KVL & Node Voltages: Verifying Ans. to I

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SUMMARY

The discussion focuses on verifying the current I in a circuit using Kirchhoff's Voltage Law (KVL) and node voltage analysis. The user initially calculated I as -4 Amps and sought confirmation through KVL in a specific mesh. The response emphasized the potential for using Kirchhoff's Current Law (KCL) instead, suggesting that solving node equations for voltage V1 could directly yield the current as V1/8. Additionally, it highlighted the importance of assigning variables to unknown voltage drops across current sources to facilitate the application of KVL.

PREREQUISITES
  • Understanding of Kirchhoff's Voltage Law (KVL)
  • Familiarity with Kirchhoff's Current Law (KCL)
  • Basic knowledge of circuit analysis techniques
  • Ability to solve simultaneous equations
NEXT STEPS
  • Learn how to apply KCL in circuit analysis for current sources
  • Study the method of node voltage analysis in detail
  • Explore the implications of assigning voltage drop variables in KVL
  • Practice solving circuits with both KVL and KCL to reinforce understanding
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Electrical engineering students, circuit designers, and anyone involved in analyzing electrical circuits using KVL and KCL methods.

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Homework Statement


[PLAIN]http://img163.imageshack.us/img163/408/circuit2w.jpg
We are asked to solve for I. I found I = -4 Amps.
I want to verify my answer using KVL in the bottom left mesh

Homework Equations


KVL

The Attempt at a Solution


I follow the convention: voltage drops are positive, which means that the first sign I hit is the sign of the voltage. Starting at the 8 ohm resistor:

- 8*(-4) + ( ? ) - ( ?) = 0

I am confused because we usually apply KVL using the voltage drops/rise between elements. For the current source and the 2 ohm resistor, I have the node voltages between these elements. Therefore, I don't know if the current source is + (v2 - v1) or - (v2 - v1). Same thing for the 2 ohm resistor, I hit the - sign first, but is it - (v2 -0) or - (0-v2) ?

I am a little bit confused.. thank you very much
 
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Since your circuit has only current sources in it, did you consider using KCL node equations instead of KVL loops? If you solved the node equations for the node V1, then the current you're looking for is just V1/8.

If you insist on using KVL loops, then one way to approach current sources is to assign variables to the unknown voltage drops across them (choose whichever polarity you want, the math will take care of the details) and then use the current sources to place constraints on the relationships between the loop currents. This gives you the extra simultaneous equations you need to solve for the variables you introduced.

So suppose that the top loop is Loop2 with current I2 flowing around it. Loop1 is your loop with current I as shown. Both currents are assumed to be running counterclockwise around their loops. Then a suitable constraint equation as imposed by the 4A current source would be I2 - I = 4A.
 

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