Question

An inductor L of inductance XL is connected in series with a bulb B and an ac source. How would brightness of the bulb change when (i) number of turn in the inductor is reduced, (ii) an iron rod is inserted in the inductor and (iii) a capacitor of reactance XC = XL is inserted in series in the circuit. Justify your answer in each case.

Answer 1

The inductive resistance (XL) is given by XL = ωL Where ω = Angular frequency of ac source L = Inductance of the inductor The net resistance of the circuit is given by Z=XL2+R2 where R = Resistance of the bulb
(i) We know that if the number of turns in the inductor decreases, then inductance L decreases. So, the net resistance of the circuit decreases and, hence, the current through the circuit increases, increasing the brightness of the bulb.
(ii) If soft iron rod is inserted in the inductor, then the inductance L increases. Therefore, the current through the bulb will decrease, decreasing the brightness of the bulb.
(iii) If the capacitor of reactance XC = XL is connected in series with the circuit, then Z=XL-XC2+R2⇒Z=R ∵XL=XC This is a case of resonance. In this case, maximum current will flow through the circuit. Hence, the brightness of the bulb will increase.

Answer 2

Hey !!

(i) Increases               XL = ωL

       As number of turns decrease,  decreases, hence current through bulb increases / voltage across bulb increases.

    Therefore, Brightness increases.

(ii) Decreases

Iron rod increases the inductance, which increases XL, hence current through the bulb decreases / voltage across bulb increases

Therefore, Brightness decreases.

(iii) Increases

Under this condition (Xc = Xi) the current through the bulb will become maximum / increase.

Therefore, Brightness increases.

Good luck !!