Question

An inductor $L$ of inductance ${X_L}$ is connected in series with a bulb B and an ac source. How would the brightness of the bulb change when
(i) the number of turns in the inductor is reduced?
(ii) an iron rod is inserted in the inductor and
(iii) a capacitor of reactance ${X_C} = {X_L}$ is inserted in series in the circuit. Justify your answer in each case.

Answer 1

First use the formula for net resistance in the circuit and then recall the formula for the inductance, from there you will get the relation between the inductance and the number of turns. Recall the conditions for what happens when a soft iron rod is inserted. Also, recall the formula for the resistance when a capacitor is inserted in series.

Complete step by step solution:
As we already know that the inductance and the number of turns in the inductor is related to the formula given by,
$L = {\mu _0}\dfrac{{{N^2}}}{l}A$ .................(1)
Where $L$ is the inductance
${\mu _0}$ is the Permeability of Free Space
$N$ is the number of turns
$A$ is the area of the inner core
$l$ is the length of the Coil
From the above formula, if the number of turns decreases, the inductance also decreases.
Now, the inductive resistance is given by,
${X_L} = \omega L$ ………………..(2)
Where $\omega$ is the angular frequency of ac source
$L$ is the inductance (in Henries)
And from this above formula, if the inductance is decreased, the inductive resistance will also decrease.
Thereby, according to the formula for the net resistance in the circuit is given by,
$z = \sqrt {{X_L}^2 + {R^2}}$ …………..(3)
As, ${X_L}$ decreases thereby, reducing the net resistance in the circuit.
Now, the current is related to the net resistance is given by,
$i = \dfrac{V}{z}$ ……………..(4)
Here, from the above formula, the current is inversely proportional to the net resistance.

So, if net resistance decreases, the current increases, and therefore, the brightness of the bulb is increased.

(ii) When a soft iron rod is inserted in the circuit, the value of inductance $L$ increases. Therefore, according to equation (2), inductive reactance also increases.

Then, according to equation (3), net resistance increases, and then according to equation (4), the flow of current in the circuit decreases.

Thus this effect in the brightness of the bulb, that is the brightness decreases.

(iii) If a capacitor of reactance ${X_C} = {X_L}$ is inserted in series with the circuit.
Then, the net resistance is given by,
$z = \sqrt {{{({X_L} - {X_C})}^2} + {R^2}}$
Where, $R$ is the resistance of the bulb.
On putting ${X_C} = {X_L}$ in the above formula, we get
$\Rightarrow z = R$
This is the condition of the resonance.
The condition of resonance means the whole voltage supply drops across the resistance. Hence, the maximum current will flow through the circuit.

Therefore, as the current flow increases, the brightness of the bulb increases.

Note: Always keep in mind that the inductance of a coil is due to the magnetic flux around it. The higher the value of the magnetic flux for a given current the greater will be the inductance and the higher the number of turns in a coil will have a higher value of inductance than a coil having few turns.