Kirchhoff's Rules Circuit Simulation
This interactive simulation demonstrates how to analyze complex circuits using Kirchhoff's junction and loop rules.
Circuit Diagram
Current in Each Branch:
Kirchhoff's Equations for This Circuit:
Loop ADCA: 7I₁ - 6I₂ - 2I₃ = 10
Loop ABCA: I₁ + 6I₂ + 2I₃ = 10
Loop BCDEB: 2I₁ - 4I₂ - 4I₃ = -5
Step-by-Step Solution
1. Identify Junctions and Currents
We identify three independent currents in the circuit:
- I₁: Current through the 1Ω resistor (CA)
- I₂: Current through the 4Ω resistor (BA)
- I₃: Current through the 2Ω resistor (BC)
All other branch currents can be expressed in terms of these three currents.
2. Apply Kirchhoff's Junction Rule
At junction A:
I₁ = I₂ + I₃ (Current entering equals current leaving)
3. Apply Kirchhoff's Loop Rule
We apply the loop rule to three independent loops:
Loop ADCA:
Starting at A, going clockwise:
10V - 4Ω(I₁ - I₂) + 2Ω(I₂ + I₃ - I₁) - 1Ω(I₁) = 0
Simplified: 7I₁ - 6I₂ - 2I₃ = 10
Loop ABCA:
Starting at A, going clockwise:
10V - 4Ω(I₂) - 2Ω(I₂ + I₃) - 1Ω(I₁) = 0
Simplified: I₁ + 6I₂ + 2I₃ = 10
Loop BCDEB:
Starting at B, going clockwise:
5V - 2Ω(I₂ + I₃) - 2Ω(I₂ + I₃ - I₁) = 0
Simplified: 2I₁ - 4I₂ - 4I₃ = -5
4. Solve the System of Equations
Solving the three equations simultaneously gives:
I₁ = 2.5A
I₂ = 0.625A
I₃ = 1.875A
5. Determine All Branch Currents
Using these values, we can find all branch currents:
- CA: I₁ = 2.5A
- BA: I₂ = 0.625A
- BC: I₃ = 1.875A
- BE: I₂ + I₃ = 2.5A
- DE: I₂ + I₃ = 2.5A
- AB: I₂ = 0.625A
- AD: I₁ - I₂ = 1.875A
- CD: I₁ - (I₂ + I₃) = 0A
6. Verify with Another Loop
Let's verify using loop BADEB:
5V + (0.625A × 4Ω) - (1.875A × 4Ω) = 0
5V + 2.5V - 7.5V = 0 ✓
This confirms our solution is correct.
Key Takeaways
- Kirchhoff's rules allow us to analyze complex circuits with multiple loops
- We need as many independent equations as unknown currents
- Always verify your solution with an additional loop
- The direction of current flow affects the sign in the equations
