Question

2(a). A transformer has a primary-to-secondary turns ratio of 100:1. The voltage on the primary side is 27.7 kv.a.c. at 60.0 Hz ( assume 60.0 Hz when treating electric power topics). A load is connected across the secondary winding. Assuming that the transformer is ideal, what voltage is developed to the load? Briefly show your work. Sketch a diagram.

2(b). The load of part a is a 2.00-ohm load. How much current does the load draw from the transformer’s secondary winding? [Answer: 138.5 amps.]

2(c). How much current must be supplied to the transformer’s primary winding?

2(d). How much power does the secondary winding deliver to the load? In view of the first law of thermodynamics and that the transformer is assumed to be ideal, how much power is drawn by the transformer’s primary winding? Show that your answer to part d is consistent with your answers to parts (b) and (c).

2(e). Referring parts (a) through (d). Suppose that the electric utility supplied 27.7 kv.d.c. rather than 27.7 kv.a.c. How much power would the transformer’s secondary winding deliver to the load in this case?

2(f). The transformer’s primary winding is supplied by a pair of wires from the electric utility. Show that, with the 2.00-ohm load connected across the secondary winding, the primary winding appears to the utility as it were a 20.0-kilohm resistor.

2(g). Briefly explain why, if the 2.00-ohm load is removed from the secondary, leaving an open circuit, the primary—though indeed connected to the utility—appears to the utility as it were an open circuit.

2(h). Referring to part h, express the voltage across and current entering the transformer’s primary winding in the time domain in the short-circuit case. That is, write expressions for v(t) and i(t). It is recommended to work in phasors and then to change to the time domain at the end of your work, but don’t forget to adjust for rms.

2(i). Returning to parts (a) through (f), in which a 2.00-ohm load is connected across the transformer’s secondary winding, suppose that the electric utility supplies the transformer’s primary winding by a pair of wires (two wires) suspended from utility poles. The wires are 50.0 miles long and each has a resistance of 1.50 ohms per mile. How much power is lost in the utility’s wires

2(j). Referring to part (i), compute the system’s efficiency.

2(k). If no transformer were available, the utility would have to supply power at 277 v.a.c. over its 50.0-mile pair of wires. Suppose that no transformer were available and that the utility did indeed supply power at 277 v.a.c. Suppose that the industrial plant’s 2.00-ohm load were directly connected across the utility’s pair of wires. Repeat parts (i) and (j) in this case.

2(k). The cost of an industrial plant’s main transformer can run to the hundreds of thousands of dollars. Seemingly, this cost could be saved if the electric utility and the plant only used the same voltage. A voltage like 27.7 kv is too high to handle safely inside a plant, but it would seen that the utility could safely use 277 v. This is indeed true: the utility could safely use 277 v. In view of part p, briefly, why doesn’t the utility use 277 v.?

Answer 1