📘 Norton’s Theorem – Definition & Full Worked Examples
🔹 Definition of Norton’s Theorem
Norton’s Theorem states that any linear, bilateral two-terminal network can be replaced by an equivalent circuit consisting of a current source (IN) in parallel with a resistance (RN) as seen from the load terminals.
🔹 Step-by-Step Solution Method
Step 2: Find Short Circuit Current across load terminals.
This current = IN (Norton Current).
Step 3: Deactivate all independent sources.
• Voltage source → Short circuit
• Current source → Open circuit
Step 4: Find equivalent resistance from terminals.
This = RN.
Step 5: Draw Norton equivalent:
Current source IN in parallel with RN.
Step 6: Reconnect RL and use Current Division Rule.
🔹 Example 1 – Basic Norton Example
Given:
- Voltage Source = 12V
- Series Resistor = 6Ω
- Load RL = 3Ω
Step 1: Remove Load
Step 2: Find Short Circuit Current (IN)
If load terminals are shorted:
IN = V / R
IN = 12 / 6
IN = 2A
Step 3: Find RN
Deactivate voltage source → Short circuit.
RN = 6Ω
Step 4: Load Current
Using Current Division:
IL = IN × (RN / (RN + RL))
IL = 2 × (6 / (6 + 3))
IL = 2 × (6/9)
IL = 1.33A
🔹 Example 2 – Voltage Divider Type Circuit
Given:
- Voltage Source = 24V
- R1 = 4Ω
- R2 = 8Ω
- Load RL connected across R2
Step 1: Remove Load
Step 2: Find Norton Current
First find Thevenin voltage:
Vth = 24 × (8 / (4+8)) = 16V
Rth = (4×8)/(4+8) = 2.67Ω
IN = Vth / Rth
IN = 16 / 2.67
IN = 6A
Step 3: RN = Rth = 2.67Ω
Step 4: Load Current
IL = 6 × (2.67 / (2.67 + RL))
(Substitute RL value if given)
🔹 Example 3 – Maximum Power Using Norton
Given:
- IN = 4A
- RN = 5Ω
Condition for Maximum Power
RL = RN
RL = 5Ω
Maximum Power
Pmax = (IN² × RN) / 4
Pmax = (4² × 5) / 4
Pmax = (16 × 5) / 4
Pmax = 20W
🎯 Important Exam Insights
- Norton and Thevenin are dual theorems
- RN = Rth always
- IN = Vth / Rth
- Current division mistakes are common traps
- Frequently asked in GATE & IES
⚡ Master Norton – You Master Parallel Networks ⚡

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