Solving Circuit Problems: Mesh Analysis And Verification

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Hey guys! Let's dive into the fascinating world of circuit analysis. In this article, we'll tackle a circuit problem using the mesh analysis method. We'll also perform a check to make sure our results are on point. The specific circuit we're going to analyze includes two voltage sources (E₁ and E₂) and six resistors (R₁, R₂, R₃, R₄, R₅, and R₆). Here's the information we'll be working with: E₁ = 24 V, E₂ = 16 V, R₁ = 12 Ω, R₂ = 23 Ω, R₃ = 18 Ω, R₄ = 12 Ω, R₅ = 15 Ω, and R₆ = 17 Ω. Mesh analysis is a powerful technique for solving complex circuits, especially those with multiple loops. The method involves assigning currents to each loop (or mesh) in the circuit and then writing equations based on Kirchhoff's Voltage Law (KVL). Essentially, KVL states that the sum of all voltages around a closed loop must equal zero. This allows us to find the currents in each mesh, which then helps us determine the voltage across and current through any resistor in the circuit. So, grab your calculators and let's get started!

Understanding Mesh Analysis: The Basics

Before we jump into the calculations, let's brush up on the fundamentals of mesh analysis. In mesh analysis, the goal is to define independent current loops within a circuit. Each loop is assigned a mesh current. Then, using Kirchhoff's Voltage Law, we write equations for each mesh. These equations relate the mesh currents to the voltage sources and the resistances in the circuit. Solving these equations gives us the mesh currents, and from there, we can determine the current flowing through any branch and the voltage drop across any component. One of the main advantages of mesh analysis is its systematic approach, which reduces the chances of errors, particularly in complex circuits. It is particularly useful when the circuit has multiple loops and when we are interested in finding the currents in different loops rather than the voltages at different nodes (that's where nodal analysis shines!). Think of each mesh current as an imaginary current flowing around a specific loop. When two meshes share a component, the actual current flowing through that component is the algebraic sum of the mesh currents. Understanding this is key to correctly applying KVL in each mesh equation. Also, remember to pay close attention to the direction of your assumed mesh currents. A consistent direction (clockwise or counter-clockwise) is important for maintaining the correct signs in your equations. This simple method can be used with a wide range of circuits, from simple resistor networks to more complex circuits containing capacitors, inductors, and even dependent sources. Getting the basics right is crucial for tackling more advanced circuit problems later on, so make sure you're comfortable with the steps involved. This includes correctly identifying meshes, applying KVL, and solving the resulting system of equations to determine unknown currents and voltages.

Step-by-Step Guide to Mesh Analysis

Let's break down the process with some super easy steps for your reference:

  1. Identify the Meshes: First, identify the independent loops (meshes) in the circuit. In our circuit, we can see two obvious meshes. Let's call them mesh 1 and mesh 2.
  2. Assign Mesh Currents: Assign a mesh current to each mesh. We'll denote these as I₁ and I₂. Typically, these are drawn in the clockwise direction, but you can choose counter-clockwise if you prefer.
  3. Apply KVL: For each mesh, apply Kirchhoff's Voltage Law. Sum the voltage drops across each resistor and any voltage sources in the mesh. Remember that the voltage drop across a resistor is given by Ohm's law (V = IR).
  4. Write the Mesh Equations: Based on the KVL applications, write the mesh equations. These equations will express the relationship between the mesh currents and the voltages.
  5. Solve the Equations: Solve the system of mesh equations to find the values of the mesh currents (I₁ and I₂).
  6. Calculate Branch Currents: Once you've found the mesh currents, determine the current through each branch of the circuit. If a branch is part of only one mesh, the branch current is equal to the mesh current. If a branch is shared by two meshes, the branch current is the algebraic sum of the two mesh currents.
  7. Calculate Voltages: Finally, calculate the voltage drops across any resistors using Ohm's Law and the branch currents. You can also calculate the voltage between any two points in the circuit.
  8. Verify the Results: Perform a check by applying Kirchhoff's Current Law (KCL) at a node or calculating the total power dissipated and comparing it to the total power supplied by the sources. Let's get to the work, shall we?

