Question

A variable capacitor with a range from 12 to 279 pF is used with a coil to form a variable-frequency LC circuit to tune the input to a radio. (a) What is the ratio of maximum frequency to minimum frequency that can be obtained with such a capacitor

Answer 1

The ratio of maximum frequency and minimum frequency is 4.9.

Calculation:

The relation between frequency, f, and conductance, C and inductance, L is given as:

$f = \frac{1}{2\pi \sqrt{LC} }$

Given,

Cmax = 279 pF

Cmin = 12 pF

The smaller value of C gives the the larger value of f, therefore,

$fmax = \frac{1}{2\pi\sqrt{LCmin} }$           ....(1)

$fmin = \frac{1}{2\pi\sqrt{LCmax} }$           .....(2)

Dividing both the equations (1) and (2):

$\frac{fmax}{fmin} =\frac{\sqrt{Cmax} }{\sqrt{Cmin} }$

$\frac{fmax}{fmin} = \frac{\sqrt{279} }{\sqrt{12} }$

        $= 4.9$

Hence, the ratio of maximum to minimum frequency is 4.9.

Learn more about the frequency calculation here:

brainly.com/question/18650584

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