# Still trying to find unknown using the given values

Discussion in 'Homework Help' started by Kid347, Sep 22, 2015.

1. ### Kid347 Thread Starter Member

Aug 14, 2015
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0
I have read through the book "Electronics for Dummies" and I am still having trouble finding the unknown values for different circuits. I have learned some things about series and parallel circuits. My point that I am stuck on now is I am given a parallel circuit which has V=12 volts I=300 mA R1=50 Ohms what is R2? and what is P1? given the formula P=I squared X R but I only have the value for R1 which is 50 Ohms. Also when I calculate 12 volts X .300 mA I get 3.6 the book gives the answer of 2.88 watts. I will upload the circuit.

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2. ### MikeML AAC Fanatic!

Oct 2, 2009
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First, find the Equivalent resistance that would draw 300mA from a 12V supply:

R = E/I = 12/0.3 = 40Ω.

R1 is given as 50Ω

What does R2 have to be to make the parallel combination of R1//R2 = 40Ω knowing R1=50 and Rp=40?

3. ### Kid347 Thread Starter Member

Aug 14, 2015
68
0
I also came up with 40 ohms, but I thought i did something wrong, because if R1=50 how can the combined resistance be 40 ohms?

4. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,500
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Because you have resistors in parallel.

5. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,500
511
Welcome to the wonderful world of resistors.

If you want bigger resistor, place resistors in series.

If you want smaller resistor, place resistors in parallel.

6. ### WBahn Moderator

Mar 31, 2012
18,079
4,917
P1 is the power being dissipated by R1 (while P2 would be the power being dissipated by R2 and P would be the total power -- i.e., the power associated with current I). This has to be inferred from the information given since it isn't stated or shown explicitly.

7. ### MikeML AAC Fanatic!

Oct 2, 2009
5,450
1,066
Have you encountered the method of combining parallel resistors to find the equivalent single resistor value in your book?

Does this look familiar:

8. ### Kid347 Thread Starter Member

Aug 14, 2015
68
0
Yes that looks very familiar except I don't know the value of both resistors. How would I use the formula of R1+R2?

9. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,990
1,115
1/40 = 1/50 + 1/x
All you need to do now is to solve for x

10. ### Kid347 Thread Starter Member

Aug 14, 2015
68
0
I have found what I am looking for. I was looking for the formula to figure out how to find the missing value for R1, the question gave me the value for R2 and for voltage and current. Waiting for the doctor I found an app on my Smart Phone that led me to this.
Finding Total Resistance (RT) when R1 and R2 Given;
1/RT = 1/R1 + 1/R2
RT = R1||R2
RT= R1 x R2 / (R1 + R2)
Finding R1 when RT and R2 Given;
R1 = R2× RT / (R2 – RT)
Finding R2 when RT and R1 Given;
R2 = R1 × RT / (R1 – RT)

Now I have a formula that I can use to find any missing R value.

11. ### WBahn Moderator

Mar 31, 2012
18,079
4,917
If you have to rely on finding an app to give you the exact formula you need instead of being able to apply basic algebra skills to solve for what you need, then you are going to have a nightmarish time in electronics. I suggest you devote some quality effort into learning at least basic algebra.

12. ### Kid347 Thread Starter Member

Aug 14, 2015
68
0
I am trying to do that, like I have said in other posts I am trying to learn this on my own at home by text books and the web. it is difficult for me, I did purchase algebra books since I have not used algebra in 30 years. I am trying to find a course locally or on line to learn electronics but I have had no luck. I agree with you 100% that I need to learn the basics, but I have not found a means of teaching myself yet. But I do learn when I get the answers and then work backwards. I then try to figure out how to get the answer.

13. ### WBahn Moderator

Mar 31, 2012
18,079
4,917
Yes, now that you mention it I do recall your efforts to find some way to learn this stuff. But before you were focusing on learning electronics. Perhaps if you focus on learning the math you might fair better. I would imagine that most community colleges are going to offer courses in algebra, pre-calc, and calc. Also, there are places like Mathnasium and Kumon Math Centers that might be able to help. I don't know how good or how expensive the latter are, though.

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14. ### Kid347 Thread Starter Member

Aug 14, 2015
68
0
I have another problem, if V = 12 and I = .300, R1 = 50 ohms R2 = 200 ohms How does the book come up with the answer that P1 =2.88 Watts? should it not be P=VI 12*.300 = 3.6 P=I squared*R =3.6 P=V squared/R = 3.6 Can someone please tell me how the book comes up with P1 = 2.88 Watts. Thank You in advance. I just don't know what I missed.

15. ### WBahn Moderator

Mar 31, 2012
18,079
4,917
You missed Post #6.

300 mA is the current flowing in the entire circuit. It is NOT the current flowing in R1. The power P1 is the power being dissipated by R1, not the entire circuit. So you need the current that is flowing through R1 and the voltage that appears across R1.

How much voltage is across R1?

How much current is flowing through R1?

What is the product of that voltage and that current?

16. ### Kid347 Thread Starter Member

Aug 14, 2015
68
0
I am happy to report that I have completed the problem in the DC Pre Test, from chapter 1 of the book "Complete Electronics Self-Teaching Guide" the questions was Vs - 12, I = 300 mA, and R1 = 50 ohms.

question what is R2- R1xRt/(R1-Rt) 50x40/(50-40) = 200 ohms

second part what is P1 - 12v /50 ohms = .24 mAmps .24mA squared x 50 = 2.88 Watts

Finally . is there a way to close this thread? And thank you to all who helped.

17. ### WBahn Moderator

Mar 31, 2012
18,079
4,917
While we CAN close a thread, we normally just let them die a natural death.

You can always unsubscribe ("unwatch") a thread by clicking the appropriate link just above the first post on a given page (at least in the old blue layout) so that you won't get notifications if someone does reply. Personally I find it not worth the hassle. I have literally thousands of threads that I have replied to and are therefore automatically watching and relatively few out-of-date threads are ever heard from again -- and often when they are it is worth seeing the new post and responding. I only unwatch a thread when it is very active AND I just am not interested in it.

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