How to do Mesh Analysis with current controlled voltage source

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SUMMARY

This discussion focuses on performing mesh analysis with a controlled voltage source represented by Vs = 100cos(2000t). The participant outlines their mesh equations using Kirchhoff's Voltage Law (KVL) and identifies discrepancies between their mesh analysis results and those obtained through nodal analysis. The corrected mesh equation incorporates the potential drop across the inductor due to mesh current I1. The final computed values for currents I1 and I2 are (1.93736+0.385512i) and (1.83193+0.694348i), respectively, leading to a voltage V(120j) of 37.06+j12.65 V, which contrasts with the nodal analysis result of 2.95+j1.126.

PREREQUISITES
  • Understanding of mesh analysis in electrical circuits
  • Familiarity with Kirchhoff's Voltage Law (KVL) and Kirchhoff's Current Law (KCL)
  • Knowledge of complex numbers in electrical engineering
  • Experience with controlled voltage sources in circuit analysis
NEXT STEPS
  • Study advanced mesh analysis techniques for circuits with inductors
  • Learn about the differences between mesh and nodal analysis
  • Explore the application of phasors in AC circuit analysis
  • Investigate the impact of reactive components on circuit behavior
USEFUL FOR

Electrical engineering students, circuit designers, and professionals involved in circuit analysis and design, particularly those focusing on AC circuits and complex impedance.

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Homework Statement


Vs = 100cos(2000t)
upload_2016-11-7_21-11-57.png


Homework Equations


KVL, KCL

The Attempt at a Solution


My mesh equations are:
-Vs + I1(30-12.5j)-I2(120j)=0
4Ix+I2(120j+20)-I1(120j)=0
Ix = (I1-I2)

Are these correct? When I solve I get a different result than when I do nodal analysis.
 
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Your first mesh equation is missing a term for the potential drop over the inductor due to mesh current ##I1##.
 
gneill said:
Your first mesh equation is missing a term for the potential drop over the inductor due to mesh current ##I1##.
-Vs + I1(30-12.5j+120j)-I2(120j)=0

i1 = (1.93736+0.385512i)
i2 = (1.83193+0.694348i)

ix = (1.93736+0.385512i)-(1.83193+0.694348i)
V(120j) = ix*120j = 37.06+j12.65 V
The solution (using nodal analysis) says V(120j) = 2.95+j1.126
 

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