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 the ratio of secondary to primary turns of the transformer? b. What power must be supplied to the transformer when the rms secondary current is 8.50 mA?
c. What current rating should the fuse in the primary circuit have?

Answer 1

Answer with Explanation:

We are given that

$V_1=120 V$

$V_2=13000 V$

$I_{2}=8.5 mA=8.5\times 10^{-3} A$

$1 mA=10^{-3} A$

a.We know that

$\frac{N_2}{N_1}=\frac{V_2}{V_1}$

$\frac{N_2}{N_1}=\frac{13000}{120}=108.3$

b.$P_{avg}=I_2V_2$

$P_{avg}=13000\times 8.5\times 10^{-3}=110.5 Watt$

c.$I_1=\frac{P_{avg}}{V_1}$

$I_1=\frac{110.5}{120}=0.92 A$

Answer 2

Answer:

Explanation:

Voltage in primary, Vp = 120 V

Voltage in secondary, Vs = 13000 V

Current in secondary, Is = 8.5 mA

(a)

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

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

Ns : Np = 108.33

(b)

For an ideal transformer

Input power = Output power

Input Power = Vs x Is = 13000 x 0.0085

Input power = 110.5 Watt

(c)

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

$I_{p}=\frac{13000}{120}\times 8.5$

Ip = 920.8 mA