Nodal Analysis w/ Controlled Sources

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SUMMARY

The discussion focuses on solving a nodal analysis problem involving controlled sources, specifically finding voltages V1, V2, and V3. Key equations utilized include KCL (Kirchhoff's Current Law) and KVL (Kirchhoff's Voltage Law), along with supernode equations. The equations derived from the analysis include Ic = v2(-G1) + v3(-G3) and the supernode equation -Hm*IC + V3 - V2 = 0. The confusion arises from the setup of these equations, particularly the relationship between the currents and voltages at the nodes.

PREREQUISITES
  • Nodal Analysis techniques
  • Understanding of Kirchhoff's Current Law (KCL)
  • Familiarity with Kirchhoff's Voltage Law (KVL)
  • Basic knowledge of controlled sources in circuit analysis
NEXT STEPS
  • Study the application of supernode analysis in circuit problems
  • Learn about the implications of controlled sources in nodal analysis
  • Explore advanced techniques for solving complex nodal equations
  • Review examples of KCL and KVL applications in circuit design
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Electrical engineering students, circuit analysts, and anyone involved in solving circuit problems using nodal analysis techniques.

bob29
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Homework Statement


Find V1, V2 and V3.

Homework Equations


Nodal Analysis
Supernode Equation
Capacitance = 1/R
KCL & KVL

The Attempt at a Solution


KCL @ node 0: Ic = v2(-G1) +v3(-G3)
Node 1: Ic + I1 +v3(G2+G3) +V2(-G3) = 0 (subbing in Ic)
v2(G1+G3) + v3(g-3) = I1....Equation (1)

Node 2: v2(G1+G3) +v3(-G3) - I1 = 0...Equation (2)

Node 3: Same as equation 1

KVL @ supernode: -Hm*IC +V3-V2=0

I'm confused on how the equations were setup in the image.
 

Attachments

  • Node Voltage Problem.png
    Node Voltage Problem.png
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The first equation in the image is the same as what you've called equation 2. It's just the application of KCL to node 2.

The second equation in the image is a constraint equation; it's setting V1 equal to zero.

The third equation in the image is found like this:

The current in the Hm*Ic source is equal to the sum of the currents in R3 and R2, and is equal to G3*(v3-V2) + G2(V3).

Now, over at node 1, we have Ic + I1 + G3*(v3-V2) + G2(V3) = 0 or:

-Ic = G3*(v3-V2) + G2(V3) + I1

Equating the voltage across the Hm*Ic source we have V3 - V1 = Hm*Ic

giving V3 - V1 = -Hm*(G3*(v3-V2) + G2(V3) + I1)

or V3 - V1 + Hm*(G3*(v3-V2) + G2(V3) = -I1*Hm

Rearranging gives equation 3 in the image.
 

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