# Finding Req in parallel circuit

1. Apr 28, 2010

### Aeden

Hi. I am having problems finding the Req in parallel circuits. I have a problem like this

R1- 12
R2- 8
R3- 8
emf- 12
Now I know Req= 1/R1, 1/R2, 1/R3 but my homework book explains that

1/12 1/8 1/8 = (2+3+3)/24 = 8/24 = 1/3 = Req is 3 coulomb

I want to know where the 2+3+3 comes from, I have absolutely no idea how to find that.

I also have another question that says
r1-5
r2-10
r3-15

R1 1/5, R2 1/10, R3 1/15

1/5= .2
1/10= .1
1/15= .066
Req= 0.37

I know this isn't right . I have tried for awhile just by myself to figure this out (I am doing this as a home class) but I cannot do it. Someone please help me out!

2. Apr 28, 2010

### thebigstar25

you have R1=12 ohm, R2=8 ohm, R3=8 ohm (I assume that all of them connected in parallel)

so: 1/Req = 1/R1 + 1/R2 + 1/R3 = 1/12 + 1/8 + 1/8 (now you have to make the denominator the same so you can add them up, and in this case it would be 24, then you have to multiply and divide the first term (1/12) with 2/2, so you get 2/24, and the second and the third terms (1/8) multiply and divide them with 3/3, so you get 3/24..

then 1/Req = 2/24 + 3/24 + 3/24 = 8/24 = 1/3 >> then Req =3 ohm ..

Regarding your second question, just follow what I did and you should get the right answer.. :)

3. Apr 28, 2010

### tiny-tim

Welcome to PF

Hi Aeden! Welcome to PF!
When adding fractions with different denominators (the number on the bottom), it helps to rewrite the fractions so that they all have the same denominator.

Here, 1/12 = 2/24; 1/8 = 3/24; so 1/12 + 1/8 + 1/8 = 2/24 + 3/24 + 3/24 = (2 + 3 + 3)/24.

(the reason why 24 is chosen is that it's the least common multiplier of 8 and 12)
erm … you forgot to 1/ it at the end!

now try it with the same method as the last one (use the least common multiplier of 5 10 and 15), and then …

get some sleep! :zzz:

4. Apr 28, 2010

### Aeden

Oh wow, that was really easy to understand. Thanks a lot, I will now try to solve the other with the instructions given but I completely understand how that works now. Thanks again!