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Semiconductor Electronics Material Devices and Simple Circuits

151371 original image
For the given logic circuit, the output $Y$ for inputs $(A = 0, B = 1)$ and $(A = 0, B = 0)$ respectively are

1

(0, 0)

2

(0, 1)

3

(1, 0)

4

(1, 1)

Explanation:

C original image
If input $A = 0, B = 1$
$∴ Y_{1} = A . B$
$Y = 0 \times 0 = 0$
$Y_{2} = A \cdot B = 1 \times 1 = 1$
$Y = Y_{1} + Y_{2}$
$Y = 0 + 1 = 1$
$Y = 1$
For input $A = 0, B = 0$
$Y_{1} = A \cdot B = 0.1 = 0$
$Y_{2} = A \cdot B = 1.0 = 0$
$∴ Y = Y_{1} + Y_{2} = 0 + 0 = 0$
Hence, the output is $(1, 0)$

WB JEE 2021

Semiconductor Electronics Material Devices and Simple Circuits

151373 In the circuit shown, inputs $A$ and $B$ are in states 1 and 0 respectively. What is the only possible stable state of the outputs $X$ and $Y$ ?
original image

1

X = 1, Y = 1

2

X = 1, Y = 0

3

X = 0, Y = 1

4

X = 0, Y = 0

Explanation:

C Given that, $A = 1, B = 0$
Let,
$X = 0 then Y = \overset{―}{X . B} = \overset{―}{0.0} = \overset{―}{0} = 1$ the possible stable state of output $x$ and $y$ respectively are 0,1 .

WB JEE 2018

Semiconductor Electronics Material Devices and Simple Circuits

151374 original image
In the given circuit, the binary inputs at $A$ and $B$ are both 1 in one case and both 0 in the next case. The respective outputs at $Y$ in these two cases will be

1 1,1

2 0,0

3 0,1

4 1,0

Explanation:

B original image
Case-(1)
When,
$A = 1, B = 1$
$Y_{1} = \overset{―}{A} \cdot B = 1 \cdot 1 = 1$
$Y_{2} = \overset{―}{A} \cdot \overset{―}{B} = \overset{―}{1} \cdot 1 = 0 \cdot 0 = 0$
$Y = \overset{―}{Y_{1} + Y_{2}} = \overset{―}{1 + 0} = \overset{―}{1} = 0$
Case-(2)
When,
$A = 0, B = 0$
$Y_{1} = A \cdot B = 0 \cdot 0 = 0$
$Y_{2} = \overset{―}{A} \cdot \overset{―}{B} = \overset{―}{0} \cdot \overset{―}{0} = 1 \cdot 1 = 1$
$Y = \overset{―}{Y_{1} + Y_{2}} = \overset{―}{0 + 1} = \overset{―}{1} = 0$ Hence, the output is $(0, 0)$

WB JEE 2017

Semiconductor Electronics Material Devices and Simple Circuits

151375 The inputs to the digital circuit are as shown below. The output $Y$ is
original image

1

A + B + \overset{―}{C}

2

(A + B) \bar{C}

3

\overset{―}{A} + \overset{―}{B} + \overset{―}{C}

4

\overset{―}{A} + \overset{―}{B} + C

Explanation:

C original image
$Y_{1} = \overset{―}{A \cdot B}$
$Y_{2} = \overset{―}{C}$
$Y = \overset{―}{A \cdot B} + \overset{―}{C} (∵ \overset{―}{A \cdot B} = \overset{―}{A} + \overset{―}{B})$
$Y = \overset{―}{A} + \overset{―}{B} + \overset{―}{C}$

WB JEE 2015

Semiconductor Electronics Material Devices and Simple Circuits

151371 original image
For the given logic circuit, the output $Y$ for inputs $(A = 0, B = 1)$ and $(A = 0, B = 0)$ respectively are

1

(0, 0)

2

(0, 1)

3

(1, 0)

4

(1, 1)

Explanation:

C original image
If input $A = 0, B = 1$
$∴ Y_{1} = A . B$
$Y = 0 \times 0 = 0$
$Y_{2} = A \cdot B = 1 \times 1 = 1$
$Y = Y_{1} + Y_{2}$
$Y = 0 + 1 = 1$
$Y = 1$
For input $A = 0, B = 0$
$Y_{1} = A \cdot B = 0.1 = 0$
$Y_{2} = A \cdot B = 1.0 = 0$
$∴ Y = Y_{1} + Y_{2} = 0 + 0 = 0$
Hence, the output is $(1, 0)$

WB JEE 2021

Semiconductor Electronics Material Devices and Simple Circuits

151373 In the circuit shown, inputs $A$ and $B$ are in states 1 and 0 respectively. What is the only possible stable state of the outputs $X$ and $Y$ ?
original image

1

X = 1, Y = 1

2

X = 1, Y = 0

3

X = 0, Y = 1

4

X = 0, Y = 0

Explanation:

C Given that, $A = 1, B = 0$
Let,
$X = 0 then Y = \overset{―}{X . B} = \overset{―}{0.0} = \overset{―}{0} = 1$ the possible stable state of output $x$ and $y$ respectively are 0,1 .

