Know this is going to be an obvious answer but...1 ohm resistors in parallel?

[Rucap09]

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I think this is one of those questions where I'm going to face palm after I hear the answer but how does the math work for resistors with 1 ohm resistance when added in parallel?

1/R1 + 1/R2= 2ohms if each is 1 ohm, but conceptually I know resistors in parallel have decreased resistance due to wider path, so what's tripping me up here?

Thanks!

 

[t5Nitro]

(1/R1) + (1/R2) = (1/R_eq)

Check this picture out; remember to take the inverse at the end.

Edit: The short cut for adding 2 resisters in parallel is R_eq = (R1)(R2)/(R1 + R2). So with this equation you get R_eq = (1)(1)/(1+1) = 1/2 ohms.

 

[Rucap09]

Haha dang it! I just realized I had the equation upside down and was rushing back to save face. Thanks for replying though.

I think I need another coffee.

 

[t5Nitro]

No worries! This is where you want to make those mistakes, not on test day. Good luck with your studying.

 

[Rucap09]

Alright, I've got one more follow up if you (or anyone else) wants to field another.

Resistivity (rho)=E(electric field)/J(current density)

E=K(q/r^2) = (N•m^2/C^2)(C/m) = (Newton)/Coulombs

Am I correct in understanding the expression of E = Volts/meters to mean this is just Newtons/Coulombs rewritten to have meters/meters added into the units? So that now you have Newton•meters/Coulombs•meters = Joules/(Coulombs•meters) = Volts/meters?