Question

Show that if two resistors $R_1$ and $R_2$ are combined and one is much greater than the other ( $R_1 >> R_2$ ): (a) Their series resistance is very nearly equal to the greater resistance $R_1$. (b) Their parallel resistance is very nearly
equal to smaller resistance $R_2$.

Final Answer

see video solution for explanation.

### Solution video

# OpenStax College Physics for AP® Courses, Chapter 21, Problem 11 (Problems & Exercises)

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Video Transcript

This is College Physics Answers with Shaun Dychko. Connecting resistance

*R1*and resistance*R2*in series means that you add the two resistances together directly. And so since*R1*we are told is much much bigger than*R2*, this total is going to essentially be*R1*since*R1*is so big*R2*has no effect on it when you add to it. So this resistance is essentially approximately equal to*R1*. When connecting parallel, the total parallel resistance will be the reciprocal of the sum of the reciprocals. And so, we have one over*R1*over*R2*all to the power negative one. And we can get a common denominator here by multiplying this by*R2*over*R2*. And this*R1*over*R1*. And we have*R1*plus*R2*on top divided by*R1*multiplied by*R2*at the bottom. And then take the reciprocal of that when get*R1R2*over*R1*plus*R2*. And since this denominator is essentially*R1*from the same logic that we have up here.*R1*as so much bigger than*R2*,*R2*has no effect when we add to it. The denominator is essentially*R1*. And then this fraction is*R2*. And there we go.