How Do You Solve a Circuit with a Voltage Dependent Current Source?

AI Thread Summary
To solve a circuit with a voltage-dependent current source, KCL (Kirchhoff's Current Law) can be effectively applied. The user attempted to set up KCL equations at nodes Va and Vb but expressed uncertainty about sign conventions and whether to use KVL (Kirchhoff's Voltage Law) or mesh analysis as well. It was suggested that the user clarify the relationships between voltages, noting that Va equals Vx and Vb equals VL. Attention to sign conventions is crucial, with a recommendation to consistently sum currents either entering or leaving a node. Properly applying these principles should help in solving the circuit effectively.
morrison.344
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Voltage dependent current source? Please help

Homework Statement


Attached the problem.


Homework Equations


??


The Attempt at a Solution


Tried to Use KCL, not sure what else to do.
 

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morrison.344 said:

Homework Statement


Attached the problem.


Homework Equations


??


The Attempt at a Solution


Tried to Use KCL, not sure what else to do.

Welcome to the PF. Show us your work so far. Show us your KCL equations.
 


I tried KCL at Va: (Vs-Va)/Rs+(Vb-Va)/100000+(Vx-Va)/2500=0
and KCL at Vb: (VL-VB)/1000-.05Vx+(Va-Vb)/100000=0,
So i have 2 equations and 4 variables assuming my work so far is correct, I am not positive on the sign conventions. Should I be using KVL, or Mesh analysis instead or in addition?
 


morrison.344 said:
I tried KCL at Va: (Vs-Va)/Rs+(Vb-Va)/100000+(Vx-Va)/2500=0
and KCL at Vb: (VL-VB)/1000-.05Vx+(Va-Vb)/100000=0,
So i have 2 equations and 4 variables assuming my work so far is correct, I am not positive on the sign conventions. Should I be using KVL, or Mesh analysis instead or in addition?

KCL should work fine. Look at the circuit again, and see that Va = Vx, and Vb = VL.

Also, be careful with your sign conventions -- either sum the currents leaving a node or going into it. I believe you were mixing signs in the first line (but the fact that Va = Vx kind of muddies that sign error anyway). I personally prefer to sum the currents leaving each node, but that's just me.
 
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