Ncert Exercises & Solutions

A truth table for the given circuit is:

1.	\(A\)	\(B\)	\(E\)	2.	\(A\)	\(B\)	\(E\)
	\(0\)	\(0\)	\(1\)		\(0\)	\(0\)	\(1\)
	\(0\)	\(1\)	\(1\)		\(0\)	\(1\)	\(0\)
	\(1\)	\(0\)	\(1\)		\(1\)	\(0\)	\(0\)
	\(1\)	\(1\)	\(0\)		\(1\)	\(1\)	\(0\)

3.	\(A\)	\(B\)	\(E\)	4.	\(A\)	\(B\)	\(E\)
	\(0\)	\(0\)	\(0\)		\(0\)	\(0\)	\(0\)
	\(0\)	\(1\)	\(1\)		\(0\)	\(1\)	\(1\)
	\(1\)	\(0\)	\(0\)		\(1\)	\(0\)	\(1\)
	\(1\)	\(1\)	\(1\)		\(1\)	\(1\)	\(0\)

Hint: The output depends on the logic of the circuit.

Step: Find the truth table for the circuit.
Here, \(C=A.B\) and \(D=\bar A.B\)
\(E=C+D=(A.B)+(\bar A. B)\)

The truth table of this arrangement of gates can be given by;

\(A\)	\(B\)	\(\bar A\)	\(C=A.B\)	\(D=\bar A.B\)	\(E=(C+D)\)
\(0\)	\(0\)	\(1\)	\(0\)	\(0\)	\(0\)
\(0\)	\(1\)	\(1\)	\(0\)	\(1\)	\(1\)
\(1\)	\(0\)	\(0\)	\(0\)	\(0\)	\(0\)
\(1\)	\(1\)	\(0\)	\(1\)	\(0\)	\(1\)

Hence, option (3) is the correct answer.