Math Problem Statement

solve using mesh

solve using mesh

Solution

To solve this circuit using mesh analysis, we will follow these steps:

  1. Label the mesh currents:

    • Assume the mesh currents as I1I_1 (left loop) and I2I_2 (right loop), flowing clockwise in both loops.
  2. Write mesh equations using Kirchhoff's Voltage Law (KVL):

    • Left loop: V1I1R1I3R3=0V_1 - I_1 R_1 - I_3 R_3 = 0
    • Right loop: I2R2+V2I3R3=0-I_2 R_2 + V_2 - I_3 R_3 = 0
    • The current through R3R_3 is I3=I1I2I_3 = I_1 - I_2.
  3. Solve the equations for I1I_1 and I2I_2.

Step 1: Circuit Parameters

  • V1=30V,V2=10VV_1 = 30 \, \text{V}, V_2 = 10 \, \text{V}
  • R1=1.2kΩ=1200ΩR_1 = 1.2 \, \text{k}\Omega = 1200 \, \Omega
  • R2=800ΩR_2 = 800 \, \Omega
  • R3=120ΩR_3 = 120 \, \Omega

Step 2: Write Mesh Equations

Left Loop: V1I1R1(I1I2)R3=0V_1 - I_1 R_1 - (I_1 - I_2) R_3 = 0 Substitute the values: 301200I1120(I1I2)=030 - 1200 I_1 - 120 (I_1 - I_2) = 0 Simplify: 301200I1120I1+120I2=030 - 1200 I_1 - 120 I_1 + 120 I_2 = 0 30 - 1320 I_1 + 120 I_2 = 0 \tag{1}

Right Loop: I2R2+V2(I1I2)R3=0-I_2 R_2 + V_2 - (I_1 - I_2) R_3 = 0 Substitute the values: 800I2+10120(I1I2)=0-800 I_2 + 10 - 120 (I_1 - I_2) = 0 Simplify: 800I2+10120I1+120I2=0-800 I_2 + 10 - 120 I_1 + 120 I_2 = 0 10 - 120 I_1 - 680 I_2 = 0 \tag{2}

Step 3: Solve the Equations

From equations (1) and (2):

  1. 1320I1120I2=301320 I_1 - 120 I_2 = 30
  2. 120I1680I2=10-120 I_1 - 680 I_2 = -10

Solve this system using substitution or elimination:

  1. Multiply equation (1) by 680: 897600I181600I2=20400897600 I_1 - 81600 I_2 = 20400
  2. Multiply equation (2) by 1320: 158400I1897600I2=13200-158400 I_1 - 897600 I_2 = -13200 Add the equations: 739200I1979200I2=7200739200 I_1 - 979200 I_2 = 7200 I2=7200979200=0.00735A(or 7.35mA)I_2 = \frac{7200}{979200} = 0.00735 \, \text{A} \, (\text{or } 7.35 \, \text{mA})

Substitute I2I_2 into equation (1): 1320I1120(0.00735)=301320 I_1 - 120 (0.00735) = 30 1320I10.882=301320 I_1 - 0.882 = 30 1320I1=30.8821320 I_1 = 30.882 I1=30.8821320=0.0234A(or 23.4mA)I_1 = \frac{30.882}{1320} = 0.0234 \, \text{A} \, (\text{or } 23.4 \, \text{mA})

Final Results

  • I1=23.4mAI_1 = 23.4 \, \text{mA}
  • I2=7.35mAI_2 = 7.35 \, \text{mA}
  • I3=I1I2=23.4mA7.35mA=16.05mAI_3 = I_1 - I_2 = 23.4 \, \text{mA} - 7.35 \, \text{mA} = 16.05 \, \text{mA}

Let me know if you’d like further clarifications or a breakdown of any step!


Related Questions:

  1. What happens if R3R_3 is removed from the circuit?
  2. How does the voltage across R3R_3 change if V2V_2 increases?
  3. What is the power dissipated in R2R_2?
  4. How does mesh analysis compare to nodal analysis for this circuit?
  5. Can this circuit be solved using Thevenin's theorem?

Tip: Always double-check units when converting resistances or currents; small mistakes can propagate through calculations.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Mesh Analysis
Kirchhoff's Voltage Law (KVL)
Electrical Circuits

Formulas

Kirchhoff's Voltage Law (KVL): ΣVoltage around a loop = 0
Current through R3: I3 = I1 - I2

Theorems

Kirchhoff's Voltage Law

Suitable Grade Level

College-level Electrical Engineering