📘 GATE Electrical – Superposition Theorem Advanced Example Problems (With Solutions)
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Superposition Theorem states that in a linear circuit with multiple independent sources,
the total response is the algebraic sum of responses due to each source acting alone.
📌 Problem 1 – Two Voltage Sources (Opposite Direction)
A circuit contains:
12V source in series with 4Ω
6V source in opposite direction in series with 8Ω
Find current through 4Ω resistor using Superposition.
✔ Step 1: Activate 12V only
Deactivate 6V source → Replace with short circuit.
Total resistance = 4Ω + 8Ω = 12Ω
I₁ = 12 / 12 = 1A
✔ Step 2: Activate 6V only
Deactivate 12V source → Replace with short circuit.
I₂ = 6 / 12 = 0.5A
Direction opposite to I₁.
✔ Step 3: Net Current
I = I₁ − I₂ = 1 − 0.5 = 0.5A
Final Answer: 0.5 A
📌 Problem 2 – Voltage and Current Source Combined
A 5Ω resistor is connected to:
10V voltage source and 2A current source.
Find voltage across resistor using superposition.
✔ Step 1: Consider Voltage Source Only
Deactivate current source → Open circuit.
I₁ = 10 / 5 = 2A
V₁ = 10V
✔ Step 2: Consider Current Source Only
Deactivate voltage source → Short circuit.
V₂ = I × R = 2 × 5 = 10V
✔ Step 3: Total Voltage
V = V₁ + V₂ = 10 + 10 = 20V
Final Answer: 20 V
📌 Problem 3 – Power Calculation
A 6Ω resistor connected across:
18V source and 12V source.
Find total power dissipated using superposition.
✔ Step 1: 18V Active
I₁ = 18 / 6 = 3A
✔ Step 2: 12V Active
I₂ = 12 / 6 = 2A
✔ Step 3: Total Current
I = 3 + 2 = 5A
Power:
P = I²R = (5)² × 6 = 25 × 6 = 150W
Final Answer: 150 W
🎯 Important Exam Tips
- ✔ Voltage source → Replace by short when deactivated
- ✔ Current source → Replace by open when deactivated
- ✔ Maintain consistent current direction
- ✔ Add algebraically (consider sign)
Practice Superposition Deeply. It Builds Strong Thevenin & Norton Concepts.
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