GATE Electrical Engineering - Complete Master Library

 

📘 Network Theory – Full Numerical Examples (Step-by-Step)

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🔹 Example 1: Superposition Theorem (Complete Numerical)

Given Circuit:

• Voltage Source V1 = 12V
• Voltage Source V2 = 6V (opposite polarity)
• Resistors: R1 = 4Ω, R2 = 8Ω (Series)

Find: Current through 4Ω resistor.

Step 1: Total Resistance

Since resistors are in series:

R_total = 4 + 8 = 12Ω

Step 2: Activate 12V Source Only

Deactivate 6V → Replace with short circuit.

I₁ = V / R = 12 / 12 = 1A

Step 3: Activate 6V Source Only

Deactivate 12V → Replace with short circuit.

I₂ = 6 / 12 = 0.5A

Direction opposite to I₁.

Step 4: Final Current

I = I₁ − I₂ = 1 − 0.5 = 0.5A

Final Answer: Current through 4Ω = 0.5A

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🔹 Example 2: Superposition with Voltage + Current Source

• Voltage Source = 10V
• Current Source = 2A
• Resistor = 5Ω

Find: Voltage across 5Ω resistor.

Step 1: Voltage Source Only

Deactivate current source → Open circuit.

I₁ = 10 / 5 = 2A

V₁ = 10V

Step 2: Current Source Only

Deactivate voltage source → Short circuit.

I₂ = 2A

V₂ = I₂ × R = 2 × 5 = 10V

Step 3: Total Voltage

V = V₁ + V₂ = 10 + 10 = 20V

Final Answer: Voltage across resistor = 20V

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🔹 Example 3: Thevenin’s Theorem (Complete Numerical)

Given:

• Voltage Source = 24V
• R1 = 6Ω (series)
• R2 = 3Ω (load resistor)

Find: Current through R2 using Thevenin’s Theorem.

Step 1: Remove Load Resistor

Open circuit R2.

Step 2: Find Open Circuit Voltage (Vth)

Since only R1 is present:

Vth = 24V (no drop because open circuit)

Step 3: Find Thevenin Resistance (Rth)

Deactivate voltage source → Short circuit.

Rth = 6Ω

Step 4: Reconnect Load

Total Resistance = Rth + R2 = 6 + 3 = 9Ω

Step 5: Find Load Current

I = Vth / (Rth + RL)

I = 24 / 9 = 2.67A

Final Answer: Current through R2 = 2.67A

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🔹 Example 4: Thevenin with Two Resistors

• Voltage Source = 18V
• R1 = 3Ω
• R2 = 6Ω
• Load RL = 9Ω

Step 1: Remove RL

Step 2: Find Vth

Voltage Divider Rule:

Vth = 18 × (6 / (3+6))

Vth = 18 × (6/9)

Vth = 12V

Step 3: Find Rth

Deactivate source → Short circuit.

Rth = (3 × 6) / (3+6)

Rth = 18 / 9 = 2Ω

Step 4: Reconnect Load

Total Resistance = 2 + 9 = 11Ω

I = 12 / 11 = 1.09A

Final Answer: Load Current = 1.09A

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🎯 Exam Tip

Superposition is best for conceptual clarity. Thevenin is best for simplification. In GATE, both appear frequently.