Question

A transformer connected to a 120-V (rms) ac line is to supply 12,500 V (rms) for a neon sign. To reduce shock hazard, a fuse is to be introduced in the primary circuit; the fuse is to blow when the rms current in the secondary circuit exceeds 8.50 mA. (Here the "primary" line is the 120 V line and the "secondary" is the 12,500 V line). Part A
Part A). What is the ratio of secondary to primary turns of the transformer? N2/N1 = ___________
Part B). What power must be supplied to the transformer when the rms secondary current is 8.50 mA? P=
Part C) What current rating should the fuse in the primary circuit have? I1=_____________A

Answer 1

Answer:

a) 104

b) 106

c) 884 mA

Explanation:

The ratio of the transformer is given by:

$N=\frac{V_{out}}{V_{in}}=\frac{12500V}{120V}=104$

We need to know the current in the primary in order to obtain the power applied.

$I_1=I_2*N=8.50mA*104=884mA\\\\P=I*V=884*10^{-3}A*120V=106W$

The current rating of the fuse is the current on the primary, 884mA as we calculated before in order to obtain the power.