Question

A transformer connected to a 120-v(rms) at line is to supply 13,000V(rms) for a neon sign.

To reduce shock hazard, a fuse is to beinserted in the primary circuit;the fuse is to blow when the rmscurrent 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

(a) 108

(b) 110.500 kW

(c) 920.84 A

Solution:

As per the question:

Voltage at primary,
`V_(p) = 120\ V` (rms voltage)

Voltage at secondary,
`V_(s) = 13000\ V` (rms voltage)

Current in the secondary,
`I_(s) = 8.50\ mA`

Now,

(a) The ratio of secondary to primary turns is given by the relation:

`(N_(s))/(N_(p)) = (V_(s))/(V_(s))`

where

`N_(p)` = No. of turns in primary

`N_(s)` = No. of turns in secondary

`(N_(s))/(N_(p)) = (13000)/(120)` ≈ 108

(b) The power supplied to the line is given by:

Power, P =
`V_(s)I_(s) = 13000* 8.50 = 110.500\ kW`

(c) The current rating that the fuse should have is given by:

`(V_(s))/(V_(p)) = (I_(p))/(I_(s))`

`(13000)/(120) = (I_(p))/(8.50)`

`I_(p) = (13000)/(120)* 8.50 = 920.84\ A`