Question about a 'basic' equivalent resistive circuit problem

In summary, the person is trying to solve a problem, and is grateful for a suggestion. They try a different approach and are able to solve the problem.
  • #1
ajovcu
4
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:
 

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  • #2
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?
 
  • #3
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
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
 
  • #5
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
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  • #6
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.
 

FAQ: Question about a 'basic' equivalent resistive circuit problem

What is a 'basic' equivalent resistive circuit?

A 'basic' equivalent resistive circuit is a simplified representation of a more complex circuit that has the same overall behavior. It is made up of resistors that are connected in a way that produces the same total resistance as the original circuit.

How do I simplify a circuit to its equivalent resistive circuit?

To simplify a circuit to its equivalent resistive circuit, you need to identify resistors that are connected in series or parallel. In series, resistors are connected end-to-end and their resistances add up. In parallel, resistors are connected side-by-side and their resistances decrease. Once you have identified these connections, you can use the appropriate equations to calculate the equivalent resistance.

What is the formula for calculating equivalent resistance in a series circuit?

The formula for calculating equivalent resistance in a series circuit is simply adding the individual resistances together. For example, if you have three resistors with values of 10 ohms, 20 ohms, and 30 ohms, the equivalent resistance would be 10 + 20 + 30 = 60 ohms.

What is the formula for calculating equivalent resistance in a parallel circuit?

The formula for calculating equivalent resistance in a parallel circuit is slightly more complex. It is the reciprocal of the sum of the reciprocals of each individual resistance. In other words, 1/Req = 1/R1 + 1/R2 + 1/R3 + ... + 1/Rn.

How can I use equivalent resistive circuits to simplify calculations?

Equivalent resistive circuits are useful because they allow you to simplify complex circuits into easier-to-calculate versions. You can use them to calculate total resistance, current, voltage, and power within a circuit. They also make it easier to analyze and understand the behavior of a circuit as a whole.

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