Question

A tube light of $60{\text{V}}$, $60{\text{W}}$ rating is connected across an AC source of $100{\text{V}}$ and $50{\text{Hz}}$ frequency. Which of the following options is/are correct?
A) An inductance of $\dfrac{2}{{5\pi }}{\text{H}}$ may be connected in series.
B) A capacitance of $\dfrac{{250}}{\pi }\mu {\text{F}}$ may be connected in series to it.
C) An inductance of $\dfrac{4}{{5\pi }}{\text{H}}$ may be connected in series.
D) A resistance of $40\Omega $ may be connected in series

Answer 1

Hint:A resistor or an inductor can be connected in series with the tube light. The power dissipated by a resistive element is the product of the voltage across that element and the current through it. The inductive reactance in terms of the voltage across the inductor will give the inductance connected in series. Ohm’s law will give the resistance connected in series.

Formulas used:
-The power dissipated by a resistive element is given by, $P = Vi$ where $V$ is the potential difference across the element and $i$ is the current through the element.
-The voltage across the inductor is given by, ${V_L} = \sqrt {{V^2} - {V_R}^2} $ where $V$ is the source voltage and ${V_R}$ is the voltage across the resistive element in the circuit.
-The inductive reactance is given by, ${X_L} = 2\pi fL$ where $f$ is the frequency of the source and $L$ is the inductance of the inductor.
-The resistance of a circuit is given by, $R = \dfrac{V}{I}$ where $V$ is the voltage across the resistor and $I$ is the current through it.

Complete step by step answer.
Step 1: List the parameters obtained from the question.
The voltage across the tube light is ${V_b} = 60{\text{V}}$ and the power consumed by the tube light is ${P_b} = 60{\text{W}}$ .
The voltage of the AC source is given to be $V = 100{\text{V}}$ and its frequency is $f = 50{\text{Hz}}$ .
Step 2: Express the current through the circuit using the relation for the power dissipated.
The power dissipated by the tube light can be expressed as ${P_b} = {V_b}i$
$ \Rightarrow i = \dfrac{{{P_b}}}{{{V_b}}}$ ------ (1)
Substituting for ${P_b} = 60{\text{W}}$ and ${V_b} = 60{\text{V}}$ in equation (1) we get, $i = \dfrac{{60}}{{60}} = 1{\text{A}}$
Thus the current through the circuit is $i = 1{\text{A}}$ .
Step 3: Express the voltage across the inductor connected in series with the tube light to obtain the inductance in the circuit.
The voltage across the inductor can be expressed as ${V_L} = \sqrt {{V^2} - {V_b}^2} $ -------- (2)
Substituting for $V = 100{\text{V}}$ and ${V_b} = 60{\text{V}}$ in equation (2) we have ${V_L} = \sqrt {{{100}^2} - {{60}^2}} = 80{\text{V}}$ .
i.e., the voltage across the inductor is ${V_L} = 80{\text{V}}$ .
Now we can also express this voltage as ${V_L} = i{X_L}$ .
$ \Rightarrow {X_L} = \dfrac{{{V_L}}}{i}$ ----------- (3)
The inductive reactance is also given by, ${X_L} = 2\pi fL$ --------- (4)
Equating R.H.S of equations (3) and (4) we get, $\dfrac{{{V_L}}}{i} = 2\pi fL$
$ \Rightarrow L = \dfrac{{{V_L}}}{{2\pi fi}}$ ---------- (5)
Substituting for $i = 1{\text{A}}$, ${V_L} = 80{\text{V}}$ and $f = 50{\text{Hz}}$ in equation (5) we get, $L = \dfrac{{80}}{{2\pi \times 50 \times 1}} = \dfrac{4}{{5\pi }}{\text{H}}$
Thus the inductance connected in series will be $L = \dfrac{4}{{5\pi }}{\text{H}}$ .
So option C is correct.
Step 4: Express the resistance which would be connected in series.
Here out of the $V = 100{\text{V}}$ of the source, ${V_b} = 60{\text{V}}$ is used by the tube light to glow. Then the remaining voltage will be dropped across the resistance connected in series.
i.e., ${V_R} = V - {V_b} = 100 - 60 = 40{\text{V}}$
Ohm’s law gives the resistance connected in series as $R = \dfrac{{{V_R}}}{i} = \dfrac{{40}}{1} = 40\Omega $
So option D is also correct.

Thus the correct options are C and D.

Note:Here, whatever be the element (inductor or resistor) it is connected in series with the tube light. So the current through each element in the circuit will be the same but the voltage across the elements in the circuit will differ. The tube light is considered as the resistive element when the inductor is connected in series with it and so we express the power dissipated by it as ${P_b} = {V_b}i$. Inductive reactance refers to the resistance offered by the inductor.