!Series and Circuit Questions : please help asap!

[dradis45]

im really baffled by these questions, im woefully horrible at physics, please help me...
1. Prove that when n identical resistors of resistance R are connected in parallel the equivalent resistance is R/n.

2. A piece of copper wire of resistance R is cut into 3 equal parts. When these 3 parts are connected in parallel, what is the resistance of the combination?

3. Say you have a series and parallel circuit. Resistors 2 and 3 are parallel to each other. Resistor 1 is in series with them. When you increase the resistance of 3, what happens to current and voltage values?

Thanks!

---

Members don't see this ad.

 

[Phlame217]

just use the formulas:

Parallel is 1/Req = 1/R1 + 1/R2 .... etc
Series is Req = r1 + r2 ... etc

So number 1:
Simply apply parallel formula

2:
apply formula again

3:
apply forumlas (first parallel to reduce two resistors to one and then series) to see change in resistances

 

[Phlame217]

one additional formula you may need is V = IR for the last problem

 

[physics junkie]

1/Req = 1/R_1 + 1/R_2

After you convert this to the harmonic mean equation you get:

Req = R_1*R_2/ (R_1+R_2)
If R_1 = R_2 and 'n' = # of identical resistors then you have

R_1^n / n*(R_1)

Obviously here you can divide the R_1 in the denominator by R_1^n in the numerator to get: R_1^(n-1)/n.

My only mistake might have been with the generalization of R_1*R_2 = R_1^n but I'm sure you could figure that out. I'm approaching this as you are a math/physics major since you are asking for a proof and this is probably not an MCAT question. That's why you get to fill in the details

Proof by induction will do fine. Assume it is true for k and then prove it for k+1.

Edit: Oops, just realized you need it generalized for 'n' resistors and I only did it for two. What you could do is realize that all you have to do to get your new resistance by adding another resistor is consider the two resistors a single resistor(with their combined equivalent resistance) and then use the formula again. Or what I would do is just look up the derivation of the harmonic mean.