Question

A coil of self-inductance L is connected in series with a bulb B and an ac source. Brightness of the bulb decreases when
(a) frequency of the ac source is decreased.
(b) number of turns in the coil is reduced.
(c) a capacitance of reactance Xc = XL in included.
(d) an iron rod is inserted in the coil.

Answer 1

A coil of self-inductance L is connected in series with a bulb B and an ac source. Brightness of the bulb decreases when

frequency of the ac source is decreased.

Answer 2

Answer:

a) frequency of ac source is decreased

c) a capacitance of Xc=XL is included

Explanation:

A bulb's brightness depends upon the power supplied by the A.C circuit

And since power is given by

$\boxed{P=\frac{E}{Impedence} }$

Our aim here is to minimise impedence, thus maximising power output

Impedence of an LCR circuit is given by :

$\boxed{Impedence=\sqrt{R^2+(XL-Xc)^2}}$

So we need to minimise (XL-Xc) term to minimise impedence for a bulb of given resistance R

In case of AC circuit with a bulb and L, it is similar to LR circuit.

The instantenous current for LR circuit is given by $\boxed{I=\frac{Eo}{L\omega}sin(\omega t+\phi) }$----(1)

OPTION (A) is correct because

From (1), since I increases on decreasing frequency (or ω)

OPTION (B) is wrong because

In such a case as L will reduce in case the number of turns are reduced.As L lowers, the current through the circuit reduces & hence the brightness of the bulb will reduce.

Hope this answer helped you :)