SATHEE: Electro Magnetic Induction And Alternating Currents Question 219

Electro Magnetic Induction And Alternating Currents Question 219

Question: The tuning circuit of a radio receiver has a resistance of $50 Ω$ , 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 $(T a k e π^{2} = 10)$

Options:

A) 2.5 pF

B) 5.0 pF

C) 25 pF

D) 50 pF

Show Answer

Answer:

Correct Answer: A

Solution:

$L = 10 m H z = 10^{- 2} H z$

$f = 1 M H z = 10^{6} H z$

$f = \frac{1}{2 π \sqrt{L C}} \Rightarrow f^{2} = \frac{1}{4 π^{2} L C}$

$\Rightarrow C = \frac{1}{4 π^{2} f^{2} L} = \frac{1}{4 \times 10 \times 10^{- 2} \times 10^{12}}$

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Electro Magnetic Induction And Alternating Currents Question 219

Question: The tuning circuit of a radio receiver has a resistance of 50Ω , 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 (Takeπ2=10)

Options:

Answer:

Solution:

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Question: The tuning circuit of a radio receiver has a resistance of $50 Ω$ , 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 $(T a k e π^{2} = 10)$