Semiconductor Electronics: Materials, Devices and Simple Circuits Test 7

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Semiconductor Electronics: Materials, Devices and Simple Circuits Test 7

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Weekly Quiz Competition

Question 1
1 / -0

An AND gate is followed by a NOT gate in series. With two inputs $$A$$ & $$B$$, the Boolean expression for the out put $$Y$$ will be :

A
$$A.B$$

B
$$A + B$$

C
$$\overline{A+B}$$

D
$$\overline{A.B}$$

Solution

Expression for AND is $$A.B$$ and if it is followed by a NOT gate then the Boolean expression will just be the complement of AND.
Question 2
1 / -0

The value of $$A.\bar{A}$$ in Boolean algebra is

A
$$0$$

B
$$1$$

C
$$A$$

D
$$\bar{A}$$

Solution

Correct answer: Option A

Hint: Complement Law – A term AND‘ed with its complement equals “0”

Step 1: Finding output

Let $$Y$$ be the output

$$Y = A.\overline A $$

If $$A = 0$$,

$$Y = 0.\overline 0$$

$$(\overline 0 = 1)$$

$$Y = 0$$

If $$A = 1$$,

$$Y = 1.\overline 1 $$

$$(\overline 1 = 0)$$

$$Y = 0$$

Truth table for$$A$$and $$Y$$

$$A$$

$$Y$$

0

0

1

0

Therefore the value of $$A.\overline A $$ in the Boolean algebra is 0
Question 3
1 / -0

In Boolean algebra $$A.B = Y$$ implies that :

A
product of $$A$$ and $$B$$ is $$Y$$

B
$$Y$$ exists when $$A$$ exists or $$B$$ exists

C
$$Y$$ exists when both $$A$$ and $$B$$ exist but not when only $$A$$ or $$B$$ exists

D
$$Y$$ exists when $$A$$ or $$B$$ exists but not both $$A$$ and $$B$$ exist.

Solution

In Boolean algebra $$A.B = Y$$ implies that $$Y$$ exists when both $$A$$ and $$B$$ exist but not when only $$A$$ or $$B$$ exists.
Question 4
1 / -0

In the Boolean algebra, the following one is wrong

A
$$1 + 0 = 1$$

B
$$0 + 1 = 1$$

C
$$1 + 1 = 1$$

D
$$0 + 0 = 1$$

Solution

In the Boolean algebra, binary addition is given as:
$$1 + 0 = 1$$
$$0 + 1 = 1$$
$$1 + 1 = 1$$
Hence, $$0 + 0 = 1$$ is wrong.
Question 5
1 / -0

In Boolean algebra, $$A + B = Y$$ implies that

A
sum of $$A$$ and $$B$$ is $$Y$$.

B
$$Y$$ exists when $$A$$ exists or $$B$$ exists or both $$A$$ and $$B$$ exist.

C
$$Y$$ exists only when $$A$$ and $$B$$ both exist.

D
$$Y$$ exists when $$A$$ or $$B$$ exist but not when both $$A$$ and $$B$$ exist.

Solution

In Boolean algebra, $$A + B = Y$$ implies that $$Y$$ exists when $$A$$ exists or $$B$$ exists or both $$A$$ and $$B$$ exist.
Question 6
1 / -0

The output of a 2-input OR gate is zero only when its

A
both inputs are $$0$$.

B
either input is $$1$$.

C
both inputs are $$1$$.

D
either input is $$0$$.

Solution

The truth table for 2-input OR gate is as shown in the figure. Thus, the output is zero only when both inputs are zero.
Question 7
1 / -0

The value of $$\bar{A} + A$$ in the Boolean algebra is

A
$$A$$

B
$$\bar{A}$$

C
$$0$$

D
$$1$$

Solution

Correct answer: Option D

Hint: Complement Law – A term OR´ed with its complement equals “1”

Step 1: Finding output

Let $$Y$$ be the output

$$Y = \overline A + A$$

If $$A = 0$$,

$$Y = \overline 0 + 0$$

$$(\overline 0 = 1)$$

$$Y = 1$$

If $$A = 1$$,

$$Y = \overline 1 + 1$$

$$(\overline 1 = 0)$$

$$Y = 1$$

Truth table for $$A$$ and $$Y$$

$$A$$

$$Y$$

0

1

1

1

Therefore the value of Boolean algebra is 1
Question 8
1 / -0

In the Boolean algebra, the following one is wrong

A
$$1.0 = 0$$

B
$$0.1=0$$

C
$$1.1 = 0$$

D
$$1.1 = 1$$

Solution

In the Boolean algebra,
$$1.0=0$$
$$0.1=0$$
$$1.1=1$$
Question 9
1 / -0

The following truth table is for :
A B Y
1 1 0
1 0 1
0 1 1
0 0 1

A
NAND gate

B
AND gate

C
XOR gate

D
NOT gate

Solution

The output is 0 only when the two inputs are 1.
The output is 1 when any of the input is zero.
This is the characteristic of a NAND gate.
Question 10
1 / -0

If $$A = B = 1$$, then in terms of Boolean algebra the value of $$A.B + A$$ is not equal to

A
$$B.A+B$$

B
$$B+A$$

C
$$B$$

D
$$\bar{A}.B$$

Solution

For $$A = B = 1$$,
$$\Rightarrow A.B+B = 1.1 + 1 = 1 + 1 = 1$$
Therefore,
(A) $$B.A+B = 1.1 + 1 = 1 + 1 = 1$$
(B) $$B+A = 1 + 1 = 1$$
(C) $$B = 1$$
(D) $$ \bar A.B=1.1+1 = \bar 1.1 = 0.1 = 0$$
Hence, option D is the correct answer.

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$$A$$	$$Y$$
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$$A$$	$$Y$$
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Semiconductor Electronics: Materials, Devices and Simple Circuits Test 7

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Semiconductor Electronics: Materials, Devices and Simple Circuits Test 7

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SHARING IS CARING

Question 1

$$A.B$$

$$A + B$$

$$\overline{A+B}$$

$$\overline{A.B}$$

Solution

Question 2

$$0$$

$$1$$

$$A$$

$$\bar{A}$$

Solution

Question 3

product of $$A$$ and $$B$$ is $$Y$$

$$Y$$ exists when $$A$$ exists or $$B$$ exists

$$Y$$ exists when both $$A$$ and $$B$$ exist but not when only $$A$$ or $$B$$ exists

$$Y$$ exists when $$A$$ or $$B$$ exists but not both $$A$$ and $$B$$ exist.

Solution

Question 4

$$1 + 0 = 1$$

$$0 + 1 = 1$$

$$1 + 1 = 1$$

$$0 + 0 = 1$$

Solution

Question 5

sum of $$A$$ and $$B$$ is $$Y$$.

$$Y$$ exists when $$A$$ exists or $$B$$ exists or both $$A$$ and $$B$$ exist.

$$Y$$ exists only when $$A$$ and $$B$$ both exist.

$$Y$$ exists when $$A$$ or $$B$$ exist but not when both $$A$$ and $$B$$ exist.

Solution

Question 6

both inputs are $$0$$.

either input is $$1$$.

both inputs are $$1$$.

either input is $$0$$.

Solution

Question 7

$$A$$

$$\bar{A}$$

$$0$$

$$1$$

Solution

Question 8

$$1.0 = 0$$

$$0.1=0$$

$$1.1 = 0$$

$$1.1 = 1$$

Solution

Question 9

NAND gate

AND gate

XOR gate

NOT gate

Solution

Question 10

$$B.A+B$$

$$B+A$$

$$B$$

$$\bar{A}.B$$

Solution

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