During the positive half cycle of the supply, diodes D1 and D2 conduct are forward biased and conduct current while diodes D3 and D4 are reverse biased and they act as an open circuit, the current flows through the load.
In the given diagram all are NOR Gate . The final output is shown in the figure. At stage 1 the output will be \overline A \& \overline B At stage 2 the output will be \overline {\overline A + \overline B } = A.B And the final output will be \overline {A.B} Hence for input A & B the output is \overline {AB} in case of Nand gate.
When variable loss becomes equal to the constant loss, efficiency is maximum. Losses = Pi + Pc Since copper loss is a variable loss therefore Losses = Pi + Pi = 2pi Thus at a maximum efficiency of this transformer total loss = 150 x 2 = 300 W
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