Given total resistance in a combination circuit find the missing value of R2
Rt= 23,250ohms
in series R1= 12,000ohms
in parallel with R2 is R3 which equals 18,000 ohms
R2 =?

 

What's giving you trouble?

 

Given total resistance in a combination circuit find the missing value of R2
Rt= 23,250ohms
in series R1= 12,000ohms
in parallel with R2 is R3 which equals 18,000 ohms
R2 =?

This is hopelessly fragmented. I can't tell which resistors are in which configurations.

Rt is the effective resistance when which resistors are configured how?

Is R1 12kohms by itself, or is the 12kohms the total resistance when something else (what?) is in series with it?

Please try to describe the circuit (or throw a picture together an post it) in clear and complete sentences. Here might be what you are trying to say (but I am really only guessing).

I have a compound resistor, Rt, that consists of three resistors, R1 (12kohms), R2 (uknown), and R3 (18kohms). The resistors are arranged as follows: R2 is in parallel with R3, and then R1 is in series with the R2,R3 combination. The task is to find the value of R2 if we know that Rt is 23,250 ohms. Here is what I have tried so far....

Please do not overlook that last sentence. We really do want and expect you to show what efforts you have made to solve the problem yourself.

 

The op needs to draw the circuit diagram.

As I read it,

R1 = 12kΩ
R2 = unknown
R3 = 18kΩ

R1 in series with R2 || R3

Equivalent resistance = 23.25kΩ

 

That would be best, but can they attach a diagram before they have 10 posts? I guess they can put it on one of those other picture posting sites and put a link here, but I am getting to the point where I am hesitant to go to those sites because most of them want to put literally dozens of tracking cookies on your machine - some of them in excess of 60 cookies.

We are reading his sparse description the same way (the way you presented it is better, IMHO, but I like encouraging the use of complete sentences wherever possible). If the OP will just confirm that we have captured his intended configuration (plus show some effort to solve it), that will work for me.

 

The problem here I think is the OP doesn't speek goodly inglish!

My guess is R2=30KΩ

Guess:
R2 and R3 are in parallel. This set is in series with R1. Total = 23250Ω



R(total) = R1 + 1/(1/R2 + 1/R3)

so

R2= 1/[1/(Rtotal-R1)-1/R3]

(where Rtotal = 23250, R1=12000, R3=18000)

so

R2 = 1/[1/(23250-12000) - 1/18000] = 30K

QED. If it right???!

 

The problem here I think is the OP doesn't speek goodly inglish!

My guess is R2=30KΩ

Guess:
R2 and R3 are in parallel. This set is in series with R1. Total = 23250Ω



R(total) = R1 + 1/(1/R2 + 1/R3)

so

R2= 1/[1/(Rtotal-R1)-1/R3]

(where Rtotal = 23250, R1=12000, R3=18000)

so

R2 = 1/[1/(23250-12000) - 1/18000] = 30K

QED. If it right???!

Not speaking good English is just fine - we can work through that. But just working the problem for someone without them having to lift a finger is highly frowned on (not to mention against the AAC guidelines for the Homework Help forum).

 

There are only 3 resistors and the English is perfect.
Here is the simple puzzle that has an answer that is extremely easy to calculate without guessing:

 

Not speaking good English is just fine - we can work through that. But just working the problem for someone without them having to lift a finger is highly frowned on (not to mention against the AAC guidelines for the Homework Help forum).

30 k is the answer i came up with also

 

30k is the answer i came up with also

 

Not speaking good English is just fine - we can work through that. But just working the problem for someone without them having to lift a finger is highly frowned on (not to mention against the AAC guidelines for the Homework Help forum).

Oops, sorry I didn't realize about the rules (and the goodly inglish was only a joke). I'll be more careful next time to let the OP show they've done a bit of homework first.

 

No problem. Live and learn. The OP appears to have gotten the right answer.

To the OP: When you post, if you have already gotten an answer (don't know if that was the case here or not, this is just for your own benefit and, mostly, for the benefit of others that run across this) be sure to post it. Feel free to qualify however makes sense,, such as saying that you think it's the right answer or you suspect its the wrong answer, or you think the right answer should be bigger, or whatever. Also, if you have 'correct' answer (such as from the back of the book), then post that as well. Basically, the more information we have about what information you already know, the freer we can be with what we say.