My problem is R1=16 R2=10 R3=? R4=20 Rt=3.582 I3=3.2A

Now I did the 1/R1 + 1/R2 etc

which gets me to .2125, so I did 1/.2125
that leads me to... 4.705882353 for an Rt, that excludes R3's resistance.... I think?

I cant figure out how to use that to calculate my R3...

I also did 1/Rt which equaled .279173, I imagine that might be necessary to find R3?

The answers are in the back of the book so I know what R3 should be, but without knowing how to get there, it does me no good haha.

This is my first post, I did try searching and didn't really find what I was looking for... Thanks for any help on this issue, great site you guys got going here, Im sure Ill be on here quite a bit throughout my apprenticeship. Hopefully I'll be able to contribute to someone else's question one of these days.

 

"picture is worth a thousand words"

 

E1_____E2________E3________ E4________ Et________
I1_______I2________I3= 3.2A____I4_________It_______
R1=16___R2=10____R3________R4=20______Rt=3.582
P1 ______P2 ______P3________P4_________P5

That's the problem in the book, once I get R3 the rest is easy...

 

Do you have any circuit diagram for this problem ? The name of the book and the page number?

 

No diagram, just the chart and fill in the blanks.
Book is Delmars Standard Textbook of Electricity, Sixth Edition, Page 171.

 

I think I understated you have four resistors connected in parallel. R1 = 16Ω; R2 = 10Ω; R3 =? R4 = 20Ω and Total resistance is Rt = 3.582Ω.
So you can use the equation for Rt and solve it for R3. Can you do that?

 

I tried using 1/R1 + 1/R2 + 1/R4 + 1/Rt then putting 1 over the sum of that, and it didn't work.

but I wonder if I do that above, but minus the 1/Rt instead of add it... hmm

Sorry if im hard to understand... im trying to explain it the best I can from what im looking at, but its not translating all that great apparently lol.

 

Here you have a tip:
Let R1 = 20Ω and R2 = ?? and Rt = 5Ω

1/Rt = 1/R1 + 1/R2

1/5Ω = 1/20Ω + 1/R2

0.2S = 0.05S + 1/R2

1/R2 = 0.2S - 0.05S ( S -- >electrical conductance, measured in Siemens)

1/R2 = 0.15S

R2 = 1/0.15S = 6.67Ω

 

1/16 + 1/10 + 1/20 - 1/3.582
.0625 + .1 + .05 = .2125 - .279 equals -.0665
1/-.0665 = -15.03759398 Which is close to what I need for R3 according to the back of the book. (15ohms) Except My answer is one the wrong side of Zero...

But Im not sure if that is the correct way to do it, or I just lucked into it.

 

Check your math. You cannot get the negative number.

 

A co worker just figured it out...

1/16 + 1/10 + 1/x + 1/20 = 1/3.582
.2791 -.0625 - .1 - .05 = 1/x
1/.0666 = 15
R3=15

Thx for the replies. Sorry I couldn't explain it better.

 

R1+R2+R3+R4 = Rt
1/16+1/10+1/?+1/20 = 3.582
1/16 = 8/2
1/10= 5/2 You need to find the Least common denominator, Basic Math.
1/20= 10/2

= 23/2 = 11.5Ω
3.582 However this does not work. In the question do they use this 3.582Ω or this 3.582KΩ or this 3.582MΩ?

Thomas
KF6BSL Extra Class Ham License

 

You need to do the remaining part of the equation, For a parallel circuit the total resistance is equal to

R1 * R2 * R3 * R4
Rt = R1 + R2 + R3 + R4 I think this is Correct

What I put above in my earlier post is even wrong.

Thomas

 

Got it figured man, but thank you... The post above yours is what I used to find R3 which was 15ohms.