Test: Logic Gates & Switching Circuits- 2 - Electrical Engineering (EE) MCQ

From given logic diagram, output is:

= A.(A + B) + B.(B + C)
= A.A + A B + BB + BC
= A + AB + B + BC
= A(1 + B) + B(1 + C)
= A + B

Given expressio n is f = πM( 0, 1, 2, 3, 4, 7)
The K-map for this expression is shown below.

The minimized expression in POS form is

The minimized expression 'f' can be implemented using two input NOR gate as shown below.

Thus, we require six NOR gates.

• An OR gate in the positive logic system becomes an AND gate in the negative logic system and vice-versa.

  • Correct. In positive logic, an OR gate outputs HIGH if any input is HIGH. In negative logic, an OR gate behaves like an AND gate because the inputs are inverted.

• The dual of an expression is obtained by changing ANDs to ORs, ORs to ANDs, 0s to 1s, 1s to 0s, and by complementing the variables.

  • Incorrect. The dual of a logic expression is obtained by changing ANDs to ORs and ORs to ANDs, and by swapping 0s and 1s, but the variables are not complemented.

• NAND and NOR are the only two available universal logic gates.

  • Correct. NAND and NOR gates are universal gates because you can construct any other logic gate using just NAND or just NOR gates.

• Universal gates do not follow commutative law.

  

• On logic diagrams, an inversion bubble at the point where the input is connected to a gate, indicates an active-HIGH input.

  • Incorrect. An inversion bubble indicates an active-LOW input, not an active-HIGH input. The bubble shows that the input or output is inverted.

Conclusion:

The statements that are not correct are 2, 4, and 5.

Thus, the correct answer is:

2. 2, 4 and 5

OR gate using NAND gate:

EX-NOR gate using NAND gate:

Thus, we require 3 NAND gates for an OR gate while 5 NAND gates for an EX-NOR gate.

Given,

or,

i.e. must be 0 and B must be 0 Hence, A = 1 and B = 0
Also, AB = AC or 1.B - 1.C
or B = C
Hence, C = 0 (∵ B = 0)
Also,
or, 1.0 + 1.1 + 0.D = 1.D
or D = 1
Thus, the values of A, 6, Cand Dare given by
A = 1, B = 0, C = 0 and D = 1.

Output = X.1 +  = X = Input

The truth table for above circuit is shown below.

Hence, the output y is equivalent to output of an AND gate having two inputs A and 6. Thus, an AND gate is equivalent to a series switching circuit.

For same inputs, A A = 0
For different inputs,

Observations:

• The output is 1 when either A is 1 and B is 0 or when A is 0 and B is 1.
• The output is 0 when both A and B are 1.

This truth table corresponds to the XOR (Exclusive OR) gate.

XOR Gate:

• The XOR gate outputs 1 when the inputs are different, and 0 when the inputs are the same.

Given the truth table:

• When A=1 and B=0, output f=1
• When A=0 and B=1, output f=1
• When A=1 and B=1, output f=0

Conclusion:

The logic gate represented by the above truth table is the XOR gate.

Truth tables

Thus, the gate can be either an EX-NOR or a NOR gate.

• When V₁ = 0 volt and V₂ = 5 volt, diode D₁ will be on.
  Using voltage division rule,
  
  
  = 0.24 volt
• When V₁ = V₂ = 5 volt, then both diodes are off, no current flows anywhere, and V₀ = 5 volt.
• When V₁ = V₂ = 0 volt, then both diodes D₁ and D₂ will be on and their two R_s's will effectively come in parallel.

Thus,
= 0.12 V

The truth table for the given condition is shown below:

The K-map for above truth table is shown below.

Thus, Y = AB + BC + CA which can be implemented using 3 AND gates and 1 OR gate (total 4 basic gates).

Let the output of second last NAND gate be X. 
Then,

Thus, 

Given circuit represents EX-NOR gate using NAND gate

A two input OR gate truth table is shown below:

From the truth table of OR gate it is clear tha output is 0’s only when all the inputs are 0’s otherwise for all other input combination outpu is always 1's. Hence, in the column for the outpu of the OR gate, the number of zeros is always 1's.

• Statement-1 is correct because any circuit can be realized using AND, OR and NOT gates.
• The two logic gates NAND and NOR can realize any logic circuit single-handedly. Hence, these two gates are called universal building blocks. AND, OR and NOT are called basic gates. Thus, statement-2 is not correct.
• Both NAND and NOR gates can perform all the three basic logic functions (AND, OR and NOT). Therefore, AND/OR/INVERT logic can be converted to NAND logic or NOR logic. Hence, statement-3 is correct.
• 
  Statement-4 is not correct.

Since three or more variable X-NOR gates do not exist, therefore when a number of variables are to be X-NORed, a number of two input X-NOR gases have to be used. Thus, reason is not a correct statement.

When n is even, 
When n is odd, 

From the given circuit, output is
(∴ (X + Y) (X + Z) = X + YZ)


Therefore, the given logic circuit can be simplified as Do shown below.

Thus, it require a AND gate and a NOT gate.

In case of ECL circuit, the floating inputs are considered as logic ‘0’. Hence output