GATE Electrical Engineering - Complete Master Library

 

📘 Norton’s Theorem – Complete Theory & Detailed Worked Examples

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🔹 1. Introduction

Norton’s Theorem is the dual of Thevenin’s Theorem. While Thevenin converts a network into a single voltage source in series with resistance, Norton converts the same network into a current source in parallel with resistance.

Norton is extremely useful for parallel circuits and current analysis problems. In GATE and IES exams, Norton is frequently combined with Thevenin.

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🔹 2. Statement

Any linear bilateral two-terminal network can be replaced by an equivalent current source (IN) in parallel with a resistance (RN) as seen from the load terminals.

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🔹 3. Relationship Between Thevenin & Norton

IN = Vth / Rth RN = Rth

This relationship is very important for fast conversion in competitive exams.

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🔹 4. Step-by-Step Procedure

Step 1: Remove load resistor (RL).

Step 2: Short the load terminals and find short-circuit current → IN.

Step 3: Deactivate independent sources.
Voltage source → Short circuit.
Current source → Open circuit.

Step 4: Find equivalent resistance from terminals → RN.

Step 5: Draw Norton equivalent circuit and reconnect RL.

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🔹 5. Worked Example 1 – Basic Norton Equivalent

Given:

• Voltage Source = 12V
• Series Resistor = 6Ω
• Load RL = 3Ω

Step 1: Remove RL

Step 2: Find Short Circuit Current

IN = 12 / 6 = 2A

Step 3: Find RN

Deactivate source → Short RN = 6Ω

Step 4: Load Current Using Current Division

IL = IN × (RN / (RN + RL)) IL = 2 × (6 / (6+3)) IL = 1.33A

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🔹 6. Worked Example 2 – Two Resistor Network

Given:

• Voltage Source = 24V
• R1 = 4Ω
• R2 = 8Ω
• Load across R2

Step 1: Convert to Thevenin First

Vth = 24 × (8/12) = 16V Rth = (4×8)/12 = 2.67Ω

Step 2: Convert to Norton

IN = Vth / Rth IN = 16 / 2.67 IN = 6A

RN = 2.67Ω

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🔹 7. Worked Example 3 – Current Source Circuit

Given:

• Current Source = 5A
• Parallel resistor = 10Ω
• Load RL = 5Ω

Step 1: Short Circuit Current

IN = 5A

Step 2: Find RN

Deactivate current source → Open circuit RN = 10Ω

Step 3: Load Current

IL = 5 × (10/(10+5)) IL = 3.33A

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🔹 8. Maximum Power Transfer Using Norton

Condition:

RL = RN

Maximum Power:

Pmax = (IN² × RN) / 4

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🔹 9. Common Exam Mistakes

• Confusing short circuit current with normal branch current
• Incorrect deactivation of sources
• Wrong application of current division rule
• Forgetting RN = Rth

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🎯 Final Summary

Norton simplifies parallel network analysis. Understanding conversion between Thevenin and Norton increases solving speed. Mastering Norton strengthens conceptual clarity for competitive exams.

Norton Mastery = Parallel Network Control