QBManager

Semiconductor Electronics Material Devices and Simple Circuits

151297 Which logic gate is represented by the following combination of logic gates?
original image

1

OR

2 NAND

3 AND

4 NOR

Explanation:

C Given that in this circuit using two NOT gate and one NOR gate
original image
Then, the Boolean expression will be-
$Y = \overset{―}{(\overset{―}{A} + \overset{―}{B})} = A \cdot B$
Truth table of $Y -$
{|c|c|c|c|c|c|}
| $Extra close brace or missing open brace$ $Extra close brace or missing open brace$ $Extra close brace or missing open brace$ $Extra close brace or missing open brace$ \)|
|---|
$001110$
$011010$
$100110$
$110001$
\(
Which is output expression of AND gate.

JCECE-2016

Semiconductor Electronics Material Devices and Simple Circuits

151298 The circuit is equivalent to
original image

1 NOR gate

2 AND gate

3 NAND gate

4 OR gate

Explanation:

C Given Circuit,
original image
${\overset{―}{Y}}_{2} = (\overset{―}{\overset{―}{A \cdot B} + \overset{―}{A \cdot B}})$
${\overset{―}{Y}}_{2} = \overset{―}{\overset{―}{A \cdot B}} \cdot \overset{―}{\overset{―}{A \cdot B}}$
${\overset{―}{Y}}_{2} = (A \cdot B) \cdot (A \cdot B)$
${\overset{―}{Y}}_{2} = AB$
$Y = \overset{―}{A \cdot B}$
(apply NOR gate)
(apply NOT gate)
So output,
$Y = \overset{―}{A \cdot B}$ Which is the expression of NAND gate.

JCECE-2015

Semiconductor Electronics Material Devices and Simple Circuits

151299 In the adjoining circuit of logic gate, the output $Y$ becomes zero if the inputs are
original image

1

A = 1, B = 1, C = 0

2

A = 0, B = 0, C = 0

3

A = 0, B = 1, C = 1

4

A = 1, B = 0, C = 0

Explanation:

B Given, circuit output $Y = 0$ then-
original image
The expression of given circuit,
$Y = \overset{―}{A \cdot B \cdot \overset{―}{C}}$ (using De, Morgen theorems)
$Y = \overset{―}{A \cdot B} + \overset{―}{\overset{―}{C}} (\overset{―}{\overset{―}{C}} = C \to Involution theorem)$
$Y = \overset{―}{A \cdot B} + C = \overset{―}{A} + \overset{―}{B} + C$
Then check option (1), $A = 1, B = 1, C = 0$
$Y = \overset{―}{1} + \overset{―}{1} + 0$
$Y = 0 + 0 + 0$
$Y = 0$ This is possible when $A = 1, B = 1$ and $C = 0$ .

JCECE-2013

Semiconductor Electronics Material Devices and Simple Circuits

151300 What is an AND gate?

1 It has not equivalence to switching circuit

2 It is equivalent to switching circuit

3 It is equivalent to parallel switching circuit

4 It is a mixture of series and parallel switching

Explanation:

B AND gate is used as a switching circuit, AND gate is only active when both inputs are high same as series. Switching circuit which is active only when both switches are closed.
The truth table of AND gate will be -
Boolean Expression,
$Y = A \cdot B$
{|l|l|l|}
| $=$ \)|
|---|
$000$
$010$
$100$
$111$
\(
When both input are high then we get high output
original image

JCECE-2011

Semiconductor Electronics Material Devices and Simple Circuits

151301 To get an output 1 from the circuit shown in the figure, the input must be
original image

1

A = 0, B = 1, C = 0

2

A = 1, B = 0, C = 0

3

A = 1, B = 0, C = 1

4

A = 1, B = 1, C = 0

Explanation:

C Given circuit when output is 1 .
original image
In this circuit using OR gate and AND gate. So Boolean expression-
$Y = (A + B) \cdot C ($ Taking value $A = 1, B = 0, C = 1)$
$Y = (1 + 0) \cdot 1$
$Y = 1 \cdot 1$
$Y = 1$

