Semiconductor Electronics: Materials, Devices and Simple Circuits Test 23

Result

Semiconductor Electronics: Materials, Devices and Simple Circuits Test 23

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The combination of the gates shown above produces

A
AND gate

B
XOR gate

C
NOR gate

D
NAND gate

Solution

Truth-table for given combination of gates is given below and is equivalent to truth-table of NAND gate.
Input Output
$$0,0$$ $$1$$
$$1,0$$ $$1$$
$$0,1$$ $$1$$
$$1,1$$ $$0$$
Question 2
1 / -0

Given above are four logic gate symbols. Those for OR, NOR and NAND are respectively

A
a, d, c

B
d, a, b

C
a, c, d

D
d, b, a

Solution

a- represent OR gate
b- represent the AND gate
c- represent, the combination of OR and NOT gate, NOR gate.
d- represent, the combination of AND and NOT gate, NAND gate
Question 3
1 / -0

For the given combination of gates, if the logic states of inputs A, B, C are as follows $$A=B=C=0$$ and $$A=B=1, C=0$$ then the logic states of output D are

A
0,0

B
0,1

C
1, 0

D
1,1

Solution

OR gate add the input signals given to it and AND gate multiply the input signals given to it.
Input (0,0) given to OR gate will yield result 0.
Now input for AND gate are (0,0) hence output will be 0.
For the second case, input (1,1) given to OR gate will yield result 1.
Now input for AND gate are (1,0) hence output will be 0.
Therefore combination of gates will yield result 0 and 0.
Question 4
1 / -0

The expression of $$Y$$ in the above circuit is

A
$$ABCD$$

B
$$A+BCD$$

C
$$A +B+C+D$$

D
$$AB+CD$$

Solution

Here, to build the logic three OR gates are used hence, all the input signals will be added to each other.
Hence, the boolean expression is: $$A+B+C+D = Y$$
Question 5
1 / -0

In the given circuit, if A and B represent two inputs and C represents the output, the circuit represents

A
NOR gate

B
AND gate

C
gate

D
OR gate

Solution

The image given shows a diode OR circuit. $$R$$ is connected from the output to ground to provide bias current for the diodes. Any positive voltage will represent a logic $$1$$ state and the summing of currents through multiple diodes does not change the logic level.
Question 6
1 / -0

The output $$Y$$ of the gate circuit shown in the figure below is

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

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

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

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

Solution

Here, the inputs are reversed first and then multiplied with each other hence the boolean expression is:
$$\bar A.\bar B = Y$$

Question 7

1 / -0

To get an output $$1$$ from the circuit shown in the figure, the input must be

$$A=0, B=1,C=0$$

$$A=1, B=0, C=0$$

$$A=1, B=0, C=1$$

$$A=1, B=1, C=0$$

Solution

Input	$$A+B = Y$$	$$Y.C = Y'$$
$$(0,1,0)$$	$$1$$	$$0$$
$$(1,0,0)$$	$$1$$	$$ 0$$
$$(1,0,1)$$	$$1$$	$$1$$
$$(1,1,0)$$	$$1$$	$$0$$

Question 8
1 / -0

Statement-1: NOT gate is also called invertor circuit.
Statement-2: NOT gate inverts the input order.

A
Statement -1 is false, statement -2 is true.

B
Statement 1-is true, statement -2 is true and statement -2 is correct explanation of statement-1.

C
Statement 1-is true, statement -2 is true but statement -2 is not correct explanation of statement-1.

D
Statement 1-is true, statement -2 is false.

Solution

NOT gate is also called invertor circuit since NOT gate inverts the input order. Hence, statement 1-is true statement-2 is true statement-2 is correct explanation of statement-1.
Question 9
1 / -0

The logic gate equivalent to the above combination of logic gates is

A
OR

B
NAND

C
AND

D
NOR

Solution

Truth table for given combination of gates is shown below and is equivalent to AND gate.
Input Output
$$(0,0)$$ $$0$$
$$(0,1)$$ $$0$$
$$(1,0)$$ $$0$$
$$(1,1)$$ $$1$$
Question 10
1 / -0

The output of the combination of the gates shown in the figure is

A
$$A + A. B$$

B
$$(A+B)A+\bar{B}$$

C
$$(A.B)+(\bar{A}.\bar{B})$$

D
$$(A+B)\overline{(A.B)}$$

Solution

Here, one of the input is shared by both the gates, and the output of AND gate is feed to the OR gate as second input hence, the boolean expression will be $$A.B+A = Y$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Input	Output
$$(0,0)$$	$$0$$
$$(0,1)$$	$$0$$
$$(1,0)$$	$$0$$
$$(1,1)$$	$$1$$

Semiconductor Electronics: Materials, Devices and Simple Circuits Test 23

Result

Semiconductor Electronics: Materials, Devices and Simple Circuits Test 23

Score

-

Rank

-

SHARING IS CARING

Question 1

AND gate

XOR gate

NOR gate

NAND gate

Solution

Question 2

a, d, c

d, a, b

a, c, d

d, b, a

Solution

Question 3

0,0

0,1

1, 0

1,1

Solution

Question 4

$$ABCD$$

$$A+BCD$$

$$A +B+C+D$$

$$AB+CD$$

Solution

Question 5

NOR gate

AND gate

gate

OR gate

Solution

Question 6

$$\overline{A.B}$$

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

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

$$\bar{A}+\bar{B}$$

Solution

Question 7

$$A=0, B=1,C=0$$

$$A=1, B=0, C=0$$

$$A=1, B=0, C=1$$

$$A=1, B=1, C=0$$

Solution

Question 8

Statement -1 is false, statement -2 is true.

Statement 1-is true, statement -2 is true and statement -2 is correct explanation of statement-1.

Statement 1-is true, statement -2 is true but statement -2 is not correct explanation of statement-1.

Statement 1-is true, statement -2 is false.

Solution

Question 9

OR

NAND

AND

NOR

Solution

Question 10

$$A + A. B$$

$$(A+B)A+\bar{B}$$

$$(A.B)+(\bar{A}.\bar{B})$$

$$(A+B)\overline{(A.B)}$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.