Explanation: When the variable copper losses of a transformer becomes equal to the fixed iron losses of a transformer then we will get maximum efficiency. From these losses we’ll get the value of current required.

Explanation: The efficiency of the transformer obtained from various experiments conducted on various loads showed the efficiency greater than 90% always. Transformer thus, can be said highly efficient device.

Explanation: Efficiency of the transformer can be calculated by the output power divided by input power. Both of these powers include power factor in their calculations while load current and load voltage is also required in calculations.

Explanation: Every device is manufactured to get maximum efficiency at the rated loads, i.e. full load. Thus, transformer will give the maximum efficiency at nearly full load. Internal losses are so adjusted to get maximum efficiency.

Explanation: The main difference between power transformer and distribution transformer is distribution transformer is designed for maximum efficiency at 60% to 70% load as these transformers normally doesn’t operate at full load all the time.

Explanation: Similar to normal transformers power transformers are also designed to get maximum efficiency at load which is near to the full load of a transformer specified. Only in the case distribution transformer maximum efficiency is achieved at 60% of full load.

Explanation: Maximum efficiency for a transformer will be achieved at full load. While in the case of power factor also every device is set to get maximum efficiency at unity power factor. Thus, one will have maximum efficiency if load is nearly equal to full load and at unity power factor.

Explanation: At heavy loads current drawn by the transformer circuit increases, as we know, variable copper losses are proportional to the square of the current. Thus, we’ll get higher copper loss in proportion to the output.

Explanation: Transformer is a highly efficient device compare to all other electrical instruments. In motor we need to add windage and friction losses along with the copper losses and iron losses thus, we’ll get lee efficiency for motor compare to transformer.

Explanation: Condition for maximum efficiency is, Copper loss= Iron loss, i.e. Pc= I2 R = Pi. transformer can be operated at any load but maximum efficiency occurs at a particular load condition only. Let x be that load factor corresponds to maximum efficiency. Given that, maximum efficiency will occur at 3/4 load. The load factor= (3/4)2.

Explanation: Efficiency of any device is equal to the ratio of output power to the input power. Here, one can write input power is equal to the addition of output power with losses. Thus, expressing all these terms mathematically will give the answer.

Explanation: Efficiency of a transformer is given by, [transformer capacity*loading/ (capacity*loading + Pi + k2*PC)]. Thus, η= 500*1/ (500 + Pi +PC) = 0.95. also from the second condition given η= 500*0.6/ (500*0.6 + Pi +0.6^2*PC) = 0.95. Thus, solving simultaneously we get 9.87 kW.

Explanation: Efficiency of a transformer is given by, [transformer capacity*loading/ (capacity*loading + Pi + k2*PC)]. Thus, η= 500*1/ (500 + Pi +PC) = 0.95. also from the second condition given η= 500*0.6/ (500*0.6 + Pi +0.62*PC) = 0.95. Thus, solving simultaneously we get an answer 16.45 kW.

Explanation: Power output= VI cosθ= 200*100*0.8 = 16000 W (Independent of lag and lead). While total losses are equal to iron loss+ k2*copper losses =120+ 300= 420 W. Efficiency is equal to 1- 420/(16000+420)= 97.44%.