Question

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. Assume that the transformer is ideal.

(a) 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?

(b). 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.

(c). 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.

(d). 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.

Answer 1