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<ac circuits >

  1. Sep 3, 2009 #1
    A)) A choke boil placed in series with an electric lamp in an AC circuit causes the lamp to become dim. Why?
    B)) A variable capacitor "added" in this circuit maybe adjusted until the lamp glows with normal brilliance. Explain how this is possible

    ATTEMPT:
    If V is the voltage of alternating source, and I is the current flowing when no inductor or capacitor is connected, then
    I 1=V/R
    If now a choke of inductive reactance Xl is placed in series with the electric lamp, the new impedance of the circuit will be
    Z 1=underroot<Rsquare + Xlsquare>
    Therefore current flowing will be
    I 2=V/[underroot<Rsquare + Xlsquare>]
    From the comparison of equations of current, we see that I 2 < I 1 and that is why the electric lamp is dimmed on placing a choke in the circuit.
    When a variable capacitor is added in series, Xc opposes Xl and thus
    Z 2=underroot<Rsquare + (Xlsquare-Xcsquare)>
    Therefore,
    I 3=V/[underroot<Rsquare + (Xlsquare-Xcsquare)>]
    If Xl = Xc, then Z 2=R
    And current becomes equal to I 1 as if there's no reactance in the circuit and hence the lamp glows with normal brilliance.

    Can you check whether this is a correct explanation?
     
  2. jcsd
  3. Sep 3, 2009 #2

    rl.bhat

    User Avatar
    Homework Helper

    Yes. Your explanation is correct.
     
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