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  • The combinations of the ‘NAND’ gates shown here in fig. are equivalent to <img alt="" src="https://cdn.entrance360.com/media/uploads/2023/08/18/2.PNG" style="height:100px; wid

The combinations of the ‘NAND’ gates shown here in fig. are equivalent to

Option: 1

an ‘OR’ gate and an ‘AND’ gate respectively
 


Option: 2

an ‘AND’ gate and a ‘NOT’ gate respectively
 


Option: 3

an ‘AND’ gate and an ‘OR’ gate respectively
 


Option: 4

an ‘OR’ gate and a ‘NOT’ gate respectively


Answers (1)

best_answer

For first case, \mathrm{\mathrm{C}_1=\overline{\overline{\mathrm{A}} \cdot \overline{\mathrm{B}}}=(\mathrm{A}+\mathrm{B})} (by Demorgan's theorem)

The truth table is shown below

\mathrm{A} \mathrm{B} \mathrm{\bar{A}}
1 0 0
0 1 1
0 0 1
1 1 0

 

\mathrm{\bar{B}} \mathrm{\bar{A}.\bar{B}} \mathrm{\overline{\overline{\mathrm{A}} \cdot \overline{\mathrm{B}}}}
1 0 1
0 0 1
1 1 0
0 0 1

 This is truth table for \mathrm{C_1=A+B} i.e. OR gate

For second case, \bar{\bar{A.B}}=A.B i.e , AND gate

 

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