Answer to Question #115638 in Electric Circuits for Jafar Abbas

Question

A Step-Up Transformer connected to a 120-V (rms) AC line is to supply 13,000V (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 8.50mA? (c) What current rating should the fuse in the primary circuit have?

Answer 1

Assumption

• Since no any efficiency value is mentioned, the transformer is assumed ideal such that input (primary) power = output (secondary) power.

Explanations & Calculations

• V_p, N_p, i_p : Voltage, number of turns, & current for the primary side.
• V_s, N_s, i_s : Voltage, number of turns, & current for the secondary side.
• Secondary side is where the requirement is ; in this case the high voltage side.

a). Induced voltage is directly proportional to the number of turns, therefore,

$\qquad\qquad \begin{aligned} \small \frac{N_s}{N_p} &= \small \frac{V_s}{V_p}\\ &= \small \frac{13000V}{120V}\\ &= \small \bold{108.33} \end{aligned}$

b). Secondary power = V_s$\times$i_s = 13000V $\times$ 8.5*10^-3 A = 110.5 W

Since secondary power = primary power, above amount should be supplied.

c). Primary current (while secondary's is 8.5mA) should be calculated to estimate the rating of the fuse since it would be the maximum possible.

$\qquad\qquad \begin{aligned} \small 110.5W &= \small V_p \times i_p\\ \small i_p &= \small \frac{110.5W}{120V}\\ &= \small 0.92083A = \bold{920.83mA} \end{aligned}$