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## Homework Statement

A farmer installs a private hydroelectric generator to provide power for equipment rated at 120 kW 240 V AC. The generator is connected to the equipment by two conductors which have a total resistance of 0.20 Ω. The system is shown schematically in Fig. 1.

(a) The equipment is operating at its rated power. Calculate: (i) the power loss in the cables, (ii) the voltage which must be developed by the generator, (iii) the efficiency of the transmission system.

(b) An engineer suggests that the farmer uses a transformer to convert the generator output to give a PD of 2400 V at the end of the transmission line, as shown in Fig. 2. A second transformer is to be used to step down this PD to 240 V.

(i) Explain briefly how a transformer makes use of electromagnetic induction to produce an output voltage several times bigger than the input voltage.

(ii) The transformers are 100 % efficient. Calculate the power loss in the new transmission system.

Answers: (a) (i) 50 kW, (ii) 340 V, (iii) 70.6 %, (b) (ii) 500 W.

**2. The attempt at a solution**

(a) (i) P = V I → I = P / V = 120 000 / 240 = 500 A. P = I

^{2}R = 500

^{2}* 0.2 = 50 kW.

(a) (ii) V = P / I = 50 000 / 500 = 100 V in the cables plus 240 V in the equipment so 340 V in total.

(a) (iii) Efficiency = 120 000 / (120 000 + 50 000) = 0.706 or 70.6 %.

(b) (ii) This one I'm not sure. I found the power: P = V

^{2}/ R = 2400

^{2}/ 0.2 = 28 800 000 W and current I = V / R = 2400 / 0.2 = 12 000 A. And also current in the equpment: I = P / V = 120 000 / 240 = 500 A. But I'm not sure whether I'm going in the right direction. The numbers are so largely different.