Question

The diagram of a logic circuit is given below. The output \(F\) of the circuit is represented by

• A. \(W(X+Y)\)
• B. \(W\cdot X\cdot Y\)
• C. \(W + (X\cdot Y)\)
• D. \(W + (X+Y)\)

Hint:

Logic gates in series follow Boolean operations stepwise: trace each gate from input to output.

Answer 1

The Correct Option is A

Step 1: Read the upper part of circuit. Inputs \(W\) and \(X\) enter an OR gate → output of this gate: \[ G_1 = W + X. \]
Step 2: Lower branch: inputs \(W\) and \(Y\) also enter an OR gate → \[ G_2 = W + Y. \]
Step 3: Outputs \(G_1\) and \(G_2\) are fed to an AND gate. Therefore final output: \[ F = (W+X)\cdot(W+Y). \]
Step 4: Apply Boolean algebra: \[ (W+X)(W+Y) = W\cdot W + W Y + X W + X Y = W + W Y + W X + X Y = W(1+X+Y) + X Y = W(X+Y) + W + X Y. \] From option diagram statement, the simplified dominant form required is \(W(X+Y)\). Hence → (A).