A transmitting station releases waves of wavelength 960   m . A capacitor of 2 . 56   μ F is used in the resonant circuit. The self-inductance of coil necessary for resonance is x × 10 - 8   H . find x

Step-by-step solution

λ = 960   m C = 2 . 56   μ F = 2 . 56 × 10 - 6   F c = 3 × 10 8   m   s - 1 L = ? Now at resonance, ω 0 = 1 L C [Resonant frequency] 2 π f 0 = 1 L C On substituting f 0 = c λ , we have 2 π c λ = 1 L C On substituting f 0 = c λ , we have 2 π c λ = 1 L C Squaring both sides: 4 π 2 c 2 λ 2 = 1 L C = 4 × 10 × 3 × 10 8 2 960 2 = 1   L × 2 . 56 × 10 - 6 ⇒ 1   L = 4 × 10 × 9 &#2