The correct answer is: A. copper loss becomes high in proportion to the output.
Copper loss is the heat generated in the transformer’s windings due to the resistance of the copper wire. The amount of copper loss is proportional to the square of the current, so it increases rapidly as the load on the transformer increases.
Iron loss is the heat generated in the transformer’s core due to eddy currents and hysteresis losses. The amount of iron loss is proportional to the frequency of the applied voltage, so it is the same for all loads.
Voltage drop in the primary and secondary windings is due to the resistance of the windings and the leakage inductance of the transformer. The voltage drop is proportional to the current, so it increases as the load on the transformer increases. However, the voltage drop is usually small compared to the output voltage of the transformer, so it does not have a significant effect on the efficiency of the transformer.
Secondary output is much less as compared to primary input is not the correct answer because the output voltage of a transformer is always greater than or equal to the input voltage. The ratio of the output voltage to the input voltage is called the turns ratio of the transformer. The turns ratio is always greater than or equal to 1, so the output voltage is always greater than or equal to the input voltage.
In conclusion, the efficiency of a transformer, under heavy loads, is comparatively low because copper loss becomes high in proportion to the output.