Two resistors have a resistances of
20 plus or minus 0.1 ohms and 80 plus or minus 0.2 ohms. If the two resistors were placed in
series, what would the uncertainty of the two resistors together be?
Okay, so in this example, we have
two resistors. We’ll say that this is our first
and then this is our second. And we’re told these two resistors
are placed in series with one another. For both, we’re given the value of
their resistances. And we see that those resistances
include an uncertainty, 0.1 ohms in one case and 0.2 ohms in the other. We can recall that when resistors
are placed in series, like they are here, their resistances add together for a total
combined resistance value. If these resistance values were
stated without uncertainties, it would be straightforward enough to add 20 ohms to
80 ohms to get a total resistance of 100 ohms. But in this case, we do have these
uncertainties that we’ll need to consider as well.
The way to do this is to recall our
rule for combining uncertainties, specifically when we’re adding two values
together. Let’s say that we have one
value. We’ll call it 𝑣 sub one. And this is equal to 𝑎 plus or
minus the uncertainty in 𝑎. And similarly, we have a second
value, 𝑣 two, which is equal to 𝑏 plus or minus the uncertainty in 𝑏. Now, if we were to add 𝑣 one to 𝑣
two, then the way we would do that is we would add 𝑎 and 𝑏 and then, with the
uncertainties, add those together as well. We can apply this rule to our
particular scenario of adding the values of these two resistors.
If we were to solve for the total
resistance, we can call it capital 𝑅, of these two resistors together, then by our
rule, that would be equal to 20 plus 80 plus or minus 0.1 plus 0.2 ohms or, in other
words, 100 plus or minus 0.3 ohms. Now, it’s not the overall
resistance 𝑅 that we want to solve for, but rather the total uncertainty of these
two resistors together. In solving for 𝑅, though, we have
found that total uncertainty. We see that it’s equal to the sum
of the individual uncertainties. That total is 0.3 ohms.