Finding unknown resistance of a resistor using a schematic drawing

[seanmcgowan]

Homework Statement

What is the value of the unknown resistor in the circuit shown if the equivalent resistance for the circuit is 21.50 ohms.

Homework Equations

Series= R1+R2+R3...
Parllel= (1/R1)+(1/R2)+(1/R3)...

The Attempt at a Solution

First off, I'm sorry for the horrendous drawing I did, but its better than nothing i guess.

I have done several of these problems and I am STILL stuck on every single one of them.

What I did was I drew several different possibilities of series and parallel circuits, but none of them came anywhere near the 21.50 ohms.

First I drew a pic, that went like this: (series) 7.5 and 9.1 then parallel 6.81 and ? then series again 9.1 and 3.6. However, just adding the series circuits together I came up with 29.3, and that was before the Parallel circuit.

Then I tried leaving the very first circuits (the first 9.1 and 7.5) and series, and making everything else parallel:
1/9.1= .11
1/3.6= .29
1/6.81= .15
.11+.29+.15= .55
1/.55= 1.82

7.5+9.1= 16.6
16.6+1.82= 18.42

Im fairly certain that since the unknown factor would be prallel you can't just subtract 18.42 from 21.50

How do I find this? And, am I even close to guessing the correct parallel and series combination? I'm clueless, an help is appreciated.

 

[Kruum]

We can't see your attachment until it's been approved. You could upload your picture to a different site and use the IMG][/IMG -code, with "[" added to the start and "]" to finish. That way we could see the circuit right away.

 

[seanmcgowan]

Which website can i use for that? I've never done it before

 

[skeptic2]

There's an error in your Relevant equations. For parallel instead of (1/R1) + (1/R2) + (1/R3)... it should be 1/((1/R1) + (1/R2) + (1/R3)...).

 

[seanmcgowan]

I just wrote it down wrong, that's what i menat but wasn't sure how to write it down.

 

[Kruum]

tinypic.com should do fine for uploading images.

 

[Kruum]

The basic rules of combining resistors are quite simple. If only one end of each resistor are connected, then they are in series. But the connecting wire must be unbranched, otherwise you can't say anything about the connection. If both ends of two resistors are connected they are paralleled. In this circuit there are a total of four series connections and one paralleled. Once you find these connections you can solve the unknown quite easily.

 

[Phrak]

seanmcgowan

You might find the problem easy if you were given the value of all the resistors and asked to find the total resistance. Try taking the problem backwards.

 

[seanmcgowan]

I thought that wa what the 21.50 ohms was about? isn't that the total resistance of all of the resistor's?

 

[Kruum]

I thought that wa what the 21.50 ohms was about? isn't that the total resistance of all of the resistor's?

Yes it is. But if you have the total resistance, you can plug it in the equation and find the unknown. For this problem you could create the equation for the total resistance of the circuit and use it to solve the unknown resistor, just as Phrak said.

 

[seanmcgowan]

Ok, I am probably going to start sounding real dumb here but I am not quite getting it. WHen figure out all of the parallel and series circuits, I come up with 18.45. If I subtract this from 21.50 i get 3.05. While this sort of fits, the missing resistor is in parallel right? and when I try to figure that out the numbers never even get close, so what's up?

 

[Kruum]

While this sort of fits, the missing resistor is in parallel right?

Nope.

 

[seanmcgowan]

oh man, so I've had the answer all along! haha, thanks. one last question though, is there an easier way to find out what's parallel and what's series? or does that just come with practice?

 

[Kruum]

Once you've truly understood the basic idea behind series and parallel connection, it all comes quite naturally, of course there could be some more difficult connections.

What did you get for the unknown resistor?

 

[seanmcgowan]

3.05 ohms. Thats right...right?

 

[Phrak]

3.05 ohms. Thats right...right?

No, it's a bit larger, based on your drawing values.