When a 60 mH inductor and resistor are connected in series with an AC source, the voltage leads the current by 60^ . If the inductor is replaced by a 0.5 F capacitor, the voltage lags behind the current by 30^ . The frequency (in Hz) of AC supply is ( 30^ = 1 3 , 60^ = 3 )

Step-by-step solution

For the LR circuit, the phase angle ₁ is given by ₁ = X_L R = L R . Given ₁ = 60^ , we have 60^ = L R 3 = L R R = L 3 . For the CR circuit, the phase angle ₂ is given by ₂ = X_C R = 1 C R . Given ₂ = 30^ , we have 30^ = 1 C R 1 3 = 1 C R R = 3 C . Equating the two expressions for R : L 3 = 3 C ^2 = 3 LC Substituting the given values L = 60 10⁻³ H and C = 0.5 10⁻⁶ F: ^2 = 3 60 10⁻³ 0.5 10⁻⁶ = 3 30 10⁻⁹ = 3 3 10⁻⁸ = 10^8 = 10^4 rad/s The frequency of the AC supply is: f = 2 = 10^4 2 Hz Answer: 10^4 2