Math Problem Statement

Find loop currents i1, i2, i3 and the value of dependent source using mesh analysis.

Solution

We can solve this problem using mesh analysis, where we will apply Kirchhoff's Voltage Law (KVL) to each loop.

Given:

  • i1i_1, i2i_2, and i3i_3 are the mesh currents for loops 1, 2, and 3 respectively.
  • There's a dependent current source given as 4i04i_0, where i0=i2i3i_0 = i_2 - i_3 (as indicated in the circuit diagram).

Step 1: Write KVL equations for each mesh.

Loop 1 (Mesh 1: containing i1i_1)

For the first loop:

  • The voltage drop across the 10Ω resistor: 10(i1i2)10(i_1 - i_2).
  • The voltage drop across the 12Ω resistor: 12i112i_1.
  • The supplied voltage: 24V.

Applying KVL to this loop: 2412i110(i1i2)=024 - 12i_1 - 10(i_1 - i_2) = 0 Simplifying: 2412i110i1+10i2=024 - 12i_1 - 10i_1 + 10i_2 = 0 24=22i110i2(Equation 1)24 = 22i_1 - 10i_2 \quad \text{(Equation 1)}

Loop 2 (Mesh 2: containing i2i_2)

For the second loop:

  • The voltage drop across the 10Ω resistor: 10(i2i1)10(i_2 - i_1).
  • The voltage drop across the 4Ω resistor: 4(i2i3)4(i_2 - i_3).
  • The voltage drop across the 24Ω resistor: 24i224i_2.

Applying KVL to this loop: 10(i2i1)+4(i2i3)+24i2=010(i_2 - i_1) + 4(i_2 - i_3) + 24i_2 = 0 Simplifying: 10i210i1+4i24i3+24i2=010i_2 - 10i_1 + 4i_2 - 4i_3 + 24i_2 = 0 (10+4+24)i210i14i3=0(10 + 4 + 24)i_2 - 10i_1 - 4i_3 = 0 38i210i14i3=0(Equation 2)38i_2 - 10i_1 - 4i_3 = 0 \quad \text{(Equation 2)}

Loop 3 (Mesh 3: containing i3i_3)

For the third loop, we have the dependent current source, which simplifies the analysis.

  • The voltage drop across the 4Ω resistor: 4(i3i2)4(i_3 - i_2).
  • The voltage of the dependent source is given as 4i04i_0, where i0=i2i3i_0 = i_2 - i_3, so the voltage is 4(i2i3)4(i_2 - i_3).

Applying KVL: 4(i3i2)4(i2i3)=04(i_3 - i_2) - 4(i_2 - i_3) = 0 Simplifying: 4i34i24i2+4i3=04i_3 - 4i_2 - 4i_2 + 4i_3 = 0 8i38i2=08i_3 - 8i_2 = 0 i3=i2(Equation 3)i_3 = i_2 \quad \text{(Equation 3)}

Step 2: Solve the system of equations

We now have three equations:

  1. 24=22i110i224 = 22i_1 - 10i_2
  2. 38i210i14i3=038i_2 - 10i_1 - 4i_3 = 0
  3. i3=i2i_3 = i_2

We can substitute i3=i2i_3 = i_2 into Equation 2 to reduce the system and solve for i1i_1 and i2i_2.

Let me calculate this for you.The loop currents are:

  • i1=3427A1.26Ai_1 = \frac{34}{27} \, \text{A} \approx 1.26 \, \text{A}
  • i2=1027A0.37Ai_2 = \frac{10}{27} \, \text{A} \approx 0.37 \, \text{A}
  • i3=1027A0.37Ai_3 = \frac{10}{27} \, \text{A} \approx 0.37 \, \text{A}

Step 3: Dependent source calculation

The value of the dependent voltage source is 4i04i_0, where i0=i2i3i_0 = i_2 - i_3. Since i2=i3i_2 = i_3, the current i0=0i_0 = 0, and hence the voltage of the dependent source is: 4i0=4×0=0V4i_0 = 4 \times 0 = 0 \, \text{V}

Thus, the dependent voltage source contributes 0 V to the circuit.

Would you like further details on the solution? Here are some additional questions to expand on this:

  1. How would the mesh equations change if the dependent source had a different multiplier?
  2. Can mesh analysis be used with circuits that contain inductors or capacitors?
  3. What is the impact of changing the 24Ω resistor on the mesh currents?
  4. How would the circuit change if we replaced the dependent source with an independent voltage source?
  5. How is mesh analysis different from nodal analysis?

Tip: Always check if dependent sources introduce additional constraints based on the relationship given for the controlling variable.

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Math Problem Analysis

Mathematical Concepts

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

Formulas

KVL Equation: ΣV = 0
Ohm's Law: V = IR
Dependent source relation: i0 = i2 - i3

Theorems

Kirchhoff's Voltage Law

Suitable Grade Level

Undergraduate Electrical Engineering