A coil of inductance 8 μH is connected to a capacitor of capacitance 0.02 μF To what wavelength is this circuit tuned?

7.54 × 10³ m

4.12 × 10² m

15.90 × 10³ m

7.54 × 10² m

Solution

7.54 × 10² m

L = 8 μH = 8 × 10^-6 H

C = 0.02 μF = 0.02 × 10^-6 F

∴ Resonant frequency

f_r = $\frac{1}{2 π \sqrt{LC}}$

= $\frac{1}{2 π \sqrt{8 \times 10^{- 6} \times 0.02 \times 10^{- 6}}}$

f_r = 3.98 × 10⁵ Hz

If c ( = 3 × 10⁸ m s^-1 ) is the velocity of the electromagnetic wave, then

Wavelength

λ = $\frac{c}{f}$

= $\frac{3 \times 10^{8}}{3.98 \times 10^{5}}$

λ = 7.54 × 10² m

For Daily Free Study Material Join wiredfaculty