In first case, A acts as the two inputs of the NOR gate and Y is the output. Hence, the output of the circuit is ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯A+A.
So, Output Y=¯A.
Thus the gate is NOT gate.
The truth table is shown in the figure.
For second circuit, A and B are the inputs and Y is the output of the given circuit. By using the result obtained
in solution (a), we can infer that the outputs of the first two NOR gates are ¯A, ¯B. ¯A and ¯B are the inputs for the last NOR gate. Hence, the output for the circuit can be
written as:
Y=¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯A+¯B=A.B
So this is a AND gate.
Truth table is shown in the figure.