Electro Magnetic Induction And Alternating Currents Question 219
Question: The tuning circuit of a radio receiver has a resistance of $ 50\Omega $ , an inductor of 10 mH and a variable capacitor. A 1 MHz radio wave produces a potential difference of 0.1 mV. The values of the capacitor to produce resonance is $ (Take{{\pi }^{2}}=10) $
Options:
A) 2.5 pF
B) 5.0 pF
C) 25 pF
D) 50 pF
Show Answer
Answer:
Correct Answer: A
Solution:
-
$ L=10mHz={{10}^{-2}}Hz $
$ f=1MHz=10^{6}Hz $
$ f=\frac{1}{2\pi \sqrt{LC}}\Rightarrow f^{2}=\frac{1}{4{{\pi }^{2}}LC} $
$ \Rightarrow C=\frac{1}{4{{\pi }^{2}}f^{2}L}=\frac{1}{4\times 10\times {{10}^{-2}}\times 10^{12}} $
