Question

A coil of self-inductance L is connected in series with the bulb B and an AC source. Brightness of the bulb decreases when:
a) frequency of the AC source is decreased
b) the number of turns of the coil is reduced.
c)a capacitance of reactance ${{X}_{L}}={{X}_{C}}$ is included in the same circuit
d) an iron rod is inserted in the coil

Answer 1

An inductor in an AC circuit basically acts as an resistor. The brightness of the bulb depends on the magnitude of the current. If once we have a relation between the inductance of the coil and the current in the circuit, we can then conclude whether the brightness of the bulb will increase or decrease depending on how the inductance in the circuit affects the electric current.
Formula used:
$i=\dfrac{{{V}_{\circ }}Sin\omega t}{\sqrt{{{R}^{2}}+{{\left( {{X}_{L}}-{{X}_{C}} \right)}^{2}}}}$

Complete answer:

Let us say we have a coil of inductance ‘L’, a bulb of resistance ‘R’ and a capacitor of capacitance ‘C’ all three connected in series. The components are further connected to an A.C. source of angular frequency $\omega$ and whose output voltage at any instant of time ‘t’ is given by,
$V={{V}_{\circ }}Sin\omega t$
The reactance in the circuit is the resistance offered due to an inductor or capacitor. The capacitive reactance in an A.C. circuit is given by,
${{X}_{C}}=\dfrac{1}{\omega C}$
Similarly the inductive reactance offered by the inductor in the circuit is given by,
${{X}_{L}}=\omega L$
If all the three i.e. the resistor, inductor and the capacitor are connected in series, than the current ‘i’ in the circuit at any instant of time is given by,
$i=\dfrac{{{V}_{\circ }}Sin\omega t}{\sqrt{{{R}^{2}}+{{\left( {{X}_{L}}-{{X}_{C}} \right)}^{2}}}}$
From the above expression we can imply that when an inductor is introduced in the circuit, the current can only reduce when the inductance of the coil increases. This is due to the fact that inductance is directly proportional to inductive reactance. When an iron rod is inserted in the coil, the inductance of the coil increases due to the increase in flux linkage. As a result the current in the circuit decreases which results in the intensity of the bulb to also decrease.

Therefore the correct answer of the above question is option d.

Note:
The angular frequency is directly proportional to the frequency of the source. Hence if the frequency is increased, then the brightness of the bulb will also increase as the current increases.
If the number of turns in the coil is reduced, the inductance will decrease and as a result the current will increase due to net decrease in the capacitive reactance.
If a capacitance of reactance ${{X}_{L}}={{X}_{C}}$ is included in the same circuit, from the equation of current we can imply,
$\begin{aligned} & i=\dfrac{{{V}_{\circ }}Sin\omega t}{\sqrt{{{R}^{2}}+{{\left( {{X}_{L}}-{{X}_{C}} \right)}^{2}}}} \\\ & \because {{X}_{L}}={{X}_{C}} \\\ & \Rightarrow i=\dfrac{{{V}_{\circ }}Sin\omega t}{\sqrt{{{R}^{2}}+{{\left( 0 \right)}^{2}}}} \\\ & \therefore i=\dfrac{{{V}_{\circ }}Sin\omega t}{R} \\\ \end{aligned}$
The current in the circuit will be maximum and hence the bulb will glow with maximum brightness.