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Solve for I through Loop Analysis, Will give good review

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Answers:

Michaelh

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Karma Points: 15,786

Date Posted: 6/11/2008 3:12:12 AM Status: Live

Asker's Rating: Lifesaver

Response:

Let the current in the upper left loop be I1
Let the current in the upper middle loop be I2
Let the current in the upper right loop be I3
Let the current in the lower left loop be I4
Let the current in the lower middle loop be I5
Let the current in the lower right loop be I6

Because we have a current source between the first and second loop, we cannot perform loop analysis on an individual basis. We need to do a super-loop, which would include both loops.

Therefore, for the super-loop on top, we have:

1(I1) + 1(I2 - I3) + 12 + i(I1 -I4) = 0
So, we have:      2I1 + I2 - I3 -I4 = -12 ..................... (1)

For the third loop:

1(I3 - I2) + I3 + 1(I3 - I6) = 0
So, we have:      -2I2 + 3I3 - I6 = 0 ..................... (2)

For the fourth loop:

1I4 + 1(I4 - I1) + 1(I4 - I5) = 0
So, we have:      -I1 + 3I4 - 1I5 = 0 ........................... (3)

We cannot do a loop analysis for the fifth and sixth loop because of the current source in the middle. We need to perform a super-loop and combine them both:

So, we have:
1(I5 - I4) - 12 + 1(I6 - I3) + 1I6 = 0
This yields:      -I3 -1I4 + 1I5 + 2I6 = 12 .........................(4)

Then, from KCL on the top:

I1 + 4mA - I2 = 0      =>         I1 - I2 = -4 ....................... (5)

From KCL on the bottom:

I6 + 2mA - I5 = 0         =>      -I5 + I6 = -2 ..................... (6)

Now, you have 6 equations with 6 unknown. I don't have my calculator with me, but if you plug in the numbers you should get the answer. I hope I didn't make any mistakes. Its 3:00AM and I can barely keep my eyes open :D

By the way, I0 is I4 - I5

MarbeL's Comment:

Thanks for the help :)

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Solve for I through Loop Analysis, Will give good review

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