# Would this be a correct way of calculating total resistance?

#### barrett50

Joined Feb 1, 2016
14
So I have here that

R1 = 470 ohm
R2 = 660 ohm
R3 = 290 ohm
Ry = 500 ohm

Would I be correct in saying that I can join these into 1 Resistor by the follow?

Rs1 = 1/((1/470) + (1/660)) = 274.51 ohm

Rs2 = 1/((1/290) + (1/500)) = 183.54 ohm

then because Rs1 and Rs2 are in a series, I can now do

Rs1 + Rs2 = Rtot
=> Rtot = 274.51 + 183.54 = 458.05 ohm

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#### WBahn

Joined Mar 31, 2012
29,521
Yes, that looks correct except for the sloppiness with your units. It should be, for example

Rs1 = 1/((1/470 Ω) + (1/660 Ω)) = 274.51 Ω
Rs2 = 1/((1/290 Ω) + (1/500 Ω)) = 183.54 Ω

Rs1 + Rs2 = Rtot
=> Rtot = 274.51 Ω + 183.54 Ω = 458.05 Ω

#### barrett50

Joined Feb 1, 2016
14
Yes, that looks correct except for the sloppiness with your units. It should be, for example

Rs1 = 1/((1/470 Ω) + (1/660 Ω)) = 274.51 Ω
Rs2 = 1/((1/290 Ω) + (1/500 Ω)) = 183.54 Ω

Rs1 + Rs2 = Rtot
=> Rtot = 274.51 Ω + 183.54 Ω = 458.05 Ω
Thanks for that. I'll be a little more organized next time I post

#### shteii01

Joined Feb 19, 2010
4,644
When dealing with resistors in parallel, normal people use the formula in this form:
$$R_{Total}=\frac{R_1*R_2}{R_1+R_2}$$

It is the same formula like you have, but it is in the form that is easier and more convenient to write and solve.