Question

A radio can tune over the frequency range of a portion of MW broadcast band: (800 kHz to 1200 kHz). If its LC circuit has an effective inductance of 200 μH, what must be the range of its variable capacitor? [Hint For tuning, the natural frequency i.e., the frequency of free oscillations of the LC circuit should be equal to the frequency of the radiowave.]

Answer 1

Answer:

88 pF to 198 pF.

Explanation:

Given,

$\textrm{Lower\ frequency\ limit},\ f_1\ =\ 800 kHz$

                                                              $=\ 800\times 10^3\ Hz$

$\textrm{higher frequency limit},\ f_2\ =\ 1200\ kHz$

                                                             $=\ 1200\times 10^3\ Hz$

inductance = 200 μH

                   $=\ 200\times 10^{-6}\ H$

for lower frequency limit

   $f_1\ =\ \dfrac{1}{2\pi\sqrt{LC_1}}$

$=>\ 800\times 10^3 Hz\ =\ \dfrac{1}{2\pi\sqrt{200\times 10^{-6}.C_1}}$

$=>\ C_1\ =\ 1.98\times 10^{-10}\ F$

              = 198 pF

for higher frequency limit

  $f_2\ =\ \dfrac{1}{2\pi\sqrt{LC_2}}$

$=>\ 1200\times 10^3\ Hz\ =\ \dfrac{1}{2\pi\sqrt{200\times 10^{-6}.C_2}}$

$=>\ C_2\ =\ 8.8\times 10^{-11}\ F$

               = 88 pF

So, the range of variable capacitor will be 88 pF to 198 pF.