Question

The carrier frequency generated by a tank circuit containing 1nF capacitor and 10 μH inductor is
(A) 1592Hz
(B) 1592MHz
(C) 1592KHz
(D) 159.2Hz

Answer 1

Hint According to the given question we have to find the frequency of the tank circuit It can be found out using the formula having relation with the frequency with the inductor and the capacitor
$f = \dfrac{1}{{2\pi \sqrt {LC} }}$

Complete step by step answer
As we have to find the frequency generated by the tank in a circuit
Where we know that resonant frequency is given by
$ω_r$ = $2\pi {f_r}$ = $\dfrac{1}{{\sqrt {LC} }}$
$ω_r$ represents the resonant frequency
Hence it can be also written as
${f_r}$= $\dfrac{1}{{2\pi \sqrt {LC} }}$
In the given question
L = 10μH = 10$^{-5}$
And C = 1nF = 10$^{-9}$
Hence putting the value in the equation
${f_r}$= \[\dfrac{1}{{2\pi \sqrt {{{10}^{ - 5}} \times {{10}^{ - 9}}} }}\]
${f_r}$= \[\dfrac{1}{{2\pi {{10}^{ - 7}}}}\]
${f_r}$= 1.591 \[ \times \] \[{10^7}\]
Hence value of the frequency is
${f_r}$= 1592 KHz

Hence the correct answer is given by
${f_r}$= 1592 KHz

Note An alternating current is that current whose magnitude changes continuously with the time and the direction reverse periodically.
Direct current flows with a constant magnitude in the same direction.
Phasors –
A rotating vector that represents a sinusoidally varying quantity is called the phasors. This vector is imagined to rotate with angular velocity equal to the angular frequency of that quantity. It represents the amplitude of the quantity and its projection upon a fixed axis gives the instantaneous value of the quantity the phase angle between two quantities is shown as the phase angle between their phasors.