How many two-input AND and OR gates are required to realize Y = CD+EF+G?a.2,3b.2,2c.3,3d.3,2
Question
How many two-input AND and OR gates are required to realize ?
a. 2, 3
b. 2, 2
c. 3, 3
d. 3, 2
Solution
The given Boolean function Y = CD + EF + G can be realized using 2 AND gates and 3 OR gates. Here's how:
Step 1: Realize the AND operations The AND operations are CD and EF. Each of these operations requires one AND gate. So, we need 2 AND gates for this.
Step 2: Realize the OR operations The OR operations are between (CD), (EF), and G. Each of these operations requires one OR gate. So, we need 3 OR gates for this.
So, the answer is 2 AND gates and 3 OR gates, which corresponds to option a.2,3.
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