Question

A transformer connected to a 120 V (rms) ac line is to supply 13,000 V (rms) for a neon sign. To reduce shock hazard, a fuse is to be inserted in the primary circuit; the fuse is to blow when the rms current in the secondary circuit exceeds 8.50 mA.

(a) What is theratio of secondary to primary turns of the transformer?
(b) Whatpower must be supplied to the transformer when the rms secondarycurrent is 8.50 mA?
(c) What current rating should the fuse in theprimary circuit have?

Answer 1

Turns ratio=108, Pin= 110.5 W, Current ratting= 920mA

Step-by-step explanation:

Vp=120 V(rms)

Vs=13,000 V (rms)

Is = 8.50*10^-3 V(rms)

a.

secondary to primary turn ration:

Ns/Np=Vs/Vp

`(Ns)/(Np)` =
`(Vs)/(Vp)`

`(Ns)/(Np)` =
`(13000)/(120)`

`(Ns)/(Np)` = 108

b.

Is=8.50 mA (rms)

As Vs= 13,000V (rms)

For ideal transformer

Pin=Pout______(1)

Pout= Is*Vs (put in equ 1)

Pi=Is*Vs

Pi=(8.50*10^-3)*(13,000) W

Pi=110.5 W

Power supplied will be 110.5 W.

c.

We have to find the current passing through primary circuit to find current rating of fuse in primary circuit.

For ideal case

Ip*Vp= Is*Vs

Ip=
`(Is*Vs)/(Vp)`

Ip= (13000*8.5*10^-3)/120

Ip=920 mA

In primary circuit fuse current rating will be 920 mA.