Final answer:
The efficiency of the 2000 VA 440/110 V transformer at full load with a power factor of 0.80 is calculated to be 93.346%.
Step-by-step explanation:
To calculate the efficiency of the transformer, we need to use the information provided from the open and short circuit tests to find the core loss (Po), copper loss (Pcu), and then the efficiency itself at full load with a power factor of 0.80.
From the open-circuit test, the core loss (Po) can be determined since the power measured is mostly due to the core loss when the transformer is on no load. Po = 70 W.
From the short-circuit test, the copper loss (Pcu) at full current can be found. The actual input power is mostly copper loss since the impedance is very low and the voltage is just 15 V. Hence, Pcu = 44 W.
At full load, the transformer's apparent power (S) is 2000 VA, and with a power factor of 0.80, the real power (P) is P = S × power factor = 2000 VA × 0.80 = 1600 W.
Total losses equal the sum of core and copper losses. Total losses = Po + Pcu = 70 W + 44 W = 114 W.
The efficiency (η) at full load is the ratio of output power to the input power, which is the output power divided by the output power plus losses: η = P / (P + Total losses) = 1600 W / (1600 W + 114 W).
Calculating this gives us η = 1600 / 1714 = 0.93346 or 93.346%