Advanced Microprocessors Easy Questions and Answers

Figure shows NOR gate S-R latch or we can say S-R Flip-Flop.

Correct Option: D

Figure shows NOR gate S-R latch or we can say S-R Flip-Flop.

∴ Ring counter is a N: 1 divider, where N = no. of bits.

a =	100k	= 10kHz
	10

MOD 20 counter is a divided by 20 counter, so

∴ b =	10kHz	= 500Hz
	20

4 bit parallel counter ⇒ 4 flip-flop used
Hence, 2⁴ = 16, stages i.e., MOD-16 counter, so.

∴ C =	500	= 31.25Hz
	16

4 bit Johnson counter is a 2N: 1 divider, where N is the no. of bit

∴ d =	31.25	= 3.9Hz
	8

Hence alternative (B) is the correct answer.

Correct Option: B

∴ Ring counter is a N: 1 divider, where N = no. of bits.

a =	100k	= 10kHz
	10

MOD 20 counter is a divided by 20 counter, so

∴ b =	10kHz	= 500Hz
	20

4 bit parallel counter ⇒ 4 flip-flop used
Hence, 2⁴ = 16, stages i.e., MOD-16 counter, so.

∴ C =	500	= 31.25Hz
	16

4 bit Johnson counter is a 2N: 1 divider, where N is the no. of bit

∴ d =	31.25	= 3.9Hz
	8

Hence alternative (B) is the correct answer.

AB + A′C + BC
= AB (C + C′) + A′C (B + B′) + BC (A + A′)
∴ C + C′ = B + B′ = A + A′ = 1
= ABC + ABC′ + A′CB + A′CB′ + BCA + BCA′
= ABC + ABC′ + A′CB + A′CB′ + BCA
= AB (C + C′) + A′C (B + B′) + BCA
= AB + A′C + BCA
= AB (C + 1) + A′C
= AB + A′C

Correct Option: B

= (A′ + B′ + C′)′
= A′′. B′′. C′′
= ABC
Hence (B) is the correct alternative.

Correct Option: B

= (A′ + B′ + C′)′
= A′′. B′′. C′′
= ABC
Hence (B) is the correct alternative.

For the given Boolean theorem
AB + A′C + BC = AB + A′C
Apply dual property, we get:
(A + B) (A′ + C) (B + C) = (A + B) (A′ + C)
Hence alternative (A) is the correct answer.

Correct Option: A

For the given Boolean theorem
AB + A′C + BC = AB + A′C
Apply dual property, we get:
(A + B) (A′ + C) (B + C) = (A + B) (A′ + C)
Hence alternative (A) is the correct answer.