### Video Transcript

A 36.0-ohm resistor, a 50.0-ohm resistor, and a 700-ohm resistor are connected together. What is the largest resistance obtainable? What is the smallest resistance obtainable?

We can call our 36.0-ohm resistor π
one, our 50.0-ohm resistor π
two, and our 700-ohm resistor π
three. We want to solve, first, for the largest resistance obtainable by combining these three in some way. Weβll call that π
sub max. And next, we want to solve for the smallest possible resistance that these three can be arranged to create. Weβll call that π
sub min.

To get started, letβs recall the two different ways that resistors can be combined in a circuit. In general resistors can be combined in a circuit either in series, that is all in a row, or in parallel, in separate branches that run alongside one another. Depending on which of these two types a set of resistors are arranged in, the way those resistors are combined mathematically changes.

In series, the total resistance, π
sub π, is equal to the linear sum of each of the individual resistors up to the last resistor, π
sub π, while in parallel one over the total resistance π
sub π equals one over the first resistor value plus one over the second resistor value and so on up to one over the last resistor value, π
sub π.

One fact about adding resistors in parallel is that the total resistance, π
sub π, is always less than any of the individual resistors involved in the parallel circuit: π
sub one, π
sub two, up to the last resistor. This means that to solve for π
sub min, the least possible total resistance of these three resistors π
one, π
two, and π
three arranged together, weβll use the parallel addition formulation.

The maximum possible resistance though, π
sub max, involves adding the resistors in series. So π
sub max equals π
one plus π
two plus π
three or 36.0 plus 50.0 plus 700 ohms. Altogether, to three significant figures, π
sub max is 786 ohms. Thatβs the highest total resistance we can create combining π
one, π
two, and π
three.

Now weβll move to solving for π
sub min, where weβll use the parallel addition formula. One over π
sub min equals one over π
one plus one over π
two plus one over π
three. If we multiply both sides of the equation by π
one times π
two times π
three and then rearrange to solve for π
sub min, we find itβs equal to the product of the three resistances divided by π
two times π
three plus π
one times π
three plus π
one times π
two.

When we plug in for these values and enter these values on our calculator, we find that, to three significant figures, π
min is 20.3 ohms. Thatβs the smallest total resistance these three resistors can be combined to make. And notice that itβs smaller than any one of them.