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Calculating the heat released in one part of an electrical network?

  1. Mar 5, 2017 #1
    1. The problem statement, all variables and given/known data
    An alternating current, with the current value of i = 3*cos314*t, is split into two parts in point A (each part with its own thermal resistance, R1 = 60Ω and R2 = 40Ω), which then connect again in point B. What is the released heat in the part between point A and point B during one period of this alternating current ?
    https://scontent.flju2-1.fna.fbcdn.net/v/t34.0-12/17141803_1652179735087925_530403472_n.jpg?oh=dde3a9b2342ccf107accb7048aefbd8c&oe=58BF4666


    2. Relevant equations
    Q = Ief2*R*t - Joules heat
    one period is the reciprical of the frequency(f=50Hz) so t/T = 0,02
    Ief=Imax/√2

    3. The attempt at a solution
    I attempted to just use the Joule's equation for heat on both resistances and then add the two up but the result I need to get is Q=2,16 J and the one I got was over 3 J...
     
    Last edited: Mar 5, 2017
  2. jcsd
  3. Mar 5, 2017 #2

    gneill

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    Staff: Mentor

    Can you show the details for how you performed your calculations? It's not clear how you dealt with the divided paths.

    Also, check your "Joules heat" equation. I'd expect to see an expression for the power dissipated by a resistor within it, and something is missing.
     
  4. Mar 5, 2017 #3
    Yes the equation was incorrectly written and I did the caluclations with the incorrect version. I edited the post and now the equation is Q = Ief2*R*t . I hope this is the expression for the power dissipated you aimed at.

    However, calculating the result with the correct equation has given me an even larger result. Here are the details:
    f=50Hz
    t=T=1/f=0,02s
    Ief=Imax/√2=2,12
    Q1=(2,12)2*60*0,02=5,39328J
    Q2=(2,12)2*30*0,02=3,59552
    Q1+Q2=8,9888J

    I'm thinking that the simple addition isn't correct since they are connected serialy ? If so how then should I calculate ?
     
  5. Mar 5, 2017 #4

    gneill

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    Staff: Mentor

    The two resistors are in parallel, not in series. They will not carry the same current; the current splits and some goes through one resistor and the remainder through the other. What you have calculated assumes that the full current passes though both resistors which is why you ended up finding such a large heat output.

    There are a few ways to approach the problem and you can take your choice. To proceed, look up the following concepts:

    1. Combining resistors in parallel
    2. Current division (or current divider)

    With (1) you can reduce the circuit to one equivalent resistance and apply your formula for Q using the total current. With (2) you can determine the current through each separate resistor, apply your formula, and then sum the results. Both approaches should give you identical results.
     
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