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(University of Calgary)
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posted by  R.Ach on 10/12/2008 2:29:54 AM  |  status: Closed  

Really Simple Mesh Analysis Problem. Will rate livesaver!

Course Textbook Chapter Problem
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Question Details:
This circuit is very easy...except I'm getting two different answers with two different methods of approaching it.

The problem is attahced - the numbers beside the resistors without units are the resistance

And here's what I want to know..

I approached the problem without combining the two independent current sources and I got it wrong doing mesh analysis. I know how to get the answer after you combine the two current sources

So the question is..

A. Is there something wrong with approaching the problem without combining the two current sources?

or B. How do you do it without combining the 1 A and -4 A current sources?

Thanks!

BTW: the answer is the current through Vr is 4.5 A


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Karma Points: 13,692
posted by Shree_vani on 10/12/2008 6:01:00 AM  |  status: Live
Asker's Rating: Lifesaver   
R.Ach's comment:
"Thank you! I was forgetting to subtract the 4 A....you saved me hours of unproductive staring at the question!"
Response Details:

 There is nothing in solving the problem with out combining the current sources

 
I1 = 1A
I2 = 4A
Applying KVL to the loop
5VR = 12(I-I1 ) + VR
4VR = 12(I - 1)
But
VR = 5(I - I2) = 5(I-4)
20 ( I - 4) = 12 (I - 1)
20I - 12I = -12 + 80
8I = 68
I = 8.5A
 
Current through 5Ω reisitor or through VR is = I - 4 = 8.5-4 = 4.5A
 
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