GATE Electrical Engineering - Complete Master Library

 

📘 Example 1 – Norton Equivalent Across Load

Given:

• Voltage Source = 20V
• Series Resistor = 10Ω
• Load Resistor RL = 5Ω

Step 1: Remove Load Resistor (RL)

We need to find Norton Current (IN).

Step 2: Short Circuit the Load Terminals

Now circuit becomes 20V in series with 10Ω only.

Short circuit current:

IN = V / R = 20 / 10 = 2 A

Step 3: Find Norton Resistance (RN)

Deactivate voltage source → Replace 20V by short circuit.

Now only 10Ω is seen from terminals.

RN = 10Ω

Final Norton Equivalent:

• Current Source = 2 A
• Parallel Resistance = 10Ω

📘 Example 2 – Norton Equivalent with Parallel Branch

Given:

• Voltage Source = 12V
• R1 = 4Ω
• R2 = 6Ω
• Load RL connected across R2

Step 1: Remove Load

Step 2: Find Short Circuit Current (IN)

Total resistance in circuit:

RT = R1 + R2 = 4 + 6 = 10Ω

Total current from source:

I = 12 / 10 = 1.2 A

Current through R2 (by current division):

IN = I × (R1 / (R1 + R2)) IN = 1.2 × (4 / 10) IN = 0.48 A

Step 3: Find RN

Deactivate source → Short 12V

Now R1 and R2 are in parallel:

RN = (R1 × R2) / (R1 + R2) RN = (4 × 6) / 10 RN = 24 / 10 = 2.4Ω

Final Norton Equivalent:

• IN = 0.48 A
• RN = 2.4Ω

📘 Example 3 – Mixed Network Problem

Given:

• Current Source = 5A
• Parallel resistor = 8Ω
• Series resistor = 4Ω
• Load connected at output

Step 1: Find Short Circuit Current

Since 5A source already exists,

IN = 5A

Step 2: Find Norton Resistance

Deactivate current source → Open circuit it.

Remaining resistors seen:

4Ω + 8Ω (series)

RN = 4 + 8 = 12Ω

Final Norton Equivalent:

• IN = 5A
• RN = 12Ω

📘 KVL – Worked Examples (Page 2)

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🔹 Example 1 – Single Loop

Voltage source = 20V Resistors: 4Ω and 6Ω in series

20 − 4I − 6I = 0 20 − 10I = 0 I = 2A

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🔹 Example 2 – Two Loop System

Loop1: 10V, 2Ω Loop2: 5Ω shared

(2+5)I1 − 5I2 = 10 −5I1 + 5I2 = 0 Solve → I1 = 1A

📘 Superposition – Worked Examples (Page 3)

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🔹 Example 1 – Two Voltage Sources

V1 = 12V V2 = 6V (opposite) R = 6Ω

Step 1: V1 only → I1 = 12/6 = 2A

Step 2: V2 only → I2 = 6/6 = 1A (opposite)

Net I = 2 − 1 = 1A

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🔹 Example 2 – Voltage + Current Source

10V source 2A source 5Ω resistor

Voltage only → 2A Current only → 2A Total = 4A

📘 Thevenin – Worked Examples (Page 4)

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🔹 Example 1 – Voltage Divider

V = 20V R1 = 4Ω R2 = 6Ω Load = 5Ω

Vth:

Vth = 20 × (6/10) = 12V

Rth:

Rth = (4×6)/10 = 2.4Ω

Load Current:

I = 12 / (2.4 + 5) = 1.62A

📘 Norton – Worked Examples (Page 5)

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🔹 Example 1

Voltage = 12V R = 6Ω Load = 3Ω

Short Circuit Current:

IN = 12/6 = 2A

RN:

RN = 6Ω

Load Current:

IL = 2 × (6/(6+3)) = 1.33A