Solve KVL and KCL for V0: Find Voltage Using Kirchhoff's Laws

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SUMMARY

The discussion focuses on solving for the voltage V0 using Kirchhoff's Voltage Law (KVL) and Kirchhoff's Current Law (KCL). The equations used include KVL: ƩV=0 and KCL: iin=iout. The user provided specific calculations for currents i1, i2, and i3, arriving at values of 0.45mA, 0.7mA, and -0.2mA respectively. The discussion emphasizes the importance of using sufficient KVL loops to cover each component and correctly applying the laws to derive the necessary equations.

PREREQUISITES
  • Understanding of Kirchhoff's Voltage Law (KVL)
  • Understanding of Kirchhoff's Current Law (KCL)
  • Basic circuit analysis techniques
  • Familiarity with Ohm's Law and current calculations
NEXT STEPS
  • Practice solving circuit problems using KVL and KCL
  • Explore advanced circuit analysis techniques such as nodal and mesh analysis
  • Learn about the impact of component values on circuit behavior
  • Investigate the use of simulation tools like LTspice for circuit verification
USEFUL FOR

Students studying electrical engineering, circuit designers, and anyone looking to deepen their understanding of circuit analysis using Kirchhoff's laws.

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Homework Statement



Use Kirchhoff's voltage and current laws to find the voltage V0

Homework Equations



KVL: ƩV=0
KCL: iin=iout

The Attempt at a Solution



http://s2.postimg.org/6b2013cbt/001.jpg

If you guys want the original circuit diagram, i'll post it.
Note: The current flowing through i2 is -0.2mA. Forgot to put the minus sign.
 
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You only need enough KVL loops to "touch" each component once. Any additional loops will just be dependent upon the one's you've already written. So you can stop with the first two loops equations that you wrote (I and II). That plus your KCL equation are sufficient.

Replace VA in equation II with i1*4k. Solve for i3.
 
I did the following:
KCL at a: i1-i2-i3=0
KVL at I: 4ki1+6ki2=6
KVL at II: 16ki1-6ki2+12ki3=0 (VA=4ki1, 4VA=16ki1)
The answers i got:
i1=0.45mA
i2=0.7mA
i3=-0.2mA
 
I'm seeing -(1/4) mA for i3. Maybe check your math?
 
Maybe yeah. The calculator is showing only one significant digit.
 
Last edited:

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