A transformer on a utility pole operates at an rms potential of 750,000 v (relat

Question

A transformer on a utility pole operates at an rms potential of 750,000 v (relative to the Earth) and supplies electric energy to a number of nearby houses at an rms potential of 120 v (relative to the Earth). Assume an ideal step-down transformer, a purely resistive load, and a power factor of one. a) Find the turns ratio N_p/N_s of the transformer. b) The average rate of energy consumption in the houses served by the transformer is 75.0 kW. Find the rms currents in the primary and secondary of the transformer. c) Find the resistive load in the secondary circuit. d) Find the resistive load in the primary circuit.

Explanation / Answer

(a)

from the transformer ratio formula

Np/Ns = Vp/ Vs = 750000/120 = 6250

(b)

ip = Pav/ Vp = 75 * 10^3 W/750 * 10^3 = 0.1 A

is = Pav/ Vs = 75 * 10^3 W/120 V = 625 A

(c)

Rs = Vs/ is = 120/625 = 0.192 ohm

(d)

Rp = Vp/iP = 750 * 10^3 /0.1 = 7500 * 10^3 ohm

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