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Current Through a Capacitor in a RC Circuit

  1. Nov 11, 2017 #1

    Drakkith

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    1. The problem statement, all variables and given/known data
    Find ##i(0^+) ## and ##i(t), t≥0^+##

    pr_7-51.jpg
    2. Relevant equations


    3. The attempt at a solution

    I'm having trouble finding the current through this capacitor. I tried using a KVL loop for ##t=0^+##, but I'm doing it incorrectly or something.

    Earlier in the problem I found: ##v_c(0^+) = -120 V##
    Sign convention for resistors: Positive on the top for the 150k resistor, positive on the right side for the other 2.

    KVL1: Starting at node ##b## and moving clockwise with current ##i_1## :
    ##-2.5ki_1+150k(i_1-i_2)-v_c=0##
    ##-2.5ki_1+150ki_1-150ki_2-(-120)=0##
    ##147.5ki_1-150ki_2=-120##

    KVL2: Starting just to the left of the 50k resistor and moving clockwise:
    ##-50ki_2+200-150k(i_2-i_1)=0##
    ##-50ki_2-150ki_2+150ki_1=-200##
    ##150ki_1-200ki_2=-200##

    Solving, I get:
    ##i_1=0.85 mA##
    ##i_2=1.6mA##

    ##i=-i_1=-0.85 mA##

    However, this is appears to be incorrect.
     
  2. jcsd
  3. Nov 11, 2017 #2

    berkeman

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    Looks correct.
    For me, the KVL equations are generally less intuitive and move complicated. Not to make extra work for you, but can you write the KCL equation for the main node in the circuit for t>0 and see if that makes it easier to solve?
     
  4. Nov 11, 2017 #3

    Drakkith

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    Sure. From the main node, ##i## goes left, ##i_1## goes down, and ##i_2## goes right.
    ##i+i_1+i_2=0##
    ##\frac{v-v_c}{2.5}+\frac{v}{150k}+\frac{v-200}{50k}=0##
    Multiplying both sides by 150k:
    ##60v-60v_c+v+3v-600=0##
    ##64v-60(-120)=600##
    ##64v+7200=600##
    ##64v=-6600##
    ##v=-103.125 V##

    ##i=\frac{v-v_c}{2500}=6.75 mA##
    Which is correct...

    ##i(t)## is then: ##i(t)=6.75e^{(1000t)} mA##

    Why in the world did one method work but not the other?
     
  5. Nov 11, 2017 #4

    berkeman

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    Yahoo! :smile:
    Both methods should work, but at least for me, the KVL equations are much less intuitive and therefore much easier to make a mistake when using them. I'll leave it to other (smarter) members (EDIT -- like vela) to find the small typo in your work with the KVL equations... :smile:
     
    Last edited: Nov 11, 2017
  6. Nov 11, 2017 #5

    vela

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    Looks like the signs aren't consistent in your KVL equations. If you're going clockwise around the left loop, for example, you should get
    $$-2500i_1 + 150000(i_2-i_1) + v_c = 0.$$ In the second KVL equation, the sign on 200 should be negative since you're moving from higher potential to lower potential as you go clockwise around the loop.
     
  7. Nov 11, 2017 #6

    Drakkith

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    Isn't that already consistent though? Moving from positive to negative voltage across a resistor (moving from higher to lower potential), the sign is +, whereas it's - when moving from negative to positive (as in the ##-2500i_1##).
     
  8. Nov 11, 2017 #7

    vela

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    Isn't what already consistent?

    Which way are you assuming the currents ##i_1## and ##i_2## flow? If it's clockwise around the loops, then according to your sign convention, the first term should be ##+2500i_1##.
     
  9. Nov 11, 2017 #8

    Drakkith

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    The way I was doing it already.

    How so? The negative terminal of the 2500 ohm resistor is on the left.
     
  10. Nov 11, 2017 #9

    vela

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    Potential drops as the current flows through the resistor. If the current enters on the left, the left end of the resistor is at a higher potential than the right end.
     
  11. Nov 12, 2017 #10

    Drakkith

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    Looking back at my homework, I can't reconcile what I did there with what I've done here or what you're telling me vela. I'll have to dig a bit deeper and maybe talk to my instructor to figure out what's going on. Thanks all.
     
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