Q. 8

Truth table for the given circuit (Fig. 14.6) is



A.



























A


B


E


0


0


1


0


1


0


1


0


1


1


1


0



B.




























A


B


E


0


0


1


0


1


0


1


0


0


1


1


1


C.




























A


B


E


0


0


0


0


1


1


1


0


0


1


1


1


D.




























A


B


E


0


0


0


0


1


1


1


0


1


1


1


0


Answer :

In the given network of gates, the output at C is due to A and B through the AND gate. The output at D is due to the signal of A through the NOT gate and B. Therefore,



and


The output at E is due to C and D through an OR gate. therefore



Therefore, the truth table can be constructed as follows,


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