The circuit shown in the question is an AND GATE. In an AND gate has two or more inputs but it has only one output. An input signal applied to a gate has only two stable states, either 1 (HIGH) or 0 (LOW). In AND gate for any input A&B the output is A.B.
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.
Admittance (Y) is the reciprocal of the impedance of a circuit. Admittance of an AC circuit is analogous to the conductance of a DC circuit. The unit of Admittance is Simen or MHO Admittance = 1/Z simen Y = Conductance ± J Susceptance Or the Admittance can be written as Y = (G ± J B) Simen Now comparing the above equation by the given equation in the question i.e Y= a + jb ∴ a = G = Conductance
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