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Circuit analysis

  1. Nov 4, 2015 #1
    1. The problem statement, all variables and given/known data
    fG2YMYN.jpg

    2. Relevant equations
    Thevenin theorem
    3. The attempt at a solution
    ##R_{60+180}=240##
    ##R_{80||240}=\frac{240*80}{240+80}=60##
    ##R_{total}=60+20=80##
    ##=\frac{480}{80000}=0.006A##
    How should I proceed?
     
  2. jcsd
  3. Nov 4, 2015 #2

    phyzguy

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    I think you are a factor of 10 off, otherwise what you have done is OK to find i. Your next step is to find v0. How much current is flowing through the 180 KOhm resistor?
     
  4. Nov 4, 2015 #3
    Sorry, fixed the current.

    I know that the current need to split twice before it will get to ##R_{180}##
     
  5. Nov 4, 2015 #4

    phyzguy

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    It only splits once. How much of the total current i goes through the 80 KOhm resistor, and how much through the 60 KOhm + 180 KOhm series combination?
     
  6. Nov 4, 2015 #5
    ##R_{80}## gets 0.0045A so ##R_{60+180}## gets 0.0015A therefore it is 360V?
     
  7. Nov 4, 2015 #6

    phyzguy

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    The 0.0015A is right. Given this, what is v0?
     
  8. Nov 4, 2015 #7
    ##V_{60}=0.0015*60000=90## so ##V_0=480-90-120=270v##?
    why could I say that because ###v_0## or ##R_{80}## is connected in parallel to the battery, it has the same Voltage?
     
  9. Nov 4, 2015 #8

    phyzguy

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    Yes, v0 = 270V is correct. This is not the same as the voltage across the 80 KOhm resistor, because some voltage is dropped across the 60 KOhm resistor (90V, as you calculated).
     
  10. Nov 5, 2015 #9
    Since both currents unite at the 20k resistor, then u got 240k.80k/320k + 20k= 80000ohms V/Rtotal= 0.006amps net current. From there on u get that 120volts fall at the 20k resistior, and the current through the 60k+180k branch is 0.0015amps. From there the rest current sjould be 0.0045amps
     
  11. Nov 5, 2015 #10
    Isnt that right mates?
     
  12. Nov 9, 2015 #11
    Yes, thanks
     
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