Question

: A 60 volt - 10-watt bulb is operated at 100 volts - 60 Hz a.c. The inductance required is?
(A) 2.56H
(B) 0.32H
(C) 0.64H
(D) 1.28H

Answer 1

Hint:- The given problem can be seen as a a.c. circuit containing resistance and inductance in which conditions are given at which this bulb operates. Taking the consideration of given conditions this problem can be solved by the procedure given below.

Complete step-by-step solution :
Step 1: As it is given in this question that the bulb is being operated for the given values –
Voltage $V = 60$V and power $P = 10$W
Now, we know that power can be given by the formula given below –
$P = \dfrac{{\mathop V\nolimits^2 }}{R}$ (1)
Where $V =$Voltage and $R =$resistance
Keeping the given values in the above equation (1) resistance can be calculated –
$R = \dfrac{{\mathop V\nolimits^2 }}{P}$
$R = \dfrac{{60 \times 60}}{{10}}$ On further solving this equation
$R = 360\Omega$
Step 2: Current that is flowing through bulb can be calculated by the given formula –
$I = \dfrac{V}{R}$ by keeping the values from above in this equation
$I = \dfrac{{60}}{{360}}$ on simplifying this equation
$I = \dfrac{1}{6}{\rm A}$
Step 3: Let L be the inductance required for the bulb to be operated in the given conditions.
We know that impedance $Z$ can be calculated by the given formula –
$Z = \dfrac{V}{I}$ by keeping the values in this equation
$Z = 100 \times \dfrac{6}{1}$ on solving this equation
$Z = 600\Omega$
But we also know that $\mathop Z\nolimits^2 = \mathop R\nolimits^2 + \mathop X\nolimits_L^2$
And on rearranging this equation, we will get $\mathop X\nolimits_L = \sqrt {\mathop Z\nolimits^2 - \mathop R\nolimits^2 }$
Now on substituting the required values in this above equation, we will get –
$\mathop X\nolimits_L = \sqrt {\mathop {(600)}\nolimits^2 - \mathop {(360)}\nolimits^2 }$
$\mathop X\nolimits_L = 480\Omega$
Step 4: Now we know that $\mathop X\nolimits_L = \omega L = 2\pi \nu L$
And frequency $\nu = 60$Hz is given so after keeping all the requires values in the above equation –
$L = \dfrac{{\mathop X\nolimits_L }}{{2\pi \nu }}$
$L = \dfrac{{480}}{{2 \times 3.14 \times 60}}$ on simplifying this
$L = 1.27388 \simeq 1.28$H
So, the correct option is (D).

Note:- While calculating the impedance we know that $\mathop Z\nolimits^2 = \mathop R\nolimits^2 + \mathop X\nolimits_L^2 + \mathop X\nolimits_C^2$ but in the case of resistance and inductance circuit $\mathop X\nolimits_C = 0$ will be used.
The phase relation between alternating voltage and current in any a.c. circuit is given by –
$\tan \phi = \dfrac{{\mathop X\nolimits_L - \mathop X\nolimits_C }}{R}$.