Why Does Adding a Choke Dim an Electric Lamp in an AC Circuit?

[uzair_ha91]

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?

 

[rl.bhat]

Yes. Your explanation is correct.