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Power dissipated by a resistor in parallel

  1. Jan 14, 2013 #1
    1. The problem statement, all variables and given/known data

    We are given a circuit as follows: V= 10 V and two resistors are in parallel. R1 is 12 ohms and R2 is 5 ohms. What is the power dissipated in the 12 ohm resistor?


    2. Relevant equations

    V=IR
    P=(I^2)R

    3. The attempt at a solution

    I know that the total resistance of the circuit is 3.54 ohms... I found this by solving (1/Rtot) = (1/12)+(1/5) .

    The total current of the circuit is 2.82 A, which I found by using V=IR, or 10 = (3.54) I. I will call this current I3.

    I will just say that the current going through R1 is named I1 and the current going through R2 is named I2. Using V = IR, I can then say that 10 = (R1*I1) + (R2*I2). I can also say that I1 + I2 = 2.82 A.

    I also know that P= V^2 / R. I can deduce that the total power dissipated is (10)^2/3.54, which is 28.2 W.

    If the total power dissipated is the sum of each resistor's power dissipation, then 28.2 = (I1^2)(R1) + (I2^2)(R2).

    I have a lot of pieces but I still am not sure how to figure out the power dissipated through a single resistor, especially since they have different resistances. Any guidance on where to go next?
     
  2. jcsd
  3. Jan 14, 2013 #2
    Based on the work you've already done, you have the system of equations
    \begin{cases}
    10 = R_1I_1 + R_2I_2 \\
    2.82 = I_1 + I_2
    \end{cases}
    Since [itex]I_1[/itex] and [itex]I_2[/itex] are the only unknown quantities, you can easily solve this system for either one.

    You've stated both that [itex]P = \frac{V^2}{R}[/itex] and V = IR - this tells you that [itex] P = I^2R [/itex]. Once you have found the necessary current, then you can use that relationship to calculate the power dissipated in the individual resistor(s).
     
  4. Jan 14, 2013 #3

    gneill

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    Staff: Mentor

    The resistors are in parallel so they will both have the same potential (10V) across them. That renders your first equation incorrect.
     
  5. Jan 14, 2013 #4
    Ah, you're right. Serves me right for not thinking.

    In that case, since you know the voltage is the same across both resistors, you know that [itex] 10 = R_1I_1 = R_2I_2 [/itex]. If you replace my first equation with this, then I think that the rest of my post is still fine.

    EDIT: However, now that I think about it, my method seems unnecessarily long. Since we know the voltage across both resistors, doesn't it suffice to use the equation [itex]P = \frac{V^2}{R} [/itex]?
     
  6. Jan 14, 2013 #5

    gneill

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    Staff: Mentor

    There are two more handy expressions for power. Do you know what they are?
    You can say the second, but not the first. The resistors are in parallel, which means that they both share the same potential difference. The potentials across them do not add, they are equal.
    Okay. That uses one of the other handy expressions for power.
    Why not use the same expression for power, P = V2/R, that you used before, but for the individual resistors? You know the potential across each...
     
  7. Jan 14, 2013 #6
    Thanks for all the responses.

    I see how 10 = (R1*I1) + (R2*I2) is not valid here because of the loop rule. An electron can only go through one resistor.

    Now, if I use P = V^2 /R, then I know that the power dissipated is 8.33 W for the 12 ohm resistor and 20 W for the 5 ohm resistor. I got this by plugging in V= 10 V and then each individual resistance. Is that it? It's that easy?!

    If this is the case, I have another small question.

    Does the 5 ohm resistor (the smaller resistance of the two) dissipate more power because it has a bigger current going through it, and therefore gets hotter?
     
  8. Jan 14, 2013 #7

    gneill

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    Staff: Mentor

    Yup. It's that easy :smile:
    Yes. The potential drop is the same but the current is higher. The other "handy" expression for power is P = VI.
     
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