Ohm's law (i.e., I forgot my Algebra)

  • Context: Undergrad 
  • Thread starter Thread starter ASEmech
  • Start date Start date
  • Tags Tags
    Algebra Law Ohm's law
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
Sign up Log in
3 replies · 6K views
ASEmech
Messages
2
Reaction score
0
I'm an ASE certified Technician specializing in Brakes and Four-wheel Alignments. I'm going to go for my cert. in Electrical systems this fall. As I began my self-study in electrical systems, I realized...I forgot my Algebra.

I'm currently studying Ohm's Law. Within Ohm's Law, I'm studying Resistance in a parallel circuit. The formula I'm given is:

R0=1/(1/R1) + (1/R2) = (R1 x R2)/(R1 + R2)

...where R0 (read R-sub0) is a combination of resistances R-sub1 and R-sub2.

I thought the sum of the resistances in any circuit was R0=R1+R2. Why is 1 divided by (1/R1) + (1/R2) and in the "denominator portion" of the formula, why is 1 divided by R1 (same question for 1 divided by R2) and then added to 1/R2? Also, how do you get from R0=1/(1/R1) + (1/R2) to (R1 x R2)/(R1 + R2)?
 

Discussion

Mathematics news on Phys.org
This thread probably belongs in the EE forum.
Anyway,

[itex]R_{tot}=R_1+R_2[/itex] is only correct if the resistors are connected in series. In
a parallell ciruit the current can flow in several branches at once which is why you need to use the [itex]1/R_{tot}=1/R_1+...[/itex] formula.
This makes sense if you think about. Let's say you have two 1K resistors connected in parallell; now the current can flow via two different paths and since the resistors have the same value each one will carry half the current; plugging two 1K resistors into the formula for parallel resistors you will see that you end up with an equivalent resistance of 500 ohm. This is what you would expect since the two paths make "half as difficult" for the current to flow; i.e. connecting resistors in parallel REDUCES the total resistance.

You can derive the formula yourself by just using ohms law and remembering that the voltage across each of the resistors must be the same since they are connected in series.

Also

1/R=1/R1+1/R2 =R2/(R1*R2)+R1/(R2*R1=(R1+R2)/(R1*R2)
i.e.
1/R=(R1+R2)/(R1*R2)
meaning R=R1*R2/(R1+R2),
 
Thanks, guys. I put this here because when I first saw this formula the first thing I thought of was Alg. 1.

Thanks again!