Loop Analysis Solving For Currents I1, I2, And I3

Electrical circuit analysis often involves determining the currents flowing through different parts of the circuit. A powerful technique for this is loop analysis, also known as mesh analysis. This method is particularly useful for circuits with multiple voltage sources and resistors arranged in a network. In this comprehensive guide, we'll delve into the application of loop analysis to a specific circuit configuration, systematically solving for the unknown currents I₁, I₂, and I₃. Understanding this process is fundamental for anyone studying electrical engineering or working with electronic circuits.

Setting up the Loop Equations

The cornerstone of loop analysis lies in Kirchhoff's Voltage Law (KVL), which states that the algebraic sum of the voltages around any closed loop in a circuit is zero. To apply loop analysis, we first identify the independent loops within the circuit. In this case, we have three distinct loops, and we assign a circulating current to each loop: I₁, I₂, and I₃. It's crucial to choose a consistent direction (clockwise or counterclockwise) for all loop currents. Now, we can systematically apply KVL to each loop, expressing the voltage drops across each element in terms of the loop currents and resistances.

Loop 1: Establishing the First Equation

Let's begin by analyzing the first loop. Tracing the path of current I₁, we encounter a series of voltage drops across the resistors. Applying KVL, we can express the sum of these voltage drops as follows:

• -5I₁: This term represents the voltage drop across the 5-ohm resistor due to the current I₁. Remember, the voltage drop is negative when the current flows in the direction we've chosen for the loop.
• -3(I₁ - I₂): This represents the voltage drop across the 3-ohm resistor. Notice that both currents I₁ and I₂ flow through this resistor. Since we're analyzing loop 1, we consider the difference between I₁ and I₂. If I₁ is greater than I₂, the current effectively flowing through the resistor in the direction of I₁ is (I₁ - I₂).
• -5: This is a constant voltage drop of 5 volts in the direction of the loop current.

Summing these voltage drops and setting the result equal to zero (according to KVL), we get our first equation:

-5I₁ - 3(I₁ - I₂) - 5 = 0

Simplifying this equation by combining like terms, we arrive at:

8I₁ - 3I₂ = 15 ... (i)

This equation establishes a relationship between the currents I₁ and I₂ in the circuit. It's a crucial step towards solving for the unknown currents.

Loop 2: Deriving the Second Equation

Now, let's move on to the second loop and apply KVL to derive the second equation. Tracing the path of current I₂, we encounter several resistors and a voltage source. The voltage drops across these elements, considering the loop currents, are:

• -4I₂: This term represents the voltage drop across the 4-ohm resistor due to the current I₂.
• +5: This is a constant voltage source of 5 volts in the direction opposite to the loop current, so it's a voltage rise, hence the positive sign.
• -2(I₂ - I₃): This represents the voltage drop across the 2-ohm resistor. Both currents I₂ and I₃ flow through this resistor. Since we're analyzing loop 2, we consider the difference between I₂ and I₃.
• +5: Another constant voltage source of 5 volts in the direction opposite to the loop current.
• +5: A third constant voltage source of 5 volts in the direction opposite to the loop current.
• -3(I₂ - I₁): This represents the voltage drop across the 3-ohm resistor, shared with loop 1. We consider the difference between I₂ and I₁ for loop 2 analysis.

Applying KVL, we sum these voltage drops (and rises) and set the result equal to zero:

-4I₂ + 5 - 2(I₂ - I₃) + 5 + 5 - 3(I₂ - I₁) = 0

Simplifying this equation by distributing and combining like terms, we obtain:

3I₁ - 9I₂ + 2I₃ = -15 ... (ii)

This equation establishes a relationship between all three currents: I₁, I₂, and I₃. It's a critical piece of the puzzle in solving the circuit.

Loop 3: Formulating the Third Equation

Finally, let's analyze the third loop to derive the third equation. Tracing the path of current I₃, we encounter resistors and a voltage source. The voltage drops across these elements, considering the loop currents, are:

• -8I₃: This term represents the voltage drop across the 8-ohm resistor due to the current I₃.
• -30: This is a constant voltage drop of 30 volts in the direction of the loop current.
• -5: This is another constant voltage drop of 5 volts in the direction of the loop current.
• -2(I₃ - I₂): This represents the voltage drop across the 2-ohm resistor, shared with loop 2. We consider the difference between I₃ and I₂ for loop 3 analysis.

