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Circuits Help

  1. Oct 4, 2004 #1
    C = 7.20 micro-Farads, R1 = 10.5ohm, R2 = 22.5ohm, V0 = 16.0V. What is the current (to within 0.01A) through R1 immediately after the switch S is closed? I created two equations to represent the voltage drops in the system but Capacitors charge up over time which is confusing me:

    16V - i1*R1 - i2*r2 = 0
    16V - i1R1 - q/C = 0

    Therefore i2*R2 = q/C subtracting the two. Unfortunately that's two unknowns. I also sort of (but not fully understand) that for a resitor and capacitor in series, q = C*Vapplied*(1-e^(-t*R*C)) so at t = 0, q = 0 and thus no voltage drop across the capacitor and therefore no current. This means that all of the current is flowing through R2 right? Therefore i2 = i1 so..

    16V - I1(R1+R2) = 0
    16V - I1(10.5+22.5) =0
    I = 0.48A?

    Am I right to apply this to this drawing? Edit 1.33 was wrong. So was 0.48. OK I'm really lost now.

    Attached Files:

    • doh.bmp
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    Last edited by a moderator: Oct 4, 2004
  2. jcsd
  3. Oct 4, 2004 #2
    New theory: Is the capacitor being initially uncharged short-circuiting the system? Then the voltage drop along R1 would be 16V meaning that 16=10.5*I, I = 1.523A. Am I any closer? Edit YESS I got it :) Maybe there's hope for me yet.
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