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📘 Network Theory – Advanced Worked Examples (Level 4)

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🔹 Example 1 – Thevenin Using Node Voltage

Given:

• Voltage Source = 24V
• R1 = 4Ω
• R2 = 8Ω
• Load RL = 6Ω (connected at output node)

Step 1: Remove RL

Step 2: Find Vth Using Node Voltage

Let output node voltage = V

Apply Voltage Divider (since simple series network):

Vth = 24 × (8 / (4 + 8))

Vth = 24 × (8/12)

Vth = 16V

Step 3: Find Rth

Deactivate voltage source → Short circuit.

Rth = (4 × 8) / (4 + 8)

Rth = 32 / 12

Rth = 2.67Ω

Step 4: Load Current

Total resistance = 2.67 + 6 = 8.67Ω

I = 16 / 8.67

I = 1.85A

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🔹 Example 2 – Superposition (Sign Confusion Type)

Given:

• V1 = 15V
• V2 = 5V (opposite polarity)
• R = 5Ω

Step 1: Total Resistance

R = 5Ω

Step 2: V1 Active Only

I₁ = 15 / 5 = 3A

Step 3: V2 Active Only

I₂ = 5 / 5 = 1A

Direction opposite → Negative sign

Step 4: Net Current

I = 3 − 1 = 2A

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🔹 Example 3 – Thevenin with Parallel Branch

Given:

• Voltage Source = 30V
• R1 = 10Ω
• R2 = 20Ω
• Load RL = 15Ω

Step 1: Remove RL

Step 2: Find Vth

Voltage Divider:

Vth = 30 × (20 / (10 + 20))

Vth = 30 × (20/30)

Vth = 20V

Step 3: Find Rth

Deactivate source → Short circuit.

Rth = (10 × 20) / (10 + 20)

Rth = 200 / 30

Rth = 6.67Ω

Step 4: Load Current

Total resistance = 6.67 + 15 = 21.67Ω

I = 20 / 21.67

I = 0.92A

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🔹 Example 4 – Power in Load

Using Example 3 Data:

I = 0.92A

RL = 15Ω

Power:

P = I²R

P = (0.92)² × 15

P = 0.846 × 15

P = 12.69W

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