Ncert Exercises & Solutions

Truth table for the given circuit is:

1. $A B E$
$\begin{matrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix}$

2. $A B E$
$\begin{matrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ 1 & 1 & 0 \end{matrix}$

3. $A B E$
$\begin{matrix} 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \end{matrix}$

4. $A B E$
$\begin{matrix} 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix}$

(c) Hint: The output depends on the logic of the circuit.

Step 1: Find the logic for the circuit.

Here, C = A . B and D = \bar{A} . B

E = C + D = (A . B) + (\bar{A} . B)

Step 2: Find the truth table for the circuit.

Explanation 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

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