# 4 resistors in series and parallel

1. Apr 18, 2009

### abruski

1. The problem statement, all variables and given/known data

I have 4 resistors R1=R2=R3=R4=40$$\Omega$$

I need to find the Equivalent Resistance when they are connected in series and when they are connected in parallel.

The problem is my answer for the parallel differs from the answer given in the book.

2. Relevant equations

Series: R=R1+R2+R3+R4
Parallel: R=$$\frac{R_{1}*R_{2}*R_{3}*R_{4}}{R_{1}+R_{2}+R_{3}+R_{4}}$$

3. The attempt at a solution

The first is easy: 4*40=160

Parallel: R=$$\frac{40^{4}}{40*4}$$=16000$$\Omega$$

in the book the answer is: 10$$\Omega$$

2. Apr 18, 2009

### Chewy0087

You're almost there, however the second equation is wrong i'm afraid, get that checked up on (the parallel one)

3. Apr 18, 2009

### Staff: Mentor

OK.
Where did you get this equation? (Look up resistors in parallel.)

4. Apr 18, 2009

### abruski

The parallel resistor formula I took from the same lesson of the same book. Very weird. Anyway, thanx guys :)

5. Apr 18, 2009

### Cyosis

I take it the formula from that lesson only had two parallel resistors? In that case $$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2} \Rightarrow R=\frac{R_1R_2}{R_1+R_2}$$. Now add a third resistor to the first equation and see that the generalization you made is not correct.

6. Apr 18, 2009

### abruski

I saw the wrongness of my ways now. Thank you