Question

A transmitting station releases waves of wavelength 960 m. A capacitor of 2.56 µF is used in the resonant circuit. The self inductance of coil necessary for resonance is _________ x10\^{-8} H.

Hint:

In resonance problems involving radio waves, speed of light ($c = 3 \times 10^8$ m/s) is the link between wavelength and the frequency needed for the $LC$ circuit formula.

Answer 1

Correct Answer: 10

Step 1: Frequency $f = \frac{c}{\lambda} = \frac{3 \times 10^8}{960}$ Hz.
Step 2: Resonance condition: $f = \frac{1}{2\pi\sqrt{LC}} \Rightarrow L = \frac{1}{4\pi^2 f^2 C}$.
Step 3: $L = \frac{\lambda^2}{4\pi^2 c^2 C} = \frac{960^2}{4 \times 10 \times (3 \times 10^8)^2 \times 2.56 \times 10^{-6}}$ (Using $\pi^2 \approx 10$).
Step 4: $L = \frac{921600}{40 \times 9 \times 10^{16} \times 2.56 \times 10^{-6}} = \frac{921600}{921.6 \times 10^{11}} = 1000 \times 10^{-11} = 10 \times 10^{-8}$ H.
Step 5: Comparing with $x \times 10^{-8}$, we get $10$.