📘 Advanced Worked Examples – Superposition & Thevenin (Level 2)
🔹 Example 5: Superposition with Three Sources
Given:
- V1 = 10V
- V2 = 5V (same direction)
- Current Source = 1A
- Resistor R = 5Ω
Find: Current through 5Ω resistor.
Step 1: V1 Active Only
Deactivate V2 (short) and current source (open).
I₁ = 10 / 5 = 2A
Step 2: V2 Active Only
I₂ = 5 / 5 = 1A
Step 3: Current Source Only
Voltage source → Short.
I₃ = 1A
Step 4: Total Current
I = I₁ + I₂ + I₃ = 2 + 1 + 1 = 4A
Final Answer: Current = 4A
🔹 Example 6: Superposition with Dependent Source
Given:
- Voltage Source = 20V
- Resistor = 10Ω
- Dependent current source = 0.2Vx
Let current through resistor = I
Step 1: Express Controlling Variable
Vx = 10I
Step 2: Dependent Source Value
0.2Vx = 0.2 × 10I = 2I
Step 3: Apply KCL
20 = 10I + (2I × 10)
20 = 10I + 20I
20 = 30I
I = 0.67A
🔹 Example 7: Thevenin with Two Branch Circuit
Given:
- Voltage Source = 30V
- R1 = 10Ω
- R2 = 5Ω
- Load RL = 15Ω
Step 1: Remove Load
Step 2: Find Vth (Voltage Divider)
Vth = 30 × (5 / (10 + 5))
Vth = 30 × (5/15)
Vth = 10V
Step 3: Find Rth
Deactivate source → Short.
Rth = (10 × 5) / (10 + 5)
Rth = 50 / 15
Rth = 3.33Ω
Step 4: Load Current
Total resistance = 3.33 + 15 = 18.33Ω
I = 10 / 18.33
I = 0.55A
🔹 Example 8: Thevenin with Loop Analysis
Given:
- 24V source
- R1 = 4Ω
- R2 = 8Ω
- RL = 6Ω
Step 1: Remove RL
Step 2: Find Vth
Voltage Divider:
Vth = 24 × (8 / (4+8))
Vth = 24 × (8/12)
Vth = 16V
Step 3: Find Rth
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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