JCECE-2010

Semiconductor Electronics Material Devices and Simple Circuits

151297 Which logic gate is represented by the following combination of logic gates?
original image

1

OR

2 NAND

3 AND

4 NOR

Explanation:

C Given that in this circuit using two NOT gate and one NOR gate
original image
Then, the Boolean expression will be-
$Y = \overset{―}{(\overset{―}{A} + \overset{―}{B})} = A \cdot B$
Truth table of $Y -$
{|c|c|c|c|c|c|}
| $Extra close brace or missing open brace$ $Extra close brace or missing open brace$ $Extra close brace or missing open brace$ $Extra close brace or missing open brace$ \)|
|---|
$001110$
$011010$
$100110$
$110001$
\(
Which is output expression of AND gate.

JCECE-2016

Semiconductor Electronics Material Devices and Simple Circuits

151298 The circuit is equivalent to
original image

1 NOR gate

2 AND gate

3 NAND gate

4 OR gate

Explanation:

C Given Circuit,
original image
${\overset{―}{Y}}_{2} = (\overset{―}{\overset{―}{A \cdot B} + \overset{―}{A \cdot B}})$
${\overset{―}{Y}}_{2} = \overset{―}{\overset{―}{A \cdot B}} \cdot \overset{―}{\overset{―}{A \cdot B}}$
${\overset{―}{Y}}_{2} = (A \cdot B) \cdot (A \cdot B)$
${\overset{―}{Y}}_{2} = AB$
$Y = \overset{―}{A \cdot B}$
(apply NOR gate)
(apply NOT gate)
So output,
$Y = \overset{―}{A \cdot B}$ Which is the expression of NAND gate.

JCECE-2015

Semiconductor Electronics Material Devices and Simple Circuits

151299 In the adjoining circuit of logic gate, the output $Y$ becomes zero if the inputs are
original image

1

A = 1, B = 1, C = 0

2

A = 0, B = 0, C = 0

3

A = 0, B = 1, C = 1

4

A = 1, B = 0, C = 0

Explanation:

B Given, circuit output $Y = 0$ then-
original image
The expression of given circuit,
$Y = \overset{―}{A \cdot B \cdot \overset{―}{C}}$ (using De, Morgen theorems)
$Y = \overset{―}{A \cdot B} + \overset{―}{\overset{―}{C}} (\overset{―}{\overset{―}{C}} = C \to Involution theorem)$
$Y = \overset{―}{A \cdot B} + C = \overset{―}{A} + \overset{―}{B} + C$
Then check option (1), $A = 1, B = 1, C = 0$
$Y = \overset{―}{1} + \overset{―}{1} + 0$
$Y = 0 + 0 + 0$
$Y = 0$ This is possible when $A = 1, B = 1$ and $C = 0$ .

JCECE-2013

Semiconductor Electronics Material Devices and Simple Circuits

151300 What is an AND gate?

1 It has not equivalence to switching circuit

2 It is equivalent to switching circuit

3 It is equivalent to parallel switching circuit

4 It is a mixture of series and parallel switching

Explanation:

B AND gate is used as a switching circuit, AND gate is only active when both inputs are high same as series. Switching circuit which is active only when both switches are closed.
The truth table of AND gate will be -
Boolean Expression,
$Y = A \cdot B$
{|l|l|l|}
| $=$ \)|
|---|
$000$
$010$
$100$
$111$
\(
When both input are high then we get high output
original image

JCECE-2011

Semiconductor Electronics Material Devices and Simple Circuits

151301 To get an output 1 from the circuit shown in the figure, the input must be
original image

1

A = 0, B = 1, C = 0

2

A = 1, B = 0, C = 0

3

A = 1, B = 0, C = 1

4

A = 1, B = 1, C = 0

Explanation:

C Given circuit when output is 1 .
original image
In this circuit using OR gate and AND gate. So Boolean expression-
$Y = (A + B) \cdot C ($ Taking value $A = 1, B = 0, C = 1)$
$Y = (1 + 0) \cdot 1$
$Y = 1 \cdot 1$
$Y = 1$

