A transformer which steps down 330 ~V to 33 ~V is to operate a device having impedance 110 . The current drawn by the primary coil of the transformer is :

Step-by-step solution

The secondary voltage V_s is 33 V and the impedance of the device connected to the secondary coil is Z = 110 . The current in the secondary coil I_s is given by Ohm's law: I_s = V_s Z = 33 110 = 0.3 A For an ideal transformer, the power input equals the power output, so V_p I_p = V_s I_s . Given the primary voltage V_p = 330 V , we have: 330 I_p = 33 0.3 I_p = 33 0.3 330 = 0.3 10 = 0.03 A Answer: 0.03 A