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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 $R_2$.
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OpenStax College Physics Solution, Chapter 21, Problem 11 (Problems & Exercises) (1:17)

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