JCECE-2010

Semiconductor Electronics Material Devices and Simple Circuits

151297 Which logic gate is represented by the following combination of logic gates?
original image

1

OR

2 NAND

3 AND

4 NOR

Explanation:

C Given that in this circuit using two NOT gate and one NOR gate
original image
Then, the Boolean expression will be-
$Y = \overset{―}{(\overset{―}{A} + \overset{―}{B})} = A \cdot B$
Truth table of $Y -$
{|c|c|c|c|c|c|}
| $Extra close brace or missing open brace$ $Extra close brace or missing open brace$ $Extra close brace or missing open brace$ $Extra close brace or missing open brace$ \)|
|---|
$001110$
$011010$
$100110$
$110001$
\(
Which is output expression of AND gate.

JCECE-2016

Semiconductor Electronics Material Devices and Simple Circuits

151298 The circuit is equivalent to
original image

1 NOR gate

2 AND gate

3 NAND gate

4 OR gate

Explanation:

C Given Circuit,
original image
${\overset{―}{Y}}_{2} = (\overset{―}{\overset{―}{A \cdot B} + \overset{―}{A \cdot B}})$
${\overset{―}{Y}}_{2} = \overset{―}{\overset{―}{A \cdot B}} \cdot \overset{―}{\overset{―}{A \cdot B}}$
${\overset{―}{Y}}_{2} = (A \cdot B) \cdot (A \cdot B)$
${\overset{―}{Y}}_{2} = AB$
$Y = \overset{―}{A \cdot B}$
(apply NOR gate)
(apply NOT gate)
So output,
$Y = \overset{―}{A \cdot B}$ Which is the expression of NAND gate.

JCECE-2015

Semiconductor Electronics Material Devices and Simple Circuits

151299 In the adjoining circuit of logic gate, the output $Y$ becomes zero if the inputs are
original image

1

A = 1, B = 1, C = 0

2

A = 0, B = 0, C = 0

3

A = 0, B = 1, C = 1

4

A = 1, B = 0, C = 0

Explanation:

B Given, circuit output $Y = 0$ then-
original image
The expression of given circuit,
$Y = \overset{―}{A \cdot B \cdot \overset{―}{C}}$ (using De, Morgen theorems)
$Y = \overset{―}{A \cdot B} + \overset{―}{\overset{―}{C}} (\overset{―}{\overset{―}{C}} = C \to Involution theorem)$
$Y = \overset{―}{A \cdot B} + C = \overset{―}{A} + \overset{―}{B} + C$
Then check option (1), $A = 1, B = 1, C = 0$
$Y = \overset{―}{1} + \overset{―}{1} + 0$
$Y = 0 + 0 + 0$
$Y = 0$ This is possible when $A = 1, B = 1$ and $C = 0$ .

JCECE-2013

Semiconductor Electronics Material Devices and Simple Circuits

151300 What is an AND gate?

1 It has not equivalence to switching circuit

2 It is equivalent to switching circuit

3 It is equivalent to parallel switching circuit

4 It is a mixture of series and parallel switching

Explanation:

B AND gate is used as a switching circuit, AND gate is only active when both inputs are high same as series. Switching circuit which is active only when both switches are closed.
The truth table of AND gate will be -
Boolean Expression,
$Y = A \cdot B$
{|l|l|l|}
| $=$ \)|
|---|
$000$
$010$
$100$
$111$
\(
When both input are high then we get high output
original image

JCECE-2011

Semiconductor Electronics Material Devices and Simple Circuits

151301 To get an output 1 from the circuit shown in the figure, the input must be
original image

1

A = 0, B = 1, C = 0

2

A = 1, B = 0, C = 0

3

A = 1, B = 0, C = 1

4

A = 1, B = 1, C = 0

Explanation:

C Given circuit when output is 1 .
original image
In this circuit using OR gate and AND gate. So Boolean expression-
$Y = (A + B) \cdot C ($ Taking value $A = 1, B = 0, C = 1)$
$Y = (1 + 0) \cdot 1$
$Y = 1 \cdot 1$
$Y = 1$

