Question

The tuning circuit of a radio receiver has a resistance of 50W, an inductor of 10 mH and a variable capacitor. 1 MHz radio wave produces a potential difference of 0.1 mV. The values of the capacitor to produce resonance is (Take p² = 10)

• A. 2.5 pF
• B. 5.0 Pf
• C. 25 pF
• D. 50 pF

Answer 1

Answer: (A)

2.5 pF

Answer 2

L = 10 mHz = 10^–2 Hz , f = 1MHz = 10⁶ Hz

f = $\frac{1}{2\pi\sqrt{LC}}$ Ž f² = $\frac{1}{4\pi^{2}LC}$Ž C = $\frac{1}{4\pi^{2}f^{2}L}$

= $\frac{1}{4 \times 10 \times 10^{- 2} \times 10^{12}}$ = $\frac{10^{- 12}}{4}$ = 2.5 pF