Question

A neon sign transformer has 500 W AC output with an rms voltage of 16 kV when connected to a normal household outlet. There are 80,000 turns of wire in the secondary coil. When the transformer is running at full power, what is the peak current in the primary coil?

Answer 1

The current in primary coil is 2.08 A.

Step-by-step explanation:

Given that,

Power = 500 W

Voltage = 16 kV

Number of turns = 80000

We need to calculate the number of current

Using formula of voltage

`(N_(1))/(N_(2))=(V_(1))/(V_(2))`

Put the value into the formula

`(N_(1))/(80000)=(120)/(16*10^(3))`

`N_(1)=(240)/(16*10^(3))*80000`

`N_(1)=1200`

We need to calculate the current in secondary coil

Using formula of current

`i=(P)/(V_(2))`

`i_(2)=(500)/(16*10^(3))`

`i_(2)=0.03125\ A`

We need to calculate the current in primary coil

`(I_(1))/(I_(2))=(N_(2))/(N_(1))`

Put the value into the formula

`(i_(1))/(0.03125)=(80000)/(1200)`

`i_(1)=(80000)/(1200)*0.03125`

`i_(1)=2.08\ A`

Hence, The current in primary coil is 2.08 A.