WB JEE 2018

Semiconductor Electronics Material Devices and Simple Circuits

151374 original image
In the given circuit, the binary inputs at $A$ and $B$ are both 1 in one case and both 0 in the next case. The respective outputs at $Y$ in these two cases will be

1 1,1

2 0,0

3 0,1

4 1,0

Explanation:

B original image
Case-(1)
When,
$A = 1, B = 1$
$Y_{1} = \overset{―}{A} \cdot B = 1 \cdot 1 = 1$
$Y_{2} = \overset{―}{A} \cdot \overset{―}{B} = \overset{―}{1} \cdot 1 = 0 \cdot 0 = 0$
$Y = \overset{―}{Y_{1} + Y_{2}} = \overset{―}{1 + 0} = \overset{―}{1} = 0$
Case-(2)
When,
$A = 0, B = 0$
$Y_{1} = A \cdot B = 0 \cdot 0 = 0$
$Y_{2} = \overset{―}{A} \cdot \overset{―}{B} = \overset{―}{0} \cdot \overset{―}{0} = 1 \cdot 1 = 1$
$Y = \overset{―}{Y_{1} + Y_{2}} = \overset{―}{0 + 1} = \overset{―}{1} = 0$ Hence, the output is $(0, 0)$

WB JEE 2017

Semiconductor Electronics Material Devices and Simple Circuits

151375 The inputs to the digital circuit are as shown below. The output $Y$ is
original image

1

A + B + \overset{―}{C}

2

(A + B) \bar{C}

3

\overset{―}{A} + \overset{―}{B} + \overset{―}{C}

4

\overset{―}{A} + \overset{―}{B} + C

Explanation:

C original image
$Y_{1} = \overset{―}{A \cdot B}$
$Y_{2} = \overset{―}{C}$
$Y = \overset{―}{A \cdot B} + \overset{―}{C} (∵ \overset{―}{A \cdot B} = \overset{―}{A} + \overset{―}{B})$
$Y = \overset{―}{A} + \overset{―}{B} + \overset{―}{C}$

WB JEE 2015

Semiconductor Electronics Material Devices and Simple Circuits

151371 original image
For the given logic circuit, the output $Y$ for inputs $(A = 0, B = 1)$ and $(A = 0, B = 0)$ respectively are

1

(0, 0)

2

(0, 1)

3

(1, 0)

4

(1, 1)

Explanation:

C original image
If input $A = 0, B = 1$
$∴ Y_{1} = A . B$
$Y = 0 \times 0 = 0$
$Y_{2} = A \cdot B = 1 \times 1 = 1$
$Y = Y_{1} + Y_{2}$
$Y = 0 + 1 = 1$
$Y = 1$
For input $A = 0, B = 0$
$Y_{1} = A \cdot B = 0.1 = 0$
$Y_{2} = A \cdot B = 1.0 = 0$
$∴ Y = Y_{1} + Y_{2} = 0 + 0 = 0$
Hence, the output is $(1, 0)$

WB JEE 2021

Semiconductor Electronics Material Devices and Simple Circuits

151373 In the circuit shown, inputs $A$ and $B$ are in states 1 and 0 respectively. What is the only possible stable state of the outputs $X$ and $Y$ ?
original image

1

X = 1, Y = 1

2

X = 1, Y = 0

3

X = 0, Y = 1

4

X = 0, Y = 0

Explanation:

C Given that, $A = 1, B = 0$
Let,
$X = 0 then Y = \overset{―}{X . B} = \overset{―}{0.0} = \overset{―}{0} = 1$ the possible stable state of output $x$ and $y$ respectively are 0,1 .

WB JEE 2018

Semiconductor Electronics Material Devices and Simple Circuits

151374 original image
In the given circuit, the binary inputs at $A$ and $B$ are both 1 in one case and both 0 in the next case. The respective outputs at $Y$ in these two cases will be

1 1,1

2 0,0

3 0,1

4 1,0

Explanation:

B original image
Case-(1)
When,
$A = 1, B = 1$
$Y_{1} = \overset{―}{A} \cdot B = 1 \cdot 1 = 1$
$Y_{2} = \overset{―}{A} \cdot \overset{―}{B} = \overset{―}{1} \cdot 1 = 0 \cdot 0 = 0$
$Y = \overset{―}{Y_{1} + Y_{2}} = \overset{―}{1 + 0} = \overset{―}{1} = 0$
Case-(2)
When,
$A = 0, B = 0$
$Y_{1} = A \cdot B = 0 \cdot 0 = 0$
$Y_{2} = \overset{―}{A} \cdot \overset{―}{B} = \overset{―}{0} \cdot \overset{―}{0} = 1 \cdot 1 = 1$
$Y = \overset{―}{Y_{1} + Y_{2}} = \overset{―}{0 + 1} = \overset{―}{1} = 0$ Hence, the output is $(0, 0)$

