Total tolerance when adding resistances in series

[UrbanXrisis]

If there were 2 wires, and one of the wire has a resistance of 50 Ohms with a tolerance of 5%, while the second wire has a resistance of 60 Ohms with a tolerance of 3%. The total resistance would be 110 Ohms but what about the total tolerance? Would it be 8%?

 

[Chen]

I'll walk you through it and then show the general formula. For the wire with resistance of 50 Ohms and tolerance of 5%, what are the lower and upper limits of its resistance? 47.5 and 52.5 Ohms. The limits of the second wire are 58.2 and 61.8 Ohms. So the lower and upper limits of the resistance of the two wires together is 105.7 and 114.3. So what is the total tolerance of the two wires together?

For resistors connected in series, you can find the total tolerance with weighted average. If T stands for tolerance then:

$$T_T = \frac{R_1T_1 + R_2T_2 + ...}{R_1 + R_2 + ...}$$

So for the given problem:

$$T_T = \frac{50*5\% + 60*3\%}{50 + 60} = 3.9\%$$

 

[Chi Meson]

UrbanX:
I have to know, what level of Physics are you doing? That is, you are in high school, obviously, but is this an advanced class, or an electronics class?

 

[Chen]

Generally speaking, you can find the tolerance of any setup of resistors like this:

$$T_T = \frac{\mbox{total resistance of resistors multiplied their tolerance}}{\mbox{total resistance of resistors alone}}$$

It gets tricky when you connect resistors in parallel:

$$T_T = \frac{\frac{1}{\frac{1}{R_1T_1} + \frac{1}{R_2T_2} + ...}}{\frac{1}{\frac{1}{R_1} + \frac{1}{R_2} + ...}} = \frac{\frac{1}{R_1} + \frac{1}{R_2} + ...}{\frac{1}{R_1T_1} + \frac{1}{R_2T_2} + ...}$$