# Simple DC circuit with battery and resistors

Discussion in 'Homework Help' started by epsilonjon, May 30, 2011.

1. ### epsilonjon Thread Starter Member

Feb 15, 2011
65
1
Question:

"The resistance R3 in the diagram is the equivalent resistance of a pressure transducer. This resistance is specified to be 200Ω ± 5%. The voltage source is a 12V ± 1% source capable of supplying 5W. Design this circuit, using 5%, 1/8W resistors for R1 and R2, so that the voltage across R3 is Vo=4V±10%."

Attempt:

I'm ignoring the error margins for the moment and just working with the specified values.

The voltage drop accross R2 must be 8V, therefore I*R2=8. Each resistor can only dissipate 1/8W, so

$\frac{8^{2}}{R_{2}}\leq\frac{1}{8}$

$R_{2}\geq512$.

The equivalent resistance of the circuit is

$R_{T}=R_{2}+\frac{R_{1}R_{3}}{R_{1}+R_{3}} = R_{2} + \frac{200R_{1}}{R_{1}+200}$.

Therefore

$12=IR_{T}$

$12=\frac{8}{R_{2}}(R_{2}+\frac{200R_{1}}{R_{1}+200})$

$12R_{2}=8R_{2}+\frac{1600R_{1}}{R_{1}+200}$

$R_{2}=\frac{400R_{1}}{R_{1}+200}$

$R_{1}+200=\frac{400R_{1}}{R_{2}}$.

So if R2>512Ω (required by the power restrictions), you are bound to require a negative value for R1. Am I doing something silly here, or does the question not work?

Thanks for any help!

Jon.

Last edited: May 30, 2011

Feb 17, 2009
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3. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,511
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For R3= 210Ω and Vo = 3.6V and 4.4V for no load
We have Ith = 3.6V/210Ω = 17.14mA
and Rth = 0.8V/17.14mA = 47Ω
Rth = R1||R2 = (R1*R2) / (R1+R2) = 47Ω
Vth = 12V * R1/(R1+R2) = 4.4V
R2/R1 = 1.72
R2 = 128Ω and R1 = 74Ω
But then the power is to big.

So i really don't know how we can meet all those requirements.

4. ### epsilonjon Thread Starter Member

Feb 15, 2011
65
1

I'm glad it's not me misunderstanding things, but rather the question which is poorly conceived. I spent £50 on this textbook too!

5. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,511
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What is the name of the text book?

6. ### epsilonjon Thread Starter Member

Feb 15, 2011
65
1
Introduction to Electric Circuits by Richard C. Dorf and James A. Svoboda.

7. ### jegues Well-Known Member

Sep 13, 2010
735
45
That text has alot of goofy problems in it.

The answers to the problems are usually horrid. (Really retarded numbers for what could be straightforward and simple computation)

8. ### TBayBoy Member

May 25, 2011
148
19
you should be able to look up how the answer was derived here

<SNIP>

just put in the chapter and problem number.

Last edited by a moderator: Jun 1, 2011
9. ### Jony130 AAC Fanatic!

Feb 17, 2009
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The answer to this problem is to using two 200Ω resistors connected in series instead of R2. And R1 treat as a open circuit.

10. ### epsilonjon Thread Starter Member

Feb 15, 2011
65
1
Unfortunately you have to sign up, and it costs money per month :-(

Last edited by a moderator: Jun 1, 2011
11. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,511
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I saw the solution. Read my previous post.

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12. ### epsilonjon Thread Starter Member

Feb 15, 2011
65
1
Ah I see now, thanks. Kind of a strange solution, but I guess it is trying to get you to think 'outside the box' or whatever. Unfortunately I just saw it as 'the question is wrong' .

Thanks!

13. ### Kingsparks Member

May 17, 2011
118
5
TBayBoy.

I don't have to worry about it, I think the Bandersnach got it under the yum yum tree. (Or was it the tum tum tree.) English Lit was a loooong time ago.

TBayBoy likes this.
14. ### dcarson7 New Member

Apr 28, 2011
7
1
http://www.pearsonhighered.com/floyd/

Some good textbooks there.

Step by step explanations, questions and answers at the back of the book.

Still using my earlier versions, some 8 years on.

15. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,912
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I may be mistaken, but isn't 1/8 of a watt 0.125 and NOT 0.16?

The constraint was using E24 one-eight watt resistors to meet the criteria of 4 Volts output +/- 10%

It's still a BS question.

Last edited: Jun 4, 2011