WB JEE 2017

Semiconductor Electronics Material Devices and Simple Circuits

151375 The inputs to the digital circuit are as shown below. The output $Y$ is
original image

1

A + B + \overset{―}{C}

2

(A + B) \bar{C}

3

\overset{―}{A} + \overset{―}{B} + \overset{―}{C}

4

\overset{―}{A} + \overset{―}{B} + C

Explanation:

C original image
$Y_{1} = \overset{―}{A \cdot B}$
$Y_{2} = \overset{―}{C}$
$Y = \overset{―}{A \cdot B} + \overset{―}{C} (∵ \overset{―}{A \cdot B} = \overset{―}{A} + \overset{―}{B})$
$Y = \overset{―}{A} + \overset{―}{B} + \overset{―}{C}$

WB JEE 2015

Semiconductor Electronics Material Devices and Simple Circuits

151371 original image
For the given logic circuit, the output $Y$ for inputs $(A = 0, B = 1)$ and $(A = 0, B = 0)$ respectively are

1

(0, 0)

2

(0, 1)

3

(1, 0)

4

(1, 1)

Explanation:

C original image
If input $A = 0, B = 1$
$∴ Y_{1} = A . B$
$Y = 0 \times 0 = 0$
$Y_{2} = A \cdot B = 1 \times 1 = 1$
$Y = Y_{1} + Y_{2}$
$Y = 0 + 1 = 1$
$Y = 1$
For input $A = 0, B = 0$
$Y_{1} = A \cdot B = 0.1 = 0$
$Y_{2} = A \cdot B = 1.0 = 0$
$∴ Y = Y_{1} + Y_{2} = 0 + 0 = 0$
Hence, the output is $(1, 0)$

WB JEE 2021

Semiconductor Electronics Material Devices and Simple Circuits

151373 In the circuit shown, inputs $A$ and $B$ are in states 1 and 0 respectively. What is the only possible stable state of the outputs $X$ and $Y$ ?
original image

1

X = 1, Y = 1

2

X = 1, Y = 0

3

X = 0, Y = 1

4

X = 0, Y = 0

Explanation:

C Given that, $A = 1, B = 0$
Let,
$X = 0 then Y = \overset{―}{X . B} = \overset{―}{0.0} = \overset{―}{0} = 1$ the possible stable state of output $x$ and $y$ respectively are 0,1 .

WB JEE 2018

Semiconductor Electronics Material Devices and Simple Circuits

151374 original image
In the given circuit, the binary inputs at $A$ and $B$ are both 1 in one case and both 0 in the next case. The respective outputs at $Y$ in these two cases will be

1 1,1

2 0,0

3 0,1

4 1,0

Explanation:

B original image
Case-(1)
When,
$A = 1, B = 1$
$Y_{1} = \overset{―}{A} \cdot B = 1 \cdot 1 = 1$
$Y_{2} = \overset{―}{A} \cdot \overset{―}{B} = \overset{―}{1} \cdot 1 = 0 \cdot 0 = 0$
$Y = \overset{―}{Y_{1} + Y_{2}} = \overset{―}{1 + 0} = \overset{―}{1} = 0$
Case-(2)
When,
$A = 0, B = 0$
$Y_{1} = A \cdot B = 0 \cdot 0 = 0$
$Y_{2} = \overset{―}{A} \cdot \overset{―}{B} = \overset{―}{0} \cdot \overset{―}{0} = 1 \cdot 1 = 1$
$Y = \overset{―}{Y_{1} + Y_{2}} = \overset{―}{0 + 1} = \overset{―}{1} = 0$ Hence, the output is $(0, 0)$

WB JEE 2017

Semiconductor Electronics Material Devices and Simple Circuits

151375 The inputs to the digital circuit are as shown below. The output $Y$ is
original image

1

A + B + \overset{―}{C}

2

(A + B) \bar{C}

3

\overset{―}{A} + \overset{―}{B} + \overset{―}{C}

4

\overset{―}{A} + \overset{―}{B} + C

Explanation:

C original image
$Y_{1} = \overset{―}{A \cdot B}$
$Y_{2} = \overset{―}{C}$
$Y = \overset{―}{A \cdot B} + \overset{―}{C} (∵ \overset{―}{A \cdot B} = \overset{―}{A} + \overset{―}{B})$
$Y = \overset{―}{A} + \overset{―}{B} + \overset{―}{C}$

WB JEE 2015

Question Bank Series

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Question Bank Series

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