Math Problem Statement

value of X is 15

Use Mesh Analysis to find V3 in the given circuit.

Solution

Let's analyze the circuit using mesh analysis to find V3V_3.

Step 1: Identify Meshes

We have three loops (meshes) in this circuit:

  • Mesh 1 (I1): Contains the 3V3V source, the 1Ω1 \Omega resistor, and the 4Ω4 \Omega resistor.
  • Mesh 2 (I2): Contains the 4Ω4 \Omega resistor, the 2Ω2 \Omega resistor, the dependent current source 1.8V31.8 V_3, and the V3V_3 voltage.
  • Mesh 3 (I3): Contains the 5V5V voltage source, the dependent current source 1.8V31.8 V_3, and the resistor X=15ΩX = 15 \Omega.

Step 2: Write Mesh Equations

We need to write mesh equations using Kirchhoff's Voltage Law (KVL) for each loop.

  1. Mesh 1 (I1): The sum of voltages around Mesh 1 is: 3VI1(1Ω)(I1I2)(4Ω)=03V - I1(1 \Omega) - (I1 - I2)(4 \Omega) = 0 Simplifying: 3=I1+4(I1I2)3 = I1 + 4(I1 - I2) 3=5I14I2(Equation 1)3 = 5I1 - 4I2 \quad \text{(Equation 1)}

  2. Mesh 2 (I2): The sum of voltages around Mesh 2 includes the V3V_3 dependent source and the resistors: (I2I1)(4Ω)+I2(2Ω)V3=0(I2 - I1)(4 \Omega) + I2(2 \Omega) - V_3 = 0 Simplifying: 4(I2I1)+2I2V3=04(I2 - I1) + 2I2 - V_3 = 0 We also know that V3=15I3V_3 = 15I3 (from the value of X=15ΩX = 15 \Omega).

    Substituting this in: 4(I2I1)+2I215I3=04(I2 - I1) + 2I2 - 15I3 = 0 4I24I1+2I215I3=04I2 - 4I1 + 2I2 - 15I3 = 0 6I24I1=15I3(Equation 2)6I2 - 4I1 = 15I3 \quad \text{(Equation 2)}

  3. Mesh 3 (I3): The sum of voltages around Mesh 3 includes the 5V5V source, the 15Ω15 \Omega resistor, and the dependent source: 515I3=05 - 15I3 = 0 Simplifying: I3=515=13(Equation 3)I3 = \frac{5}{15} = \frac{1}{3} \quad \text{(Equation 3)}

Step 3: Solve the System of Equations

We now have the following system:

  1. 3=5I14I23 = 5I1 - 4I2
  2. 6I24I1=15×13=56I2 - 4I1 = 15 \times \frac{1}{3} = 5
  3. I3=13I3 = \frac{1}{3}

Let's solve these equations to find V3V_3.

I will calculate it step by step.The solutions for the mesh currents are:

  • I1=197A2.71AI_1 = \frac{19}{7} \, \text{A} \approx 2.71 \, \text{A}
  • I2=3714A2.64AI_2 = \frac{37}{14} \, \text{A} \approx 2.64 \, \text{A}

Additionally, the voltage V3V_3 is:

V3=5VV_3 = 5 \, \text{V}

Would you like further details or clarifications on any part of this solution?

Here are 5 related questions to further expand on this:

  1. How would the analysis change if we had a different value for the dependent current source?
  2. What happens if the value of XX (the resistor) changes?
  3. Can you explain the significance of dependent sources in mesh analysis?
  4. How would you analyze this circuit using nodal analysis instead of mesh analysis?
  5. How does the mesh current method compare with the superposition theorem in solving such circuits?

Tip: When dealing with dependent sources, always express the controlling variable (e.g., V3V_3) in terms of mesh or node variables to simplify the equations.

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Math Problem Analysis

Mathematical Concepts

Mesh Analysis
Kirchhoff's Voltage Law (KVL)
Dependent Sources

Formulas

KVL: ΣV around a closed loop = 0
Ohm's Law: V = IR
Mesh current relationships between resistors and voltage sources

Theorems

Kirchhoff's Voltage Law
Ohm's Law

Suitable Grade Level

Undergraduate (Electrical Engineering)