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This section includes 223 Mcqs, each offering curated multiple-choice questions to sharpen your Digital Electronics knowledge and support exam preparation. Choose a topic below to get started.
151. |
Consider the following circuit.How many minimum numbers of two inputs NAND gates are required to design the above circuit? |
A. | 6 |
B. | 4 |
C. | 5 |
D. | 3 |
E. | 2 |
Answer» C. 5 | |
152. |
In the circuit shown, the switch is momentarily closed and then opened. Assuming the logic gates to have equal non-zero delay, at steady state, the logic states of X and Y are |
A. | X is latched, Y toggles continuously |
B. | X and Y are both latched |
C. | Y is latched, X toggles continuously |
D. | X and Y both toggle continuously |
Answer» E. | |
153. |
Logic circuits are |
A. | analog circuits |
B. | digital circuits |
C. | hybrid circuits |
D. | None of the above |
Answer» C. hybrid circuits | |
154. |
Four statements are given below. Identify the correct statement. |
A. | XOR is universal gate |
B. | XNOR is a basic gate |
C. | XOR is derived gate |
D. | XOR is a basic gate |
Answer» D. XOR is a basic gate | |
155. |
Match the following Lists:List-I List-II(a) (i) (b) (ii) (c) (iii) (d) (iv) Correct codes are:Code: |
A. | a-ii, b-i, c-iv, d-iii |
B. | a-iii, b-iv, c-ii, d-i |
C. | a-ii, b-iv, c-i, d-iii |
D. | a-iii, b-iv, c-i, d-ii |
Answer» D. a-iii, b-iv, c-i, d-ii | |
156. |
De Morgan’s theorem states that |
A. | (X + Y)’ = Y’ + X’ |
B. | (X . Y)’ = X’ + Y’ |
C. | (X . Y)’ = Y’ . X’ |
D. | (X + Y)’ = X’ + Y’ |
Answer» C. (X . Y)’ = Y’ . X’ | |
157. |
A gate whose inputs are inverted is termed as |
A. | Drop Gate |
B. | Bubbled Gate |
C. | Converter Gate |
D. | Destructor Gate |
Answer» C. Converter Gate | |
158. |
An electric power generating station supplies power to three loads A, B, and C. Only a single generator is required when anyone load is switched on. When more than one load is on, an auxiliary generator must be started. The Boolean equation for the control of switching of the auxiliary generator will be |
A. | AA + BB + CC |
B. | ABC + BCA + CAB |
C. | AB + AC |
D. | AB + AC + BC |
Answer» E. | |
159. |
In the logic circuit shown in the figure, Y is given by |
A. | Y = ABCD |
B. | Y = (A + B) (C + D) |
C. | Y = A + B + C + D |
D. | Y = AB + CD |
Answer» E. | |
160. |
Logic expressions can be simplified by using |
A. | Boolean Algebra Method |
B. | Karnaugh-map Method |
C. | Tabulation Method |
D. | any of the these |
Answer» E. | |
161. |
A 4 × 1 MUX is used to implement a 3-input Boolean function is as shown above. The Boolean function F(A, B, C) implemented is: |
A. | F(A, B, C) = ∑(1, 2, 4, 6) |
B. | F(A, B, C) = ∑(1, 2, 6) |
C. | F(A, B, C) = ∑(2, 4, 5, 6) |
D. | F(A, B, C) = ∑(1, 5, 6) |
Answer» B. F(A, B, C) = ∑(1, 2, 6) | |
162. |
In the circuit shown below, X and Y are digital inputs, and Z is a digital output. The equivalent circuit is a |
A. | NAND gate |
B. | NOR gate |
C. | XOR gate |
D. | XNOR gate |
Answer» D. XNOR gate | |
163. |
Applying DeMorgan's theorem to the expression , we get ________. |
A. | |
B. | |
Answer» B. | |
164. |
By applying De Morgan's theorem to a NOR gate, two identical truth tables can be produced. |
A. | 1 |
B. | |
Answer» B. | |
165. |
CPLD software can be used to design original circuits that prove the Boolean rules and laws. |
A. | 1 |
B. | |
Answer» B. | |
166. |
Which of the following is true for a 5-variable Karnaugh map? |
A. | There is no such thing. |
B. | It can be used only with the aid of a computer. |
C. | It is made up of two 4-variable Karnaugh maps. |
D. | It is made up of a 2-variable and a 3-variable Karnaugh map. |
Answer» D. It is made up of a 2-variable and a 3-variable Karnaugh map. | |
167. |
A Karnaugh map is similar to a truth table because it presents all the possible values of input variables and the resulting output of each value. |
A. | 1 |
B. | |
Answer» B. | |
168. |
Boolean algebra simplifies logic circuits. |
A. | 1 |
B. | |
Answer» B. | |
169. |
A Karnaugh map is similar to a truth table. |
A. | 1 |
B. | |
Answer» B. | |
170. |
In the commutative law, in ORing and ANDing of two variables, the order in which the variables are ORed or ANDed makes no difference. |
A. | 1 |
B. | |
Answer» B. | |
171. |
The systematic reduction of logic circuits is performed using Boolean algebra. |
A. | 1 |
B. | |
Answer» B. | |
172. |
When four 1s are taken as a group on a Karnaugh map, the number of variables eliminated from the output expression is ________. |
A. | 1 |
B. | 2 |
C. | 3 |
D. | 4 |
Answer» C. 3 | |
173. |
The binary value of 1010 is converted to the product term . |
A. | 1 |
B. | |
Answer» C. | |
174. |
The process of reduction or simplification of combinational logic circuits increases the cost of the circuit. |
A. | 1 |
B. | |
Answer» C. | |
175. |
