The combinations of the ‘NAND’ gates shown here in fig. are equivalent to
Answer & explanation
Correct answer: option 1
For first case, $C_1 =\overline{\overline A.\overline B} = (A+ B)$ (by Demorgan's theorem)
The truth table is shown below
This is truth table for $C_1 = A+ B$ i.e. OR gate
For second case, $C_2 = \overline{\overline{A.B}} = A.B$ i.e., AND gate.