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How many AND and OR gates are required to realise Y = AB + BC + CD?
- a)2,3
- b)2,1
- c)3,1
- d)2,2
Correct answer is option 'C'. Can you explain this answer?
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How many AND and OR gates are required to realise Y = AB + BC + CD?a)2...
AND GATE:
The AND operation is performed the same as the ordinary multiplication of 1s and 0s.
The output of AND gate becomes 1 only when all the inputs are 1.
Hence AND gate is also called an all or nothing gate.
The AND operation is performed the same as the ordinary multiplication of 1s and 0s.
The output of AND gate becomes 1 only when all the inputs are 1.
Hence AND gate is also called an all or nothing gate.
OR GATE:
The output of the OR gate becomes 1 even if one of its inputs is 1.
Hence OR gate is also called an any or all gate.
The output of the OR gate becomes 1 even if one of its inputs is 1.
Hence OR gate is also called an any or all gate.
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How many AND and OR gates are required to realise Y = AB + BC + CD?a)2...
To realize the expression Y = AB BC CD, we first need to simplify it using Boolean algebra.
Y = AB BC CD
Y = ABCD
Therefore, we need to implement a circuit that performs the AND operation on A, B, C, and D.
To implement this circuit, we can use three AND gates and two OR gates.
Implementation:
Step 1: Use two AND gates to multiply A and B, and B and C.
Step 2: Use another AND gate to multiply the output of the first AND gate with D.
Step 3: Use two OR gates to combine the output of the second AND gate.
Therefore, the total number of gates required is three AND gates and two OR gates, which is option C.
Y = AB BC CD
Y = ABCD
Therefore, we need to implement a circuit that performs the AND operation on A, B, C, and D.
To implement this circuit, we can use three AND gates and two OR gates.
Implementation:
Step 1: Use two AND gates to multiply A and B, and B and C.
Step 2: Use another AND gate to multiply the output of the first AND gate with D.
Step 3: Use two OR gates to combine the output of the second AND gate.
Therefore, the total number of gates required is three AND gates and two OR gates, which is option C.
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