An AC generator having 400 turns and an area of cross section of 2 10⁻³ m^2 rotates with an angular speed of 200 rad s⁻¹ in a uniform magnetic field of strength 0.4 T. The generator is connected to the primary of an ideal transformer having 500 turns in the primary and 2000 turns in the secondary. The secondary is connected to a 400 resistive load. What is the rms current in the secondary of the transformer? Assume,

Step-by-step solution

The peak emf generated by the AC generator is given by E₀ = NBA . Substituting the given values, E₀ = 400 0.4 (2 10⁻³) 200 = 64 V . The rms voltage across the primary coil of the transformer is V_p = E₀ 2 = 64 2 V . For an ideal transformer, the relation between secondary and primary voltages is V_s V_p = N_s N_p . Substituting N_s = 2000 and N_p = 500 , we get V_s = V_p ( 2000 500 ) = 4V_p . V_s = 4 64 2 = 256 2 V . The rms current in the secondary circuit is I_s = V_s R . I_s = 256 2 400 = 256 400 2 = 0.64 2 A .