Question

Hello guys ;.


Answer the attachment!


Dont spam ​

Answer 1

Explanation:

GIVEN:-

=>Lower tuning = F1 = 800× 10^3Hz.

=>Upper tuning = F2 = 1200 × 10^3Hz.

L = 200 × 10^-6H

TO FIND:-

=>The range of its variable capacitor.

UNDERSTANDING THE CONCEPT:-

According to the question,

=>Capacitance of the variable capacitor for F1 is,

c1 and 1/(ω2 1L), where w is 2πF1

$\huge\tt\fbox\purple{Answer}$

Let's do it!!

>>w is 2πF1, So

C1 = 198.1pF

Similarly C2 = 88.04pE

Therefore, C∈[88.04 pF,198.1 pF].

Answer 2

Vishal

if you don't mind can I ask you one thing