Question : An inductor of self-inductance L and reactance XL is connected in series with a bulb B to an ac source. The bulb glows with some brightness. Predict the change in the brightness of the bulb when
a) number of turns of the inductor is reduced.
b) an iron rod is inserted inside the inductor.
Doubt by Lakshita
Solution :
The current in the ac circuit will be given by
I = ε/z — (1)
The Impedance will be given as
z=√(R²+XL²) — (2)
The Inductive reactance will be given by
XL=ωL — (3)
And
Inductance of Inductor will be given by
L = µ₀N²A/l — (4)
I = ε/z — (1)
The Impedance will be given as
z=√(R²+XL²) — (2)
The Inductive reactance will be given by
XL=ωL — (3)
And
Inductance of Inductor will be given by
L = µ₀N²A/l — (4)
a) When the number of turns in the inductor is reduced then
Inductance will decreases [See eq 4]
Inductive reactance will decreases [see eq 3]
Impedance will decrease [see eq 2]
Inductance will decreases [See eq 4]
Inductive reactance will decreases [see eq 3]
Impedance will decrease [see eq 2]
Current in the circuit will increase [see eq 1]
Brightness of the bulb will increase.
b) When an iron rod is inserted inside the inductor then
Permeability will increases
Permeability will increases
Inductance will increases [See eq 4]
Inductive reactance will increases [see eq 3]
Impedance will increases [see eq 2]
Inductive reactance will increases [see eq 3]
Impedance will increases [see eq 2]
Current in the circuit will decreases [see eq 1]
Brightness of the bulb will decreases.
