✅ UNIT I – BASIC CIRCUIT ANALYSIS
(Apply, Analyze, Evaluate, Create)
1. Apply
Q1: Apply Ohm’s law to calculate the current in a 10 Ω resistor connected across a 5 V battery.
A1:
2. Analyze
Q2: Analyze the effect of removing a resistor in a parallel circuit.
A2: Removing a parallel resistor increases the equivalent resistance and decreases the total current drawn from the source.
3. Evaluate
Q3: Evaluate which configuration (series or parallel) is better to maintain a constant voltage across loads.
A3: Parallel configuration is better, as it maintains constant voltage across each branch/load.
4. Create
Q4: Design a simple circuit using 3 resistors (2 Ω, 4 Ω, and 6 Ω) such that the total resistance is 12 Ω.
A4: Connect all three resistors in series:
5. Apply
Q5: Apply KVL in a loop with a 12 V battery and two resistors (2 Ω and 4 Ω in series) to find current.
A5:
6. Analyze
Q6: Analyze how node analysis is useful in solving complex circuits.
A6: Node analysis reduces the number of equations by focusing on voltages at nodes instead of individual component currents.
7. Evaluate
Q7: Compare mesh analysis and node analysis. Which is better for circuits with fewer loops?
A7: Mesh analysis is better for circuits with fewer loops, as it requires fewer equations compared to node analysis in such cases.
8. Create
Q8: Construct a circuit where star-delta transformation simplifies the analysis.
A8: Any three resistors connected in a star can be converted to delta to reduce loops and apply mesh analysis easily.
9. Apply
Q9: Use source transformation to simplify a 10 V voltage source in series with a 2 Ω resistor.
A9: Equivalent current source: in parallel with 2 Ω resistor.
10. Evaluate
Q10: Justify the use of delta-star transformation in three-phase power systems.
A10: It allows analysis and connection of loads/generators with different configurations, improving compatibility and simplification.
✅ UNIT II – NETWORK THEOREMS
(Apply, Analyze, Evaluate, Create)
11. Apply
Q11: Apply Superposition Theorem to calculate current due to one source in a dual-source circuit.
A11: Turn off other sources (voltage = short, current = open), and analyze circuit with only one active source.
12. Analyze
Q12: Analyze how Thevenin’s Theorem helps in simplifying complex circuits.
A12: It replaces an entire network with a single voltage source and resistance, making load analysis easier.
13. Evaluate
Q13: Evaluate the benefit of Norton over Thevenin in current-sensitive circuits.
A13: Norton’s form (current source) directly provides insight into current behavior, useful for current-driven loads.
14. Create
Q14: Create an equivalent Thevenin model for a 10 V source in series with 2 Ω resistor.
A14: Thevenin voltage = 10 V; Thevenin resistance = 2 Ω.
15. Apply
Q15: Apply Maximum Power Transfer Theorem to find load resistance for max power when source resistance is 5 Ω.
A15: For max power,
16. Analyze
Q16: Analyze the power delivered to the load when .
A16: Power decreases as load deviates from source resistance; either too high or too low reduces power delivery.
17. Evaluate
Q17: Compare Superposition and Thevenin’s theorem for solving circuits with multiple sources.
A17: Superposition solves step-by-step using each source; Thevenin gives an overall simplified equivalent circuit—more efficient for load analysis.
18. Create
Q18: Design a circuit where Norton and Thevenin equivalents are clearly evident.
A18: A voltage source with series resistance and a load branch—convert to Norton (current source and parallel resistor) and vice versa.
19. Apply
Q19: Use Norton’s theorem to find equivalent current if voltage source is 12 V and resistance is 6 Ω.
A19: Norton current = ; parallel resistance = 6 Ω.
20. Evaluate
Q20: Justify using Thevenin’s theorem in circuit design and testing.
A20: Simplifies testing and modification of the load section without disturbing the source or rest of the network.
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