Calculating Combined Resistance in Parallel Circuits

[hl_world]

I was trying to work out a way to calculate the combined resistance of more than 2 resistors in parallel. We all know the MAD rule for 2 resistors but does this equation work:

combined resistance = sum of resistance * # of resistors^-2

I checked wikipedia and there was nothing there above 2 resistors.

 

[Redbelly98]

http://en.wikipedia.org/wiki/Series_and_parallel_circuits#Parallel_circuits

In particular, look at the equation that begins

1/R_total = ...​

 

[ravioli]

Ultimately the way you figure this out is $$\frac{1}{R_{equivalent}}=\frac{1}{R_1}+\frac{1}{R_2}...\frac{1}{R_n}$$. (After messing around with Latex, someone else posted this... I'll keep the rest of the post as is)

A way to derive the equation that's been posted is to realize that the voltage across each resistor is the same. Since we know the basic V=IR, we can figure out the current going to each resistor. The sum of the currents is going to be the same as the current of an equivalent resistance for the given voltage. So you get...

$$\frac{V}{R_1}+\frac{V}{R_2}+...+\frac{V}{R_n}=\frac{V}{R_{equivalent}}$$. From there, just divide out the voltage V.

 

[hl_world]

The sum of the inverse resistance values. I didn't think of that. I went to the resistor page on Wikipedia. The equation I posted still works though, right?

 

[ravioli]

The sum of the inverse resistance values. I didn't think of that. I went to the resistor page on Wikipedia. The equation I posted still works though, right?

Only if all of the resistors are the same. Even then, your equation reduces down to...

$$R_{equivalent}=\frac{R}{n}$$ where n is the number of resistors.

 

[hl_world]

Wait a minute, something's wrong:
1^-1+2^-1+1^-1 = 2.5

How can a path with 2 1Ω resistors give a higher value just by adding another path of resistance albeit double resistance?

 

[dancavallaro]

It's the *inverse* of the sum of the inverses. So your equivalent resistance is not 2.5 Ω but 1/2.5 = 0.4 Ω.

 

[hl_world]

Oh right; I see now. Thanks