Question

Question 32 (6 points) 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? Ω Format Β Ι Ο ..

Answer 1

N1 = 500

a)

$\frac{N_{1} }{N_{2} } = \frac{V_{1} }{V_{2} }$

$N_{2} =\left ( \frac{V_{2} }{V_{1} } \right )N_{1}$

$N_{2} =\left ( \frac{15\times 10^{3}V }{120 V } \right )\times 500$

b)

Power P = IV

$I_{2} = \frac{P}{V_{2}}$

$I_{2} = \frac{450 W}{15\times 10^{3}V}$

the current in the secondary coill is

the current in the primary coill is

$I_{1} =\left ( \frac{N_{2} }{N_{1} } \right )I_{2}$

$I_{1} =\left ( \frac{62500 }{500 } \right )\times 0.03 A$

the current in the primary coill is

c)

the peak current in the primary coill is $I_{1peak} = \sqrt{2}I_{1}$

$I_{1peak} = \sqrt{2}\times 3.75 A$