sense to me. It seems as if current would have to be given in the problem statement.

    I calculated the total resistance of the parallel resistors to be 20/5 = 4 Ohms. I also know that one 10 Ohm resistor is 1/4 of the total resistance given in that parallel resistor circuit (Since it accounts for 25% of the total resistance). So shouldn't the current drained from that one resistor be (1/4) * (20/5) = (20/20) = 1.

    I don't think that's right. I'm missing something.
  5. Oct 22, 2012 #4


    User Avatar
    Gold Member

    That is not correct
    You can calculate total circuit current by following my procedure.
    That is correct.
    It doesn't quite work that way. As you said, the total resistance of the parallel resisters is 4 ohms. 10 ohms is not 1/4 of 4 ohms.
    You are correct, that's not right.

    You really need to find the total circuit current at this point.
  6. Oct 22, 2012 #5
    No because that 12V is applied across the 2 Ohm resistor *and* the parallel combination of three resistors.

    You want to transform your circuit so that it is more easily solvable. This is the purpose of finding an equivalent resistance for series and parallel resistors.

    Any parallel set of resistors has the same voltage applied across them. This means the total current through the parallel set can be found by adding V/R of each resistor. If you then look at the voltage applied (V) and total current through the parallel resistors (sum of all V/R), a relationship is found between V and total current through the parallel branch: V/I = 1/(1/R1 + .... + 1/Rn). This looks like Ohm's Law with an effective resistance of 1/(1/R1+...+1/Rn). This means you can *replace* your parallel set of resistors with a single resistance without affecting the rest of the circuit. Further, the current you find in this replacement resistor will be the total current through the parallel resistors. You can determine how that current divides between the resistors from their resistance ratios.

    The same can be done with series resistances, with the key insight being the same current that flows through one resistor must also flow through a series resistor in order to get from point A to B.

    Your homework problem is then to transform your relatively complicated circuit into a simple one of one resistor connected to a battery to find the circuit current and then work from there by unravelling the circuit transformations you made to compute currents and/or voltages across the original resistances that were made part of the effective resistances in the simplified circuit.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook