WBCHSE Class 12 Physics Digital Electronics & Logic Gates Short Answer Questions

Digital Electronics & Logic Gates Short Questions And Answers

Question 1. Convert the number (120)3 into decimal.
Answer:

(120)₂ = (1 × 3²) + (2 × 3¹) + (0 × 3⁰)

= 9 + 6 + 0 = (15)₁₀

Question 2. Assume that the first and the second digits of any binary system are 7 and 6, respectively. Convert (76)₁₀ into this binary system.
Answer:

Conceptual Questions on Sequential Circuits

(76)₁₀ = (1001100)₂

Here, we have to take 0 → 7 and 1→ 6.

Therefore, according to the given binary process,

(76)₁₀ = (6776677)₂

Question 3. Express 39 as a binary number
Answer:

∴ (39)₁₀ = (10011)₂

Question 4. Why is it called a universal gate?
Answer:

NOR gate can be used to obtain all the possible gates by using it as a basic building. This is why it is called the universal gate. It may be used to realize the basic logic functions OR, AND, and NOT

WBBSE Class 12 Digital Electronics Short Q&A

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Question 5. What is the decimal equivalent of the binary number 100117
Answer:

(1 × 2⁴) + (0 × 2³)  (1 × 2²) (1 × 2¹) +(1 × 2⁰)

= (16 + 0 + 0 + 2+ 1)

= (19)₁₀

Question 6. If two Inputs of n NAND gate arc are joined, what type of gate is formed
Answer:

If the joining of two inputs implies that the inputs are short-circuited, then the two inputs are generally the same. If the input is A, then the output of the NAND gate

Y = \(\overline{A \cdot A}=\bar{A}\)

Hence, in this case, a NOT gate Is formed

Question 7. Write down the value of \((\bar{X}+X) \text { and }(X \cdot \bar{X})\) in Boolean algebra
Answer:

⇒ \(\bar{X}+X\)= 1

⇒ \(X \cdot \bar{X}\) = 0

Question 8.  The Input waveforms A and B to a logic gate. 

Answer:

Short Answer Questions on Digital Circuits

Question 9.  The figure shows the Input waveforms A and B for the AND gate. 

Output waveform will ho as follows: