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.

 

Your equations looks good.

 

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.

 

Thanks for the swift reply.

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

 

What is the name of the text book?

 

Introduction to Electric Circuits by Richard C. Dorf and James A. Svoboda.

 

Introduction to Electric Circuits by Richard C. Dorf and James A. Svoboda.

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)

 

you should be able to look up how the answer was derived here

<SNIP>

just put in the chapter and problem number.

 

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

 

you should be able to look up how the answer was derived here

<SNIP>

just put in the chapter and problem number.

Unfortunately you have to sign up, and it costs money per month :-(

 

I saw the solution. Read my previous post.

 

I saw the solution. Read my previous post.

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!

 

you should be able to look up how the answer was derived here

just put in the chapter and problem number.

TBayBoy.

In reference to your signature; "Have your Tribble spayled or neutered."

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.

 

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.

 

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.