Two coils P and S have a mutual inductance of ( ) mH. The secondary coil S has resistance 4 , and self inductance (60/ ) mH. If the current in the primary is I_p = 12 (50 t) , then the maximum value of the current induced in coil S is [Take ^2 = 10 ]

Step-by-step solution

The induced emf in the secondary coil is given by e_s = -M dI_p dt . Given I_p = 12 (50 t) , differentiating with respect to time gives dI_p dt = 600 (50 t) . Substituting M = 10⁻³ H, we get: e_s = -( 10⁻³) 600 (50 t) = -600 ^2 10⁻³ (50 t) Using ^2 = 10 , the maximum induced emf is E₀ = 600 10 10⁻³ = 6 V. The angular frequency of the AC source is = 50 rad/s. The inductive reactance of the secondary coil is X_L = L_s = 50 60 10⁻³ = 3 , . The impedance of the secondary coil is Z = R^2 + X_L^2 = 4^2 + 3^2 = 5 , . The