Question about a 'basic' equivalent resistive circuit problem

Click For Summary

Discussion Overview

The discussion revolves around a circuit problem where participants are trying to determine the appropriate resistance value (R) to achieve a specific voltage (VL = 5 V) across a component in the circuit. The conversation includes attempts at using different circuit analysis techniques and clarifications regarding circuit behavior.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • A new member presents a circuit problem and expresses frustration with solving it.
  • One participant suggests using Kirchhoff's Current Law (KCL) as a preferred method for analysis, indicating it is more intuitive than other methods like source transformations.
  • A participant corrects a previous post by clarifying the units in a KVL expression related to the circuit analysis.
  • Another participant points out that the circuit contains a voltage divider and notes that a specific resistor will not affect the output voltage due to no current flowing through it.
  • A participant expresses gratitude for the clarification regarding the circuit's open nodes and their interpretation, indicating a struggle with the problem prior to the discussion.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus on the best approach to solve the problem, as multiple methods are suggested and discussed. The discussion remains unresolved regarding the final solution for R.

Contextual Notes

There are indications of missing assumptions regarding the circuit configuration and the role of certain components, as well as unresolved mathematical steps in the analysis.

ajovcu
Messages
4
Reaction score
0
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:
 

Attachments

Physics news on Phys.org
ajovcu said:
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?
 
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
 
berkeman said:
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
 
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:
 

Attachments

  • Fig1.gif
    Fig1.gif
    4 KB · Views: 850
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.
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
6K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 16 ·
Replies
16
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 5 ·
Replies
5
Views
4K
Replies
10
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 31 ·
2
Replies
31
Views
5K