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Question 1 of 5
A combinational logic circuit is one where the output:
A A. depends on the inputs that have been in the past
B B. depends only on the current state of the inputs
C C. is usually analogue
D D. requires at least three inputs
Why: In combinational logic circuits, the output depends solely on the present inputs and not on previous inputs or internal states. This distinguishes them from sequential circuits which have memory elements like flip-flops. Option B correctly states 'depends only on the current state of the inputs'. Options A describes sequential circuits, C is incorrect as outputs are digital (binary), and D is false as combinational circuits can have any number of inputs including one or two.[2]
Question 2 of 5
A logic circuit with three inputs needs a truth table with:
A A. 3 rows
B B. 4 rows
C C. 8 rows
D D. 16 rows
Why: For a logic circuit with n inputs, the truth table requires \(2^n\) rows to cover all possible input combinations. For 3 inputs (A, B, C), total combinations = \(2^3 = 8\). The truth table will have 8 rows excluding the header. Option C (8 rows) is correct. Option A (3) is too few, B (4) is for 2 inputs, D (16) is for 4 inputs.[2]
Question 3 of 5
Complete the truth table for this logic gate:
Input AInput BOutput
00?
010
100
111
Why: The given truth table corresponds to an OR gate (\(A + B\)).

Truth table for OR gate:
ABA+B
00\(0\)
01\(1\)
10\(1\)
11\(1\)


The missing output for A=0, B=0 is 0 (false + false = false). However, based on the pattern matching standard gates, the complete table confirms OR gate behavior where output is 1 only when at least one input is 1.[6]
Question 4 of 5
Consider the logic circuit shown below. Complete its truth table.
Why: The circuit shows an XOR gate where Q = X XOR Y = \(X \overline{Y} + \overline{X}Y\).

• When X=0, Y=0: \(0 \cdot 1 + 1 \cdot 0 = 0\)
• When X=0, Y=1: \(0 \cdot 0 + 1 \cdot 1 = 1\)
• When X=1, Y=0: \(1 \cdot 1 + 0 \cdot 0 = 1\)
• When X=1, Y=1: \(1 \cdot 0 + 0 \cdot 1 = 0\)

Intermediate signals: A = X, C = Y. Final output Q follows XOR truth table exactly.[7]
Question 5 of 5
Write the canonical sum-of-products expression for the logic function F(A,B,C) = Σ(0,2,3).
Why: Canonical SOP form requires expressing the function as sum of minterms where F=1. For the given minterms:
• m0 = \( \overline{A}\overline{B}\overline{C} \) (all inputs false)
• m2 = \( \overline{A}B\overline{C} \) (B true, others false)
• m3 = \( \overline{A}BC \) (B and C true, A false)

The final expression \( F = \overline{A}\overline{B}\overline{C} + \overline{A}B\overline{C} + \overline{A}BC \) can be simplified using K-map to \( F = \overline{A} \) but canonical form requires un-simplified minterms. The truth table confirms correctness of all three terms.