Question

A tuned amplifier circuit is used to generate a carrier frequency of ${\text{2 MHz}}$for the amplitude modulation. The value of $\sqrt {LC}$ is?
A. $\dfrac{1}{{2\pi \times {{10}^6}}}$
B. $\dfrac{1}{{2 \times {{10}^6}}}$
C. $\dfrac{1}{{3\pi \times {{10}^6}}}$
D. $\dfrac{1}{{4\pi \times {{10}^6}}}$

Answer 1

Hint : In this question, first write the expression for the frequency of the oscillator in L-C circuit and convert the given frequency from ${\text{MHz}}$ to ${\text{Hz}}$ and then substitute the given values in the expression in the formula to obtain the asked value.

Complete step-by-step solution :
In this question, a tuned amplifier is used to carry ${\text{2 MHz}}$ of frequency for the amplification of amplitude modulation. In this question, the value of $\sqrt {LC}$ needs to be calculated.

First, write the expression for the frequency of the transistor as an oscillator as,
$f = \dfrac{1}{{2\pi \sqrt {LC} }}$ ..… (1)
Where, the capacitance of the capacitor is $C$ and the inductance of the inductor is $L$.

Let us convert the frequency in hertz, as we know that $1\;{\text{MHz}}$ is equal to ${10^6}\;{\text{Hz}}$, so the frequency of the oscillator become,

$$f = {\text{2 MHz}} \\\ f = {\text{2 \times 1}}{{\text{0}}^{\text{6}}}{\text{ Hz}} \\\$$

In order to calculate the value of $\sqrt {LC}$ the value of frequency is to be substituted in the equation (1) as,
$\sqrt {LC} = \dfrac{1}{{2\pi f}} \\\ \sqrt {LC} = \dfrac{1}{{2\pi \times 2 \times {{10}^6}}} \\\ \sqrt {LC} = \dfrac{1}{{4\pi \times {{10}^6}}} \\\$
Therefore the value of $\sqrt {LC}$ is $\dfrac{1}{{4\pi \times {{10}^6}}}$.

Therefore, the correct option is D.

Note:-
Make sure that the units of the quantities should be in the same unit system. Be careful about the formula of the angular velocity and the frequency that is the unit of angular velocity is rad/s and the unit of the frequency is Hz.