Why This Topic is Important for ECET
Boolean Algebra is the foundation of Digital Electronics. All logic circuits, simplification techniques, and gate-level designs are built on Boolean laws.
In ECET 2026, questions often come as:
- Simplify Boolean expressions
- Apply theorems
- Solve logic design MCQs
Mastering Boolean Algebra ensures speed and accuracy in solving logic problems.
📘 Concept Notes
1. Basics
- Boolean Algebra deals with binary variables (0,1).
- Three primary operations:
- AND (·) → multiplication
- OR (+) → addition
- NOT (‾) → complement
Example:
- A⋅B=1A \cdot B = 1A⋅B=1 only if both A = 1 and B = 1.
- A+B=1A + B = 1A+B=1 if either A or B = 1.
- A‾=1\overline{A} = 1A=1 if A = 0.
2. Fundamental Laws
- Identity: A+0=A, A⋅1=AA + 0 = A, \, A \cdot 1 = AA+0=A,A⋅1=A
- Null: A+1=1, A⋅0=0A + 1 = 1, \, A \cdot 0 = 0A+1=1,A⋅0=0
- Idempotent: A+A=A, A⋅A=AA + A = A, \, A \cdot A = AA+A=A,A⋅A=A
- Complement: A+A‾=1, A⋅A‾=0A + \overline{A} = 1, \, A \cdot \overline{A} = 0A+A=1,A⋅A=0
- Commutative: A+B=B+A, A⋅B=B⋅AA + B = B + A, \, A \cdot B = B \cdot AA+B=B+A,A⋅B=B⋅A
- Associative: A+(B+C)=(A+B)+CA + (B + C) = (A + B) + CA+(B+C)=(A+B)+C
- Distributive: A⋅(B+C)=A⋅B+A⋅CA \cdot (B + C) = A \cdot B + A \cdot CA⋅(B+C)=A⋅B+A⋅C
3. Duality Principle
- Every Boolean expression has a dual obtained by:
- Replacing + ↔ ·
- Replacing 0 ↔ 1
Example:
- A+0=AA + 0 = AA+0=A → Dual = A⋅1=AA \cdot 1 = AA⋅1=A.
4. De Morgan’s Theorems
Example:
A⋅B⋅C‾=A‾+B‾+C‾\overline{A \cdot B \cdot C} = \overline{A} + \overline{B} + \overline{C}A⋅B⋅C=A+B+C.
5. Simplification of Expressions
Use Boolean laws + theorems to reduce logic circuits.
Example:
Simplify: 
→ Apply Absorption Law → 
6. Canonical Forms
- SOP (Sum of Products): OR of multiple AND terms.
- POS (Product of Sums): AND of multiple OR terms.
Example:
- SOP:
- POS:
⚙️ Formulas
🔟 10 MCQs
Q1. The dual of 
is:
a) 
b) 
c) 
d) 
Q2. Simplify: 
.
Q3. Which law is represented by 
?
a) Identity
b) Null
c) Complement
d) Distributive
Q4. Apply De Morgan’s theorem: 
.
Q5. The SOP of F = 1 for minterms (1, 2, 3) with variables (A,B):
a) 
b) 
c) 
d) None
Q6. Which of the following is true?
a) 
b) 
c) 
d) 
Q7. How many canonical SOP terms exist for 3 variables?
a) 4
b) 6
c) 8
d) 12
Q8. Simplify 
.
Q9. Which law is applied in: 
?
Q10. If F = AB + A\overline{B}, then F simplifies to:
a) A
b) B
c) AB
d) A + B
✅ Answer Key
| Q No | Answer |
|---|---|
| Q1 | b |
| Q2 | A |
| Q3 | c |
| Q4 |  |
| Q5 | a |
| Q6 | a |
| Q7 | c |
| Q8 | A |
| Q9 | Distributive Law |
| Q10 | A |
🧠 Explanations
- Q1: Dual of A+0=A is A·1=A → (b).
- Q2: A+AB → A(1+B)=A → simplified result A.
- Q3: This is the Complement Law.
- Q4: By De Morgan’s theorem → A+B‾=A‾⋅B‾\overline{A+B} = \overline{A} \cdot \overline{B}A+B=A⋅B.
- Q5: Minterms (1,2,3) in SOP → AB‾\overline{B}B+AB+A‾B\overline{A}BAB.
- Q6: Absorption Law → A+AB=A.
- Q7: For 3 variables → 2³ = 8 minterms.
- Q8: Apply distributive → A + BB‾\overline{B}B = A + 0 = A.
- Q9: This is distributive expansion.
- Q10: AB + AB‾\overline{B}B = A(B+B‾\overline{B}B) = A.
🎯 Motivation / Why Practice Matters
Boolean Algebra problems in ECET 2026 require quick simplification skills.
Practicing regularly will:
- Improve speed in reducing expressions
- Build strong fundamentals for K-maps and logic design
- Help in scoring direct formula-based marks with high accuracy
In competitive exams like ECET, Boolean simplification is a sure-shot scoring area if mastered.
📲 CTA
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