Find current using KVL (with dependent source)

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SUMMARY

The discussion focuses on solving a circuit problem using Kirchhoff's Voltage Law (KVL) with a dependent source. The user initially calculated the current I_x as 2 mA, but the correct value is 1.5 mA. The error was identified in the conversion of I_x from milliamperes to amperes, leading to the adjustment of the equation to x/1000 = -c. The correct KVL equations were confirmed as a(6000) + (a-b)(2000) + 5x - 10=0, -5x + (b-a)(2000) + b(3000) + (b-c)(4000) = 0, and 10 + (c-b)(4000) + c(2000) = 0.

PREREQUISITES
  • Understanding of Kirchhoff's Voltage Law (KVL)
  • Basic knowledge of circuit analysis
  • Familiarity with dependent sources in electrical circuits
  • Ability to manipulate equations involving current and voltage
NEXT STEPS
  • Review the principles of Kirchhoff's Voltage Law in detail
  • Explore dependent sources and their applications in circuit analysis
  • Practice solving circuit problems involving multiple loops
  • Learn about converting units in electrical calculations, specifically between milliamperes and amperes
USEFUL FOR

Electrical engineering students, circuit analysts, and anyone involved in solving complex circuit problems using KVL.

iharuyuki
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Homework Statement


upload_2015-4-15_22-24-13.png


Homework Equations


V = IR
Kirchoff's Voltage Law

The Attempt at a Solution


[/B]
Top loop - a
Bottom right - b
Bottom left - c
Ix = x

KVL:
top: a(6000) + (a-b)(2000) + 5x - 10=0
bottom right: -5x + (b-a)(2000) + b(3000) + (b-c)(4000) = 0
bottom left : 10 + (c-b)(4000) + c(2000) = 0
x = -c

x comes out to 2 mA when it should be 1.5. Where did I make a mistake?

Thank you very much.
 
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How did you account for the given detail that ##I_x## is in mA?
 
got it! That was the error. The proper equations were this:

a(6000) + (a-b)(2000) + 5x - 10=0

-5x + (b-a)(2000) + b(3000) + (b-c)(4000) = 0

10 + (c-b)(4000) + c(2000) = 0

x/1000 = -c

Thank you very much!
 

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