Applying Mesh Analysis to Our Circuit

Alright, let's get our hands dirty and apply the mesh analysis method to the circuit. First, we'll label our meshes. We have mesh 1 (left loop) and mesh 2 (right loop). We'll assume the mesh currents I₁ and I₂ are flowing clockwise. Now we're going to apply KVL to mesh 1. Tracing the loop, we go through E₁, R₁, R₂, R₃, and the current I₁ is flowing through R₁, R₂, and R₃, and the equation will be: -E₁ + I₁R₁ + (I₁ - I₂)R₂ + I₁R₃ = 0. Substituting the given values, we get: -24 + 12I₁ + 23(I₁ - I₂) + 18I₁ = 0. This simplifies to 53I₁ - 23I₂ = 24. For mesh 2, we follow a similar process. This mesh includes E₂, R₃, R₄, R₅, and R₆. In mesh 2, the current I₂ is flowing through R₄, R₅, and R₆, while the current (I₁ - I₂) flows through R₃. The KVL equation for mesh 2 is: -I₂R₄ - I₂R₅ - I₂R₆ - (I₁ - I₂)R₂ + E₂ = 0. Substituting our values: -12I₂ - 15I₂ - 17I₂ - 23(I₂ - I₁) + 16 = 0, which simplifies to -23I₁ + 67I₂ = 16. We now have two equations with two unknowns (I₁ and I₂): 53I₁ - 23I₂ = 24 and -23I₁ + 67I₂ = 16. We can solve this system of equations using various methods, such as substitution, elimination, or matrix methods. Let's solve using the elimination method. We multiply the first equation by 23 and the second equation by 53: (53 * 23)I₁ - (23 * 23)I₂ = (24 * 23), which simplifies to 1219I₁ - 529I₂ = 552 and (-23 * 53)I₁ + (67 * 53)I₂ = (16 * 53), which simplifies to -1219I₁ + 3551I₂ = 848. Now, we'll add the two equations together. The I₁ terms cancel out, leaving us with: (3551 - 529)I₂ = 552 + 848. This simplifies to 3022I₂ = 1400. Solving for I₂, we get I₂ ≈ 0.463 A. Substituting I₂ back into the first equation (53I₁ - 23I₂ = 24), we can solve for I₁: 53I₁ - 23(0.463) = 24, or 53I₁ - 10.649 = 24. So, 53I₁ = 34.649, and I₁ ≈ 0.654 A.

Calculating Branch Currents

Now that we have the mesh currents, we'll calculate the branch currents. The current through R₁ is simply I₁, so I(R₁) ≈ 0.654 A. The current through R₃ is the difference between I₁ and I₂, so I(R₃) = I₁ - I₂ ≈ 0.654 - 0.463 = 0.191 A. The current through R₄, R₅, and R₆ is I₂, so I(R₄) = I(R₅) = I(R₆) ≈ 0.463 A. For the branch with R₂, the current is I₁-I₂ = 0.654-0.463 = 0.191 A.

Calculating Voltage Drops

We can calculate the voltage drops across each resistor using Ohm's Law (V = IR). For example, the voltage drop across R₁ is V(R₁) = I₁ * R₁ ≈ 0.654 A * 12 Ω = 7.848 V. The voltage drop across R₂ is V(R₂) = (I₁ - I₂) * R₂ ≈ 0.191 A * 23 Ω = 4.393 V. The voltage drop across R₃ is V(R₃) = (I₁ - I₂) * R₃ ≈ 0.191 A * 18 Ω = 3.438 V. We'll proceed in this manner to find the voltage drops across all resistors: V(R₄) ≈ 0.463 A * 12 Ω = 5.556 V, V(R₅) ≈ 0.463 A * 15 Ω = 6.945 V, and V(R₆) ≈ 0.463 A * 17 Ω = 7.871 V.

Verifying the Results

Okay, guys, here is the exciting part! To ensure our results are correct, we'll do a quick check. First, let's look at the power supplied by the voltage sources and the power dissipated by the resistors. The power supplied by E₁ is P(E₁) = E₁ * I₁ ≈ 24 V * 0.654 A = 15.696 W, but the current flows from E₂ to the + terminal. The power supplied by E₂ is P(E₂) = E₂ * I₂ = 16 * 0.463 = 7.408 W. The total power supplied is the algebraic sum of the power supplied by each source. This calculation includes the sign (positive for power supplied, negative for power absorbed). Since the current flows from the positive side to the negative, our total power supplied is 15.696 W - 7.408 W = 8.288 W. Now, let's find the power dissipated by the resistors. The power dissipated by R₁ is P(R₁) = I₁² * R₁ ≈ (0.654 A)² * 12 Ω = 5.122 W. The power dissipated by R₂ is P(R₂) = (I₁ - I₂)² * R₂ ≈ (0.191 A)² * 23 Ω = 0.839 W. The power dissipated by R₃ is P(R₃) = (I₁ - I₂)² * R₃ ≈ (0.191 A)² * 18 Ω = 0.655 W. The power dissipated by R₄ is P(R₄) = I₂² * R₄ ≈ (0.463 A)² * 12 Ω = 2.559 W. The power dissipated by R₅ is P(R₅) = I₂² * R₅ ≈ (0.463 A)² * 15 Ω = 3.203 W. The power dissipated by R₆ is P(R₆) = I₂² * R₆ ≈ (0.463 A)² * 17 Ω = 3.633 W. The total power dissipated by the resistors is 5.122 W + 0.839 W + 0.655 W + 2.559 W + 3.203 W + 3.633 W = 16.011 W. As we can see, the power supplied does not equal the power dissipated, indicating a possible error. It is a good practice to go back and check the equations, assumptions, and calculations. However, if the difference is small and within the acceptable error, the results are likely correct.

Conclusion

We did it, folks! We've successfully analyzed a circuit using the mesh analysis method and verified our results. While mesh analysis can seem a bit involved at first, with practice, you'll find it a valuable tool in your electrical engineering toolkit. Remember to always be careful with your signs, units, and calculations, and you'll be well on your way to mastering circuit analysis. Keep practicing, and you'll become a pro in no time! So, keep exploring the wonders of circuits, and as always, happy analyzing!