Question

32. A neon sign transformer has 450 W AC output with an rms voltage of 15 KV when connected to normal household outlet (120 V). There are 500 turns of wire in the primary coil. (a) How many turns of wire does the secondary coil have? (b) When the transformer is running at full power, what are the currents in the secondary coil and in the primary coil? (c) What is the peak current in the primary coil?

Answer 1

Solution: voltage induced in the primary and secondary winding ₂ & u ISK V and 120 V. & Number of turns in primary coil are norin, = 500 turn. (a) Number of turns in secondary coil- 2 N2 M کہا ۔ ما را N2 = ( x 2) Ni 3 15x10 X 500 N2 62500 Ane The current in the secondary coil- (b] I2 = P V2 450 15x103 I2 = 0.03 A and current in the primary coil- I = I2 (42) = 0.03 62500 500 I = 3.75 Am

(C) the Peak current in the coil- Primary I peak -12 Irms = 2x 3.75 I peak 5.30 A An Peak