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📘 Thevenin’s Theorem – Advanced Worked Examples (Level 3)

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🔹 Example 1 – Mixed Source Circuit

Given:

• Voltage source = 18V
• Series resistor = 6Ω
• Parallel branch resistor = 3Ω
• Load RL = 9Ω

Step 1: Remove RL

Step 2: Find Vth

Using Voltage Divider:

Vth = 18 × (3 / (6 + 3))

Vth = 18 × (3/9)

Vth = 6V

Step 3: Find Rth

Deactivate source → Short circuit.

Rth = (6 × 3) / (6 + 3)

Rth = 18 / 9

Rth = 2Ω

Step 4: Load Current

Total Resistance = 2 + 9 = 11Ω

I = 6 / 11

I = 0.545A

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🔹 Example 2 – Current Source Based Circuit

Given:

• Current Source = 4A
• Parallel resistor = 8Ω
• Series resistor = 2Ω
• Load RL = 6Ω

Step 1: Convert Current Source to Voltage Form

V = I × R = 4 × 8

V = 32V

Step 2: Find Vth

Voltage divider across 2Ω:

Vth = 32 × (2 / (8 + 2))

Vth = 32 × (2/10)

Vth = 6.4V

Step 3: Find Rth

Deactivate current source → Open circuit.

Rth = 8 + 2

Rth = 10Ω

Step 4: Load Current

Total Resistance = 10 + 6 = 16Ω

I = 6.4 / 16

I = 0.4A

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🔹 Example 3 – Maximum Power Transfer

Given:

• Vth = 15V
• Rth = 5Ω

Find RL for Maximum Power

Condition:

RL = Rth

RL = 5Ω

Maximum Power

Pmax = Vth² / (4Rth)

Pmax = 15² / (4 × 5)

Pmax = 225 / 20

Pmax = 11.25W

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🔹 Example 4 – Conceptual GATE Trap

If Rth = 0Ω, what happens?

• Thevenin equivalent becomes ideal voltage source
• Load current depends only on RL
• Infinite current possible if RL = 0 (short circuit)

In GATE, ideal source conditions often test conceptual clarity.

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