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Multi-Loop Circuit Analysis

Multi-loop circuit analysis involves applying Kirchhoff's Laws systematically to solve complex circuits with multiple loops and junctions. This method allows us to find currents, voltages, and power in each component of the circuit.

Systematic Analysis Method

🎯 Systematic Approach

Multi-loop analysis uses Kirchhoff's Laws to create a system of equations that can be solved simultaneously.

This systematic approach prevents errors and ensures all circuit elements are properly analyzed.

Step 1: Identify Circuit Elements

Step 2: Assign Current Variables

Step 3: Apply Kirchhoff's Voltage Law

Step 4: Apply Kirchhoff's Current Law (if needed)

Step 5: Solve the System of Equations

Example: Three-Loop Circuit

Problem: Find the currents in a circuit with three loops, multiple resistors, and two batteries (12 V and 3 V).

Three-Loop Circuit
Complex circuit with three independent loops and two batteries

Step 1: Assign Current Variables

Let \( I_1 \) be the current in the left loop (clockwise), \( I_2 \) be the current through the middle vertical branch (downward), and \( I_3 \) be the current in the right loop (clockwise).

Step 2: Write KVL Equations

Apply Kirchhoff’s Voltage Law (KVL) to each loop, using the passive sign convention and Ohm’s Law.

$$\text{Loop 1 (left loop): } 12 - 3I_1 - 2(I_1 - I_2) - 1I_1 = 0$$ $$\text{Loop 2 (vertical middle branch): } 2(I_1 - I_2) + 4I_2 - 2(I_3 - I_2) = 0$$ $$\text{Loop 3 (right loop): } 5I_3 + 2(I_3 - I_2) + 3I_3 - 3 = 0$$

Step 3: Simplify the Equations

$$\text{Loop 1: } 12 = 6I_1 - 2I_2$$ $$\text{Loop 2: } 0 = 2I_1 - 8I_2 + 2I_3$$ $$\text{Loop 3: } 3 = 10I_3 - 2I_2$$

Step 4: Solve the System of Equations

Solving this system (by substitution or matrix methods) gives:

$$I_1 = \frac{231}{104} \approx 2.22\,\text{A}$$ $$I_2 = \frac{69}{104} \approx 0.66\,\text{A}$$ $$I_3 = \frac{45}{104} \approx 0.43\,\text{A}$$

Answer

The currents are:

Common Circuit Patterns

Bridge Circuits

Circuits with cross-connections require careful current assignment:

Delta-Wye Transformations

For complex resistor networks:

Superposition Method

For circuits with multiple sources:

Power Analysis

⚡ Power Calculations

Once currents are found, calculate power in each element:

Verification

Power Analysis

⚡ Power Calculations

Once currents are found, calculate power in each element:

Verification

Example: Power Verification

Problem: Verify power conservation in the three-loop circuit.

Step 1: Calculate Power in Each Element

$$P_{R1} = I_1^2 R_1 = \left(\frac{231}{104}\right)^2 \cdot 3 \approx 14.25\,\text{W}$$ $$P_{R2} = I_2^2 R_2 = \left(\frac{69}{104}\right)^2 \cdot 4 \approx 1.76\,\text{W}$$ $$P_{R3} = (I_1 - I_2)^2 R_3 = \left(\frac{162}{104}\right)^2 \cdot 2 \approx 4.85\,\text{W}$$ $$P_{R4} = I_3^2 R_4 = \left(\frac{45}{104}\right)^2 \cdot 5 \approx 0.93\,\text{W}$$ $$P_{R5} = (I_2 - I_3)^2 R_5 = \left(\frac{24}{104}\right)^2 \cdot 1 \approx 0.05\,\text{W}$$

Step 2: Calculate Power Supplied

$$P_{B1} = I_1 \cdot 12 = \frac{231}{104} \cdot 12 \approx 26.65\,\text{W}$$ $$P_{B2} = I_2 \cdot 6 = \frac{69}{104} \cdot 6 \approx 3.98\,\text{W}$$ $$P_{B3} = I_3 \cdot 3 = \frac{45}{104} \cdot 3 \approx 1.30\,\text{W}$$

Step 3: Verify Conservation

$$\text{Power Supplied} \approx 26.65 + 3.98 + 1.30 = 31.93\,\text{W}$$ $$\text{Power Dissipated} \approx 14.25 + 1.76 + 4.85 + 0.93 + 0.05 = 31.84\,\text{W}$$ $$\text{Net Power Difference} \approx 31.93 - 31.84 = 0.09\,\text{W (would be 0 using more significant figures)}$$

For More Explanation/Practice:

Key Takeaways