in this video we quickly derive the formulas for equivalent resistance
for series and parallel combinations we make a quick note on the comparison between the
equivalent resistance and each of the individual resistors in the combination
and then we finish each type with a simple example so if i check out our series combination real
quick i'm going to mark two nodes on this just to emphasize that what we're doing here is saying
if we clip out r1 and r2 out of this circuit and then replace it with req we should get exactly
the same behavior in other words the same total current through the circuit that's what we mean by
an equivalent resistance so for one thing i know that the total current in the circuit is going to
be the same when i make this replacement second i can see that the total voltage drop across the
series combination r1 and r2 is v it's the same as the voltage drop across the equivalent resistor
but that voltage drop across r1 and r2 can be split into two voltage drops one for each
of the individual resistors and those two voltage drops better add up to v and that's
where we begin the derivation so i'm going to apply ohm's law to each of these resistors v
equals ir and on the left hand i have the equivalent resistance so i have i
times req on the right hand side i have the same i again the same total current flows
either way we do this circuit times r1 plus i times r2 and because those eyes are the same in
both circuits that cancel out and i immediately obtain my formula for the equivalent resistance
of two resistors wired in series now just a quick note on comparisons here because
these are important sometimes for conceptualizing what a circuit is going to do
i can tell immediately that the equivalent resistance
is bigger than either one of the participant resistors and then our quick example i want to get
the equivalent resistance of this simple series circuit i have a 5 volt battery two resistors
220 and 100 ohms respectively and so our eq is so simple that i don't really need to show any of the
math here you just add them and i get 320 ohms so now i'm visualizing the circuit in this simplified
form it's just the battery and the equivalent resistor and i can find the current by saying
i equals v over r so i'm just turning around ohm's law and v is 5 volts and r equivalent was 320 ohms
i shifted the decimal place on this answer to get 15.6 milliamps all right now let's check out our
parallel combination the thing that's the same for each resistor now is the potential difference
across it i can see the high side of r1 is connected to the positive terminal of the battery
the low side to the negative terminal and that's voltage of v it's also going to be a voltage of v
over r2 and over the equivalent resistor that i replaced them with in that equivalent circuit i'll
go ahead and show the total current and that same total current flows in this branch of our original
circuit but it gets split into two pieces when i hit that node now since charge is not allowed
to build up anywhere i know that i is equal to i1 plus i2 and that's where we start the derivation
then i use ohm's law and i can say i equals v over r
over each of these resistors so looking at the equivalent resistance i have i equals v over req
again that v is the same thing for all three of these resistors i1 is going to be v over r1
and i2 is going to be v over r2 the v's are the same for all three and
those cancel and i immediately arrive at my formula for the equivalent resistance
of a parallel combination so far our equivalent resistance formulas are
easily extended to three or more resistors i just keep adding on terms that look like this
for series resistors i would just go r one plus r two plus r three and so on for parallel resistors
one over the equivalent is one over r one plus one over r two plus one over r three and so on
but there's some advantage in reworking the parallel formula to actually solve for r
equivalent in general because it allows us to compare the equivalent resistance
to the resistance of the individual resistors so what i'm going to do to solve for req
is take the reciprocal of both sides of the equation then to get rid of
the complex fraction we multiply the top and bottom by the least common denominator r1r2
and i end up with an r1 r2 in the numerator and when i distribute r1 r2 to 1 over r1 the
r1s cancel and i just get r2 when i distribute it to the second term the r2s cancel and i get
r1 so this is a popular reworking of the parallel equivalent resistance formula
i'm going to put it in order and call it r1 plus r2 and now i'm going to find my
comparisons with the original resistors and the idea is this i can say that req
is definitely less than r1 r2 over r1 because r1 r2 over r1 has a smaller denominator
than r1 r2 over r1 plus r2 so i just removed r2 from it made the denominator smaller
it must be a larger quantity so now it's bigger than req and i just showed that req
is less than r2 itself repeat this sort of idea removing r1 instead of r2
and i find that r equivalent is less than r1 so the parallel equivalent resistance is less than
the resistance of each of the resistors that are in the combination and we'll see that in action
in the example my parallel equivalent resistance should be less than 100 ohms less than either one
of these so i'm going to go back to my original parallel equivalent resistance formula because
it's the easiest one for me to remember so i'm just trying to be honest about the way
i really do these problems and 1 over req is going to be 1 over 220 plus 1 over 100.
my units are ohms in every one of those denominators so the way i normally handle
this is i will do 1 over 220 plus 1 over 100 in my calculator and then i say 1 over that
results 1 over the previous answer and i get my equivalent resistance
into 3 sig figs on this i get 68.8 ohms and it checks out that that equivalent resistance is less
than either one of the resistors in the original combination finally i'm asked for the total
current flowing through the circuit so i equals v over r again i'm thinking about the equivalent
circuit now that just has one resistor i equals v over r equivalent and that's five volts over
68.8 ohms and since i use milliamps the first time around i'll just do it again i get 72.7 milliamps
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