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.

 

[tomcue]

Yes, your explanation is correct. The addition of a choke in an AC circuit causes a decrease in the current flowing through the circuit due to the increased impedance. This decrease in current results in a dimming of the electric lamp. On the other hand, the addition of a variable capacitor with the same reactance as the choke cancels out the effect of the choke, resulting in a decrease in impedance and allowing the current to flow at its original level, causing the lamp to glow with normal brilliance. This phenomenon is known as resonance and is commonly used in AC circuits to control the flow of current.