Applying KVL, we sum these voltage drops and set the result equal to zero:

-8I₃ - 30 - 5 - 2(I₃ - I₂) = 0

Simplifying this equation by distributing and combining like terms, we arrive at:

2I₂ - 10I₃ = 35 ... (iii)

This equation provides the final relationship needed to solve for the three unknown currents. We now have a system of three linear equations with three unknowns.

Solving the System of Equations

With the three loop equations established, we now have a system of linear equations that we can solve to determine the values of I₁, I₂, and I₃. There are several methods for solving such systems, including:

• Substitution: Solve one equation for one variable and substitute that expression into the other equations. This process can be repeated until you have a single equation with one unknown.
• Elimination: Multiply one or more equations by constants so that the coefficients of one variable are opposites. Then, add the equations together to eliminate that variable. Repeat this process until you have a single equation with one unknown.
• Matrix Methods: Represent the system of equations in matrix form and use techniques like Gaussian elimination or matrix inversion to solve for the unknowns. This method is particularly efficient for larger systems of equations.

Choosing a Solution Method

For this particular system of equations, substitution or elimination are suitable methods. The choice often depends on personal preference and the specific structure of the equations. Matrix methods become more advantageous for systems with four or more equations. Let's demonstrate the solution using the substitution method.

Solving by Substitution

1. Solve equation (i) for I₁: From 8I₁ - 3I₂ = 15, we get: I₁ = (15 + 3I₂) / 8

2. Substitute this expression for I₁ into equation (ii): 3((15 + 3I₂) / 8) - 9I₂ + 2I₃ = -15 Simplifying, we get: 45 + 9I₂ - 72I₂ + 16I₃ = -120 -63I₂ + 16I₃ = -165 ... (iv)

3. Now we have two equations with two unknowns: equation (iii) and equation (iv): 2I₂ - 10I₃ = 35 -63I₂ + 16I₃ = -165

4. Solve equation (iii) for I₂: I₂ = (35 + 10I₃) / 2

5. Substitute this expression for I₂ into equation (iv): -63((35 + 10I₃) / 2) + 16I₃ = -165 Simplifying, we get: -2205 - 630I₃ + 32I₃ = -330 -598I₃ = 1875 I₃ = -1875 / 598 ≈ -3.135 Amperes

6. Substitute the value of I₃ back into the equation for I₂: I₂ = (35 + 10(-3.135)) / 2 I₂ = (35 - 31.35) / 2 I₂ = 3.65 / 2 I₂ ≈ 1.825 Amperes

7. Finally, substitute the value of I₂ back into the equation for I₁: I₁ = (15 + 3(1.825)) / 8 I₁ = (15 + 5.475) / 8 I₁ = 20.475 / 8 I₁ ≈ 2.559 Amperes

Therefore, the currents in the three loops are approximately:

• I₁ ≈ 2.559 Amperes
• I₂ ≈ 1.825 Amperes
• I₃ ≈ -3.135 Amperes

Interpreting the Results

The negative sign for I₃ indicates that the actual direction of the current in that loop is opposite to the direction we initially assumed. This is a common occurrence in loop analysis and doesn't affect the validity of the solution. The magnitudes of the currents tell us the amount of current flowing in each loop, which can be used to calculate voltage drops and power dissipations in various parts of the circuit.

Key Takeaways

• Loop analysis is a powerful technique for solving complex circuits with multiple loops and sources.
• Kirchhoff's Voltage Law (KVL) is the fundamental principle behind loop analysis.
• The process involves setting up a system of linear equations based on KVL and then solving for the unknown loop currents.
• Methods like substitution, elimination, and matrix methods can be used to solve the system of equations.
• The signs of the calculated currents indicate their actual directions relative to the assumed directions.

Conclusion

This comprehensive guide has demonstrated the application of loop analysis to a specific circuit, showcasing the step-by-step process of setting up the loop equations and solving for the unknown currents. Mastering loop analysis is essential for electrical engineers and anyone working with electronic circuits. By understanding the principles and techniques outlined in this guide, you'll be well-equipped to tackle a wide range of circuit analysis problems. Remember, practice is key to proficiency, so work through various examples to solidify your understanding of loop analysis.