JCECE-2010

Semiconductor Electronics Material Devices and Simple Circuits

151297 Which logic gate is represented by the following combination of logic gates?
original image

1

OR

2 NAND

3 AND

4 NOR

Explanation:

C Given that in this circuit using two NOT gate and one NOR gate
original image
Then, the Boolean expression will be-
$Y = \overset{―}{(\overset{―}{A} + \overset{―}{B})} = A \cdot B$
Truth table of $Y -$
{|c|c|c|c|c|c|}
| $Extra close brace or missing open brace$ $Extra close brace or missing open brace$ $Extra close brace or missing open brace$ $Extra close brace or missing open brace$ \)|
|---|
$001110$
$011010$
$100110$
$110001$
\(
Which is output expression of AND gate.

JCECE-2016

Semiconductor Electronics Material Devices and Simple Circuits

151298 The circuit is equivalent to
original image

1 NOR gate

2 AND gate

3 NAND gate

4 OR gate

Explanation:

C Given Circuit,
original image
${\overset{―}{Y}}_{2} = (\overset{―}{\overset{―}{A \cdot B} + \overset{―}{A \cdot B}})$
${\overset{―}{Y}}_{2} = \overset{―}{\overset{―}{A \cdot B}} \cdot \overset{―}{\overset{―}{A \cdot B}}$
${\overset{―}{Y}}_{2} = (A \cdot B) \cdot (A \cdot B)$
${\overset{―}{Y}}_{2} = AB$
$Y = \overset{―}{A \cdot B}$
(apply NOR gate)
(apply NOT gate)
So output,
$Y = \overset{―}{A \cdot B}$ Which is the expression of NAND gate.

JCECE-2015

Semiconductor Electronics Material Devices and Simple Circuits

151299 In the adjoining circuit of logic gate, the output $Y$ becomes zero if the inputs are
original image

1

A = 1, B = 1, C = 0

2

A = 0, B = 0, C = 0

3

A = 0, B = 1, C = 1

4

A = 1, B = 0, C = 0

Explanation:

B Given, circuit output $Y = 0$ then-
original image
The expression of given circuit,
$Y = \overset{―}{A \cdot B \cdot \overset{―}{C}}$ (using De, Morgen theorems)
$Y = \overset{―}{A \cdot B} + \overset{―}{\overset{―}{C}} (\overset{―}{\overset{―}{C}} = C \to Involution theorem)$
$Y = \overset{―}{A \cdot B} + C = \overset{―}{A} + \overset{―}{B} + C$
Then check option (1), $A = 1, B = 1, C = 0$
$Y = \overset{―}{1} + \overset{―}{1} + 0$
$Y = 0 + 0 + 0$
$Y = 0$ This is possible when $A = 1, B = 1$ and $C = 0$ .

JCECE-2013

Semiconductor Electronics Material Devices and Simple Circuits

151300 What is an AND gate?

1 It has not equivalence to switching circuit

2 It is equivalent to switching circuit

3 It is equivalent to parallel switching circuit

4 It is a mixture of series and parallel switching

Explanation:

B AND gate is used as a switching circuit, AND gate is only active when both inputs are high same as series. Switching circuit which is active only when both switches are closed.
The truth table of AND gate will be -
Boolean Expression,
$Y = A \cdot B$
{|l|l|l|}
| $=$ \)|
|---|
$000$
$010$
$100$
$111$
\(
When both input are high then we get high output
original image

JCECE-2011

Semiconductor Electronics Material Devices and Simple Circuits

151301 To get an output 1 from the circuit shown in the figure, the input must be
original image

1

A = 0, B = 1, C = 0

2

A = 1, B = 0, C = 0

3

A = 1, B = 0, C = 1

4

A = 1, B = 1, C = 0

Explanation:

C Given circuit when output is 1 .
original image
In this circuit using OR gate and AND gate. So Boolean expression-
$Y = (A + B) \cdot C ($ Taking value $A = 1, B = 0, C = 1)$
$Y = (1 + 0) \cdot 1$
$Y = 1 \cdot 1$
$Y = 1$

