What Is the Correct Approach to Solve This Loop Analysis Problem?

AI Thread Summary
The discussion revolves around solving a loop analysis problem involving multiple currents and voltage calculations. The user initially set up equations for the loops but struggled with verifying the results, particularly the polarity of Vx and the direction of current i2. After several attempts and corrections, including recognizing a mistake in the equation i1-i2=2mA, the user recalculated and found a new value for i2. Ultimately, the user expressed gratitude for the assistance received in resolving the confusion and arriving at a more accurate solution. The conversation highlights the importance of careful equation setup and verification in circuit analysis.
ee_ess_ee-kid
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Homework Statement



http://i.imgur.com/pVIMw.png"

Homework Equations


The Attempt at a Solution



I called the top left loop i1, the top right loop i2, the bottom right loop i3, and the bottom left loop i4. I have all of these currents going clockwise in my work. An attempt to do loop analysis has left me with the following equations:

i4=(2Vx)/(1K)
i1-i2=2mA
i3=-4mA
-i2(4kΩ)=Vx

With this, I used KVL along the top loop going clcokwise, which got me:

-2(i1-i4) - 4i1 - 4i2 - 1i2 - 8(i2-i3) = 0
Solving this got me:
i2=(-25/44)mA

Then, V0=(8kΩ)(-25/44 mA)=-1.76V

Can someone please help me figure it out? When I go and try to check it, my work gets pretty weird...Pretty sure this is wrong. Can someone please help me find my mistake?
 
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Verify the polarity of Vx.
 
I was confused by that.

Should it be i2(4κΩ)=Vx?

Redoing my work gets me V0=11V, which looks better. I'll check to see if that answer makes more sense
 
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Current through a resistor causes a voltage drop (+ to -) in the direction that the current flows. What's the assumed direction of i2?
 
gneill said:
Current through a resistor causes a voltage drop (+ to -) in the direction that the current flows. What's the assumed direction of i2?

I have all the currents going clockwise.

I have i2 going from left to right on that resistor.
 
ee_ess_ee-kid said:
I have all the currents going clockwise.

I have i2 going from left to right on that resistor.

So then...
 
gneill said:
So then...

So then it should be Vx=i2(4κΩ)
 
ee_ess_ee-kid said:
So then it should be Vx=i2(4κΩ)

Correct.
 
If I use that, I get the following:

i1=13mA
i2=11mA
i3=-4mA
i4=88mA

Using these values, I fail to get a correct check using KVL on the top loop. I can't get ƩV=0...
 
  • #10
ee_ess_ee-kid said:
If I use that, I get the following:

i1=13mA
i2=11mA
i3=-4mA
i4=88mA

Using these values, I fail to get a correct check using KVL on the top loop. I can't get ƩV=0...

I get a different value for i2 (I didn't check the other currents) when I solve your equations. You'll have to post your work in detail if we're to see what's happened.
 
  • #11
gneill said:
I get a different value for i2 (I didn't check the other currents) when I solve your equations. You'll have to post your work in detail if we're to see what's happened.

Actually, I found it myself. Turns out I did the following:

i1-i2=2mA → i1=2-i2

Which is wrong :p

Fixing this error, I got a value of i2=-14.6667=-44/3, if i2 were going clockwise.
 
  • #12
Thanks for all of your help, I really appreciate it!
 
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