A transformer used in an electronic power supply has a nominal rating of 120/12 volts and is connected to a 120-Vac source. Its equivalent impedance as seen from the primary is 10 + j10 Ohm.

Q1. What is the magnitude of the load voltage if the load is 5 ohms resistive?
Q2. Determine the regulation.

Q1. My steps:
1. ZL = 5 Ohm
2. Turns Ratio, n = 120/12 = 10
3. Reflected Load Impedance Zeq = n^2 * ZL = 500
4. Total impedance, Ztot = 10 + j10 + 500 = 510 + j10 = 510.1, phase 1.123º
5. Reflected Load Voltage, VeqL = (500/Ztot)*120 = 117.624V, phase -1.123º
6. Load Voltage = VeqL/n = 11.76V, phase -1.123º

Q2.
Vnl is Secondary Voltage at No Load = 120V/n = 12V
Vfl is Secondary Voltage at Full Load = 11.76V
Regulation = (Vnl-Vfl)/Vfl% = (12-11.76)/11.76% = 2%

The book says the right answer is 1.8V with 2% regulation but this seems clearly wrong (or am I going mad?)

Can anybody please confirm this?

 

this should be in the homework section...

 

Sounds ok to me, 1.8V in the book could be just a typo for 11.8V which is approx. equal to your 11.76V !