Question
Question: A transistor- oscillator using a resonant circuit with an inductor \( L \) (of negligible resistance...
A transistor- oscillator using a resonant circuit with an inductor L (of negligible resistance) and a capacitor C in series produce oscillations of frequency f , If L is doubled and C is changed to 4C , the frequency will be
\left( A \right)\dfrac{f}{4} \\\
\left( B \right){F_2} = \dfrac{{{F_1}}}{{2\sqrt 2 }} \\\
\left( C \right)\dfrac{f}{{2\sqrt 2 }} \\\
\left( D \right)\dfrac{f}{2} \\\
Solution
Hint : In order to solve this question, we are going to take the expression for the frequency for a transistor- oscillator and then, by finding the ratio of the two frequencies f1 and f2 , and putting the changes in the values of inductance and the capacitance , the frequency f2 can be found in terms of f1 .
The frequency for a transistor – oscillator is given by
f=2π1LC1
The ratio of two frequencies f1 and f2
f2f1=L1C1L2C2
Complete Step By Step Answer:
It is given in the question that a transistor- oscillator using a resonator circuit with an inductor L (of negligible resistance) and a capacitor C in series produce oscillations of frequency f
We need to find the frequency when L is doubled and C is changed to 4C
The frequency for a transistor – oscillator is given by
f=2π1LC1
The ratio of two frequencies f1 and f2
f2f1=L1C1L2C2
Putting the values of the inductance and capacitance,
Here L is doubled and C is changed to 4C
\dfrac{{{f_1}}}{{{f_2}}} = \dfrac{{\sqrt {2L \times 4C} }}{{\sqrt {L \times C} }} \\\
\Rightarrow \dfrac{{{f_1}}}{{{f_2}}} = \sqrt 8 = 2\sqrt 2 \\\
Now, if we find the frequency f2 in terms of the frequency f1
We get f2=22f1
Hence, option (C)22f is the correct answer.
Note :
A transistor can be operated as an oscillator for producing continuous undamped oscillations of any desired frequency if oscillatory and feedback circuits are properly connected to it. The frequency here is the resonant frequency that depends upon the values of inductance and the capacitance.