Most Boolean reductions result in an equation in only one form. |
A. | 1 |
B. | |
Answer» C. | |
176. |
In Boolean algebra, the word "literal" means ________. |
A. | a product term |
B. | all the variables in a Boolean expression |
C. | the inverse function |
D. | a variable or its complement |
Answer» E. | |
177. |
The VHDL editor provided with a schematic editor development system will produce a file with the extension .vhd, which can be used by the simulator to test the output of the logic design. |
A. | 1 |
B. | |
Answer» B. | |
178. |
Five-variable Karnaugh maps are impossible. |
A. | 1 |
B. | |
Answer» C. | |
179. |
If you need a NAND, an AND, and an inverter you would purchase a 7400 IC. |
A. | 1 |
B. | |
C. | 1 |
D. | |
Answer» B. | |
180. |
Use Boolean algebra to find the most simplified SOP expression for F = ABD + CD + ACD + ABC + ABCD. |
A. | F = ABD + ABC + CD |
B. | F = CD + AD |
C. | F = BC + AB |
D. | F = AC + AD |
Answer» B. F = CD + AD | |
181. |
The commutative law of addition and multiplication indicates that: |
A. | we can group variables in an AND or in an OR any way we want |
B. | an expression can be expanded by multiplying term by term just the same as in ordinary algebra |
C. | the way we OR or AND two variables is unimportant because the result is the same |
D. | the factoring of Boolean expressions requires the multiplication of product terms that contain like variables |
Answer» D. the factoring of Boolean expressions requires the multiplication of product terms that contain like variables | |
182. |
The Boolean expression is equal to ________. |
A. | C |
B. | D |
C. | C + D |
D. | 1 |
Answer» D. 1 | |
183. |
The product-of-sums (POS) is basically the ORing of ANDed terms. |
A. | 1 |
B. | |
Answer» C. | |
184. |
The OR function is Boolean multiplication and the AND function is Boolean addition. |
A. | 1 |
B. | |
Answer» C. | |
185. |
Which of the examples below expresses the distributive law of Boolean algebra? |
A. | (A + B) + C = A + (B + C) |
B. | A(B + C) = AB + AC |
C. | A + (B + C) = AB + AC |
D. | A(BC) = (AB) + C |
Answer» C. A + (B + C) = AB + AC | |
186. |
How many gates would be required to implement the following Boolean expression after simplification? XY + X(X + Z) + Y(X + Z) |
A. | 1 |
B. | 2 |
C. | 4 |
D. | 5 |
Answer» C. 4 | |
187. |
The symbol shown below is for a 2-input NAND gate. |
A. | 1 |
B. | |
C. | 1 |
D. | |
Answer» C. 1 | |
188. |
Occasionally, a particular logic expression will be of no consequence in the operation of a circuit, such as a BCD-to-decimal converter. These result in ________terms in the K-map and can be treated as either ________ or ________, in order to ________ the resulting term. |
A. | don't care, 1s, 0s, simplify |
B. | spurious, ANDs, ORs, eliminate |
C. | duplicate, 1s, 0s, verify |
D. | spurious, 1s, 0s, simplify |
Answer» B. spurious, ANDs, ORs, eliminate | |
189. |
When are the inputs to a NAND gate, according to De Morgan's theorem, the output expression could be: |
A. | X = A + B |
B. | |
Answer» B. | |
190. |
For the SOP expression , how many 1s are in the truth table's output column? |
A. | 1 |
B. | 2 |
C. | 3 |
D. | 5 |
Answer» D. 5 | |
191. |
The NAND or NOR gates are referred to as "universal" gates because either: |
A. | can be found in almost all digital circuits |
B. | can be used to build all the other types of gates |
C. | are used in all countries of the world |
D. | were the first gates to be integrated |
Answer» C. are used in all countries of the world | |
192. |
Which Boolean algebra property allows us to group operands in an expression in any order without affecting the results of the operation [for example, A + B = B + A]? |
A. | associative |
B. | commutative |
C. | Boolean |
D. | distributive |
Answer» C. Boolean | |
193. |
How many gates would be required to implement the following Boolean expression before simplification? XY + X(X + Z) + Y(X + Z) |
A. | 1 |
B. | 2 |
C. | 4 |
D. | 5 |
Answer» E. | |
194. |
In Boolean algebra, A + 1 = 1. |
A. | 1 |
B. | |
Answer» B. | |
195. |
Determine the binary values of the variables for which the following standard POS expression is equal to 0. |
A. | (0 + 1 + 0)(1 + 0 + 1) |
B. | (1 + 1 + 1)(0 + 0 + 0) |
C. | (0 + 0 + 0)(1 + 0 + 1) |
D. | (1 + 1 + 0)(1 + 0 + 0) |
Answer» B. (1 + 1 + 1)(0 + 0 + 0) | |
196. |
SOP stands for sum-of-powers. |
A. | 1 |
B. | |
C. | 1 |
D. | |
Answer» C. 1 | |
197. |
When grouping cells within a K-map, the cells must be combined in groups of ________. |
A. | 2s |
B. | 1, 2, 4, 8, etc. |
C. | 4s |
D. | 3s |
Answer» C. 4s | |
198. |
A truth table for the SOP expression has how many input combinations? |
A. | 1 |
B. | 2 |
C. | 4 |
D. | 8 |
Answer» E. | |
199. |
AC + ABC = AC |
A. | 1 |
B. | |
Answer» B. | |
200. |
The commutative law of Boolean addition states that A + B = A √ó B. |
A. | 1 |
B. | |
Answer» C. | |