A transformer has 400 turns in its primary winding and 800 turns in its secondary winding. The primary voltage is 20 ~V and the load in the secondary is 4 ohm. The current in the primary, assuming it to be an ideal transformer, is

Step-by-step solution

For an ideal transformer, the ratio of voltages is equal to the ratio of the number of turns: V_s V_p = N_s N_p Given N_p = 400 , N_s = 800 , and V_p = 20 V , we have: V_s 20 = 800 400 = 2 V_s = 20 2 = 40 V The current in the secondary winding I_s is given by Ohm's law with the load resistance R_s = 4 : I_s = V_s R_s = 40 4 = 10 A For an ideal transformer, the power input equals the power output, so V_p I_p = V_s I_s : 20 I_p = 40 10 20 I_p = 400 I_p = 400 20 = 20 A Answer: 20 A