JCECE-2010

Semiconductor Electronics Material Devices and Simple Circuits

151297 Which logic gate is represented by the following combination of logic gates?
original image

1

OR

2 NAND

3 AND

4 NOR

Explanation:

C Given that in this circuit using two NOT gate and one NOR gate
original image
Then, the Boolean expression will be-
$Y = \overset{―}{(\overset{―}{A} + \overset{―}{B})} = A \cdot B$
Truth table of $Y -$
{|c|c|c|c|c|c|}
| $Extra close brace or missing open brace$ $Extra close brace or missing open brace$ $Extra close brace or missing open brace$ $Extra close brace or missing open brace$ \)|
|---|
$001110$
$011010$
$100110$
$110001$
\(
Which is output expression of AND gate.

JCECE-2016

Semiconductor Electronics Material Devices and Simple Circuits

151298 The circuit is equivalent to
original image

1 NOR gate

2 AND gate

3 NAND gate

4 OR gate

Explanation:

C Given Circuit,
original image
${\overset{―}{Y}}_{2} = (\overset{―}{\overset{―}{A \cdot B} + \overset{―}{A \cdot B}})$
${\overset{―}{Y}}_{2} = \overset{―}{\overset{―}{A \cdot B}} \cdot \overset{―}{\overset{―}{A \cdot B}}$
${\overset{―}{Y}}_{2} = (A \cdot B) \cdot (A \cdot B)$
${\overset{―}{Y}}_{2} = AB$
$Y = \overset{―}{A \cdot B}$
(apply NOR gate)
(apply NOT gate)
So output,
$Y = \overset{―}{A \cdot B}$ Which is the expression of NAND gate.

JCECE-2015

Semiconductor Electronics Material Devices and Simple Circuits

151299 In the adjoining circuit of logic gate, the output $Y$ becomes zero if the inputs are
original image

1

A = 1, B = 1, C = 0

2

A = 0, B = 0, C = 0

3

A = 0, B = 1, C = 1

4

A = 1, B = 0, C = 0

Explanation:

B Given, circuit output $Y = 0$ then-
original image
The expression of given circuit,
$Y = \overset{―}{A \cdot B \cdot \overset{―}{C}}$ (using De, Morgen theorems)
$Y = \overset{―}{A \cdot B} + \overset{―}{\overset{―}{C}} (\overset{―}{\overset{―}{C}} = C \to Involution theorem)$
$Y = \overset{―}{A \cdot B} + C = \overset{―}{A} + \overset{―}{B} + C$
Then check option (1), $A = 1, B = 1, C = 0$
$Y = \overset{―}{1} + \overset{―}{1} + 0$
$Y = 0 + 0 + 0$
$Y = 0$ This is possible when $A = 1, B = 1$ and $C = 0$ .

JCECE-2013

Semiconductor Electronics Material Devices and Simple Circuits

151300 What is an AND gate?

1 It has not equivalence to switching circuit

2 It is equivalent to switching circuit

3 It is equivalent to parallel switching circuit

4 It is a mixture of series and parallel switching

Explanation:

B AND gate is used as a switching circuit, AND gate is only active when both inputs are high same as series. Switching circuit which is active only when both switches are closed.
The truth table of AND gate will be -
Boolean Expression,
$Y = A \cdot B$
{|l|l|l|}
| $=$ \)|
|---|
$000$
$010$
$100$
$111$
\(
When both input are high then we get high output
original image

JCECE-2011

Semiconductor Electronics Material Devices and Simple Circuits

151301 To get an output 1 from the circuit shown in the figure, the input must be
original image

1

A = 0, B = 1, C = 0

2

A = 1, B = 0, C = 0

3

A = 1, B = 0, C = 1

4

A = 1, B = 1, C = 0

Explanation:

C Given circuit when output is 1 .
original image
In this circuit using OR gate and AND gate. So Boolean expression-
$Y = (A + B) \cdot C ($ Taking value $A = 1, B = 0, C = 1)$
$Y = (1 + 0) \cdot 1$
$Y = 1 \cdot 1$
$Y = 1$

JCECE-2010

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