Q: When all the inputs of a NAND gate are connected together, the resulting circuit is:

The equation of NAND gate is \$Y=\bar{A}.\bar{B}\$When both the terminals are connected together.\$A=B=A\$\$Y=\bar{A}.\bar{A} =\bar{A}. \bar{A}=\bar{A}\$Which is nothing but the equation of NOT gate.\$Y=\bar{A}\$Option \$\textbf A\$ is the correct answer

Q: The Boolean expression for NAND gate is -

The Boolean algebra for \${\text{AND}}\$ gate is-\$Y = A.B\$So, the Boolean algebra for \${\text{NAND}}\$ gate is-\$Y = \overline {A.B} \$

Q: In the circuit below, \$A\$ and \$B\$ represent two inputs and \$C\$ represents the output. The circuit represents

Given, \$A\$ and \$B\$ are inputs and \$C\$ is output.Assume that some voltage is applied across \$AB,\$If the voltage is positive, then diode corresponding to \$A\$ will be on and the other diode will be off\$\Rightarrow C=A\$If the voltage is negative then diode corresponding to \$B\$ will be on and the other diode will be off\$\Rightarrow C=B\$Therefore we can say that \$C=A or B\$Therefore it acts like \$OR\$ gate.So the correct option is \$D.\$

Q: Name the gate, which represents the Boolean expression \$Y = \overline{A\cdot B}\$

Boolean equation for NAND gate is \$Y = \overline{A.B}\$

Question 1

Write the logic symbol and truth table of NAND gate.

Accepted Answer

In digital electronics, a NAND gate which is a negative-AND gate, is a logic gate which produces an output which is false only if all its inputs are true. Thus its output is complement to that of the AND gate.

Question 2

How will you obtain OR, AND gates from the NAND and NOR gates? Write symbol, Boolean formula and truth table.

Accepted Answer

(i) To obtain AND gates from NAND gates: For obtaining AND gate from NAND gate the output of NAND gate is connected to the input (obtained from NAND gate by joining the two inputs).For obtaining OR gate from NAND gate the inputs $A$ and $B$ are connected to the two NOT gates obtained from NAND gates. The input get inverted as $\overline { A }$ and $ \overline { B }$. These inputs are then fed to the NAND gate as shown in fig. drawn below.Boolean formual: $y=\overline { y } =\overline { \overline { A\cdot B }  } =A\cdot B$Input  OutputA B y y01  01 0  0 1 1 1 1 1 0 0 0 0  1(ii) To obtain OR gate from NAND gate: For obtaining OR gate from NAND gate the input $A$ and $B$ are connected to the two NOT gates obtained from NAND gates. The input gate inverted as $\overline { A }$ and $ \overline { B }$. These inputs are then fed to the NAND gate as shown in fig, drawn belows.Boolean formula: $ y=\overline { \overline { A }  } \overline { \overline { B }  } =\overline { \overline { A }  } +\overline { \overline { B }  } =A+B$Input OutputA0 1    01 B0011 $\overline { A }$ 1 0 1 0$ \overline { B }$  1  1  0  0 y 0 1 1 1(iii) To obtain OR gate from NOR gate: To obtain OR gate from NOR gate the inputs are connected to NOR gate which in turn is again connected to NOR gate.Input  Output   A      $\overline { A }$   0   1       1       0(iv) To obtain AND gate from NOR gate: To obtain AND gate from NOR gate the inputs are connected to NOT gate obtained from NOR gate giving inverted $ A\left( \overline { A }  \right) $ and inverted $B\left( \overline { B }  \right)$.Which is inturn are again connected is NOR gate.Boolean formula: $ \overline { y } =\overline { A } +\overline { B } =\overline { \overline { A } +\overline { B }  } =A\cdot B$

Question 3

The logic circuit shown below has the input waveforms 'A' and 'B' as shown. Pick out the correct output waveform.

Accepted Answer

From the figure we can see the logic gate table of NOT and NOR gate$A$                                       $NOT \ gate(X)$1                                                      00                                                     10                                                     11                                                      01                                                      00                                                     10                                                     1$B$                                      $NOT \ gate(Y)$1                                                     00                                                    11                                                     00                                                    11                                                     00                                                    11                                                     0$X$                               $Y$                     $NOR \ gate$0                                0                                   11                                 1                                    01                                 0                                   00                                1                                    00                                0                                   11                                 1                                    0  1                                 0                                   0

Question 4

The Boolean expression for NOR gate is:

Accepted Answer

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Question 5

Explain why NAND gate is known as universal gate.

Accepted Answer

NAND Gate is called Universal Gates because all the other gates (NOT, AND, OR and NOR)can be created by using this gate.

Question 6

Which logic gates is represented by the following combination of logic gates?

Accepted Answer

This is a case of AND gate. Input and output are shown below$\therefore$ $y=\overline { \overline { A } +\overline { B }  } =\overline { \overline { A } .\overline { B }  } =AB$ (since $\overline { A } +\overline { B } =\overline { A } .\overline { B } $)

Question 7

When all the inputs of a NAND gate are connected together, the resulting circuit is:

Accepted Answer

The equation of NAND gate is $Y=\bar{A}.\bar{B}$When both the terminals are connected together.$A=B=A$$Y=\bar{A}.\bar{A} =\bar{A}. \bar{A}=\bar{A}$Which is nothing but the equation of NOT gate.$Y=\bar{A}$Option $\textbf A$ is the correct answer

Question 8

The Boolean expression for NAND gate is -

Accepted Answer

The Boolean algebra for ${\text{AND}}$ gate is-$Y = A.B$So, the Boolean algebra for ${\text{NAND}}$ gate is-$Y = \overline {A.B} $

Question 9

In the circuit below, $A$ and $B$ represent two inputs and $C$ represents the output. The circuit represents

Accepted Answer

Given, $A$ and $B$ are inputs and $C$ is output.Assume that some voltage is applied across $AB,$If the voltage is positive, then diode corresponding to $A$ will be on and the other diode will be off$\Rightarrow C=A$If the voltage is negative then diode corresponding to $B$ will be on and the other diode will be off$\Rightarrow C=B$Therefore we can say that $C=A or B$Therefore it acts like $OR$ gate.So the correct option is $D.$

Question 10

Name the gate, which represents the Boolean expression $Y = \overline{A\cdot B}$

Accepted Answer

Boolean equation for NAND gate is $Y = \overline{A.B}$

Introduction to Logic Gates and Boolean Expressions

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Basic Knowledge of Logic Gates: Basic Gates

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Logic Gates and their Combinations

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