Answer :

The given circuit is as follows:


The first NOR gate gives the output (A + B)’. The second NOR gate gives the output ((A + B)’ + (A + B)’)’.


Using De Morgan’s law,


((A + B)’ + (A + B)’)’ = ((A + B)’)’∙((A + B)’)’ = (A + B)∙(A + B) = A + B (Since, A∙A = A)


So, the output is Y = A + B and the circuit acts as an OR gate.


The truth table for the given circuit is as follows:



NOTE: NOR gate is a universal gate because it can be used to implement any Boolean function without using any other gates.


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