## Digital Electronics & Logic Gates Short Questions And Answers

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

**Answer:**

(120)_{2} = (1 × 3^{2}) + (2 × 3^{1}) + (0 × 3^{0})

= 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:**

(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

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

**Answer:**

(1 × 2^{4}) + (0 × 2^{3}) (1 × 2^{2}) (1 × 2^{1}) +(1 × 2^{0})

= (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:**

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

**Output waveform will ho as follows:**