The secondary coil of an ideal neon sign transformer provides 7500 V at 10.0 mA. The primary coil operates on 120 V. What current does the primary draw?
A) 0.160 A
B) 0.625 mA
C) 0.625 A
D) 1.66 A

Question

Quizplus · Accepted Answer

For an ideal transformer, power in the primary coil equals power in the secondary coil. Thus, Vp * Ip = Vs * Is. Given Vs = 7500 V, Is = 10.0 mA, and Vp = 120 V, we solve for Ip: Ip = (Vs * Is) / Vp = (7500 V * 0.010 A) / 120 V = 0.625 A.

The Secondary Coil of an Ideal Neon Sign Transformer Provides