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Wired Circuits and Power

  1. Apr 21, 2006 #1

    o= blank space (ignore)
    battery= 9 V

    Determine the power dissipated in the R4 resistor in the circuit shown in the drawing.

    R1=3 ohms
    R2=2 ohms
    R3=1 ohms
    R4=6 ohms
    R5=1 ohms

    So here is what i know so far. The resistance of R4+R5=7. The resistance of R2+R3=3. So the resistance of R2345=1/(1/7+1/3)=2.1 ohms. The total resistance through the entire circuit is R1+R2345=3+2.1=5.1 ohms.
    So I use this total resistance to calculate the total current I=V/R so
    I=9V/5.1ohms=1.76A. Now I can use this total current to calculate the voltage across R4 and R5. V=IR=1.76(2.1)=3.7 V across R4 and R5 since Voltage across parallel circuits are the same. I know power=IR^2=V^2/R.
    But i cannot just plug in 3.7V^2/6ohms. This isnt giving me the right answer. I need to find the voltage across R4 only. I am not sure how to do this since the current is different passing through each resistor. please help and correct me if any of the previous steps i did were wrong. thanks.
    Last edited: Apr 21, 2006
  2. jcsd
  3. Apr 21, 2006 #2


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    I have no idea where your 6 ohms comes from!!
    You kno wthe voltage across the combined R45. Now find the current through both by using I = V/R45 = 3.7V/3 ohms. That's the current through R4. Now use Power = R_4 I^2
  4. Apr 21, 2006 #3
    sry, lil typo, it is fixed now.
    R4=6 ohms
    and I got the right answer. thanks man.
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