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Question about a 'basic' equivalent resistive circuit problem

  • Engineering
  • Thread starter ajovcu
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  • #1
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Hello all, I'm a new member. I've decided banging my forehead until its red will not make me any more able to solve my practice problems. I hope I haven't posted this in the wrong forum but here is the question in concern:

Select R in the circuit so that VL= 5 V

(see circuit in attachment below)




Hopefully you will be able to read this, I wrote it in OneNote- here is my attempt at a solution:
 

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Answers and Replies

  • #2
berkeman
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Hello all, I'm a new member. I've decided banging my forehead until its red will not make me any more able to solve my practice problems. I hope I haven't posted this in the wrong forum but here is the question in concern:

Select R in the circuit so that VL= 5 V

(see circuit in attachment below)




Hopefully you will be able to read this, I wrote it in OneNote- here is my attempt at a solution:
Welcome to the PF.

I would try using the KCL on the original circuit. I usually find that KCL works best for me, and is more intuitive than using source transformations, etc.

Can you give the KCL approach a try?
 
  • #3
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ay carumba, sorry for not posting the units in that last KVL expression... should've read something like:

-50/(7+R) V + (17 + 3R)/(7 + R)kOhm * ImA + 5 V = 0
 
  • #4
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Welcome to the PF.

I would try using the KCL on the original circuit. I usually find that KCL works best for me, and is more intuitive than using source transformations, etc.

Can you give the KCL approach a try?
absolutely!, I appreciate the suggestion. I'll post back my results
 
  • #5
gneill
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20,781
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If you take another look at your second diagram (after the first source transformation) you might notice that there is a clear example of a simple voltage divider. The 1K resistor will not affect the output voltage, as no current will flow through it.

attachment.php?attachmentid=37756&stc=1&d=1312428494.gif


25V source -- voltage divider -- 5V out. Should be able to solve for R with little pain :smile:
 

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  • #6
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wow thanks a bunch gneil, I assumed that those open nodes were there to represent some device or external circuit that absorbs 5V and could be even represented by a voltage source of 5V. Thanks a lot, I'd literally been racking my brains over the last day and a half on this problem.
 

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