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Amplitude in the resistor

  1. Oct 23, 2006 #1
    A resistor with a resistance of 500 ohms and a capacitor with with a capacitance of 2 x 10^-6 f are connected in parrallel to an ac generator that supplies an rms voltage of 260 v at an angular frequency of 377 rad /s find the current amplitude in the resistor (Ir)peak.

    I know I have to convert 377 from rad / s to hz when I do I get 60 hz. From there I am unsure of how to calculate the rest. I know in a traditional series lrc circuit you calculate Z = [(Xl -Xc)^2 + R^2]^.5 but I and with z you can find the i which would be V/Z.
     
  2. jcsd
  3. Oct 23, 2006 #2

    Andrew Mason

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    Since V is the rms voltage then I = V/Z is the rms current. The peak voltage is [itex]\sqrt{2}V_{rms}[/itex] So peak current is [itex]V_{peak}/Z[/itex]

    AM
     
  4. Oct 23, 2006 #3
    would you calculate Z the same way with the capaicitor and reisitor in parrallel?
     
  5. Oct 24, 2006 #4

    Andrew Mason

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    No. The voltages are the same across R and C and the total current is the sum of the R and C currents (except there is a 90 degree phase difference between the two currents, of course). It is similar to parallel circuits with only resistance.

    [tex]V_R = V_C = V; I_R = V/R; I_C = V/X_C[/tex]

    [tex]I_R + I_C = I = V/Z = V/R + V/X_C[/tex] so:

    [tex]\frac{1}{Z} = \frac{1}{R} + \frac{1}{X_c} = \frac{1}{R} + j\omega C} [/tex]

    AM
     
    Last edited: Oct 24, 2006
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