Question

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

Answer 1

Answer:

(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:

$\frac{N_{s}}{N_{p}} = \frac{V_{s}}{V_{s}}$

where

$N_{p}$ = No. of turns in primary

$N_{s}$ = No. of turns in secondary

$\frac{N_{s}}{N_{p}} = \frac{13000}{120}$ ≈ 108

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

Power, P = $V_{s}I_{s} = 13000\times 8.50 = 110.500\ kW$

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

$\frac{V_{s}}{V_{p}} = \frac{I_{p}}{I_{s}}$

$\frac{13000}{120} = \frac{I_{p}}{8.50}$

$I_{p} = \frac{13000}{120}\times 8.50 = 920.84\ A$