Why Is My Calculation of i1 Using KCL and Ohm's Law Incorrect?

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SUMMARY

The discussion focuses on the calculation of current i1 using Kirchhoff's Current Law (KCL) and Ohm's Law in a circuit analysis problem. The user incorrectly assigned signs to potential changes in their equations for V1, leading to confusion in determining the correct value of i1. Specifically, the potential drop across the 7Ω resistor and the effects of the voltage sources were misrepresented. Correctly applying KCL and understanding the direction of current flow are crucial for accurate circuit analysis.

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  • Proficiency in Ohm's Law (V=IR)
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Students studying electrical engineering, circuit designers, and anyone looking to improve their skills in circuit analysis and problem-solving using KCL and Ohm's Law.

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Homework Statement


12.jpg


Find i1

Homework Equations


KCL
V=IR

The Attempt at a Solution


See picture. What am I doing wrong?

Also, the current through 7 ohms = 2+i1.
So for V1 (left loop), can it be written as V1=-5-0.5i1+7(2+i1)?
Then V1=-5-0.5i1+14+7i1=9+6.5i1
For V1 (right loop), can it be written as V1=5-10i1?

Many thanks in advance!
 
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xlu2 said:

Homework Statement


View attachment 59056

Find i1

Homework Equations


KCL
V=IR


The Attempt at a Solution


See picture. What am I doing wrong?
I don't understand the node equation in your diagram; Why doesn't V1 appear in it? V1 should be the unknown potential at the node that determines the currents in the branches.
Also, the current through 7 ohms = 2+i1.
So for V1 (left loop), can it be written as V1=-5-0.5i1+7(2+i1)?
It doesn't look like you've assigned the correct signs to the potential changes with respect to the reference node. i1+2 flowing through the 7Ω resistor in the direction indicated should cause a potential drop, but the 5V source must cause a +5V potential rise. Similarly, the controlled source makes another potential rise of 0.5*i1 on the way from the reference node to V1.
Then V1=-5-0.5i1+14+7i1=9+6.5i1
For V1 (right loop), can it be written as V1=5-10i1?
Nope. Pay attention to the direction with which i1 passes through the 10Ω resistor. What's the resulting polarity of the potential change?
 

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