A set of resistors, the calculation of effective resistance, and
replacing it with a single resistor
As we mentioned in Section 10.4, we can assign a resistance to
any system with two terminals: set up a potential difference \(\Dl V\)
between the terminals, measure the current \(I\) that runs through the
system, and the resistance is \(R={\Dl V\over I}\). In particular, a
combination of resistors with two endpoints have what is called an
equivalent resistance \(R_{eq}\) (or sometimes effective
resistance). One can replace any set of resistors with a single
resistor \(R_{eq}\), and the rest of the circuit won't know the
difference.
One can find the equivalent resistance for any combination of
resistors with two terminals, but there are two types of
combinations for which the formula is easy.
Resistors in Series
We say that two resistors are in series if they are
connected together so that all the current through one, passes
through the other as well. If you can find a wire from one resistor
to the other with no junctions in between, they are in series.
Suppose we have three resistors connected in series, as shown.
The current through all three is the same, \(I\). According to the
tally method, the potential drop across all three is
$$\Dl V=IR_1+IR_2+IR_3$$
and so the equivalent resistance of the three is
$$R_{eq}={\Dl V\over I}=R_1+R_2+R_3$$
Generally speaking,
the equivalent resistance of resistors in series is the sum of
their resistances
$$R_{eq}=R_1+R_2+R_3+\dots$$
Resistors in Parallel
We say that two resistors are in parallel if they are
connected so that they have the same potential drop across them.
In this picture, the group of wires highlighted in purple are all at
one potential \(V_{purple}\), and the group of wires in yellow are
all at another potential \(V_{yellow}\), so the potential drop
across each resistor is the same: \(\Dl V=V_{purple}-V_{yellow}\).
If a current \(I\) passes into the system of parallel resistors shown,
the current has to move from the purple wire to the yellow wire
through one of the three resistors. The wire thus splits into three
parts, and
$$I=I_1+I_2+I_3$$
The current through resistor \(n\) is \(I_n=G_n\Dl V\) where
\(G_n=1/R_n\) is the conductance (Section 10.3), and so
$$I=G_1\Dl V+G_2\Dl V+G_3\Dl V$$
The equivalent conductance of the three is
$$G_{eq}={I\over \Dl V}=G_1+G_2+G_3$$
Generally speaking,
the equivalent conductance of resistors in parallel is the sum of
their conductances
$$G_{eq}=G_1+G_2+G_3+\dots$$
or
$$\frc{R_{eq}}=\frc{R_1}+\frc{R_2}+\frc{R_3}+\dots$$
We can summarize these results as
Resistances add in series, conductances add in parallel.
We can understand this by remembering that resistance is a measure
of how hard it is to force current through a wire, and
conductance is a measure of how easy it is. Suppose we have
a castle with a single road into it, and we have two possible guard
animals: a dragon and an angry cat. Obviously, choosing the
dragon will make it a lot harder to get into the castle: the
dragon has a high resistance and a low conductance. However, if
we add the angry cat to the path, even though it has low
resistance, its addition necessarily makes it harder to get into
the castle, even just by a little bit. Resistances add in series.
Now suppose we start with the cat guarding the castle alone, but
then we add a second road into the castle. Even if this second
road is guarded by the dragon, the addition of the road
necessarily makes it easier to get into the castle. (Perhaps a
knight has a magic dragon-slaying sword and an allergy to cat
hair.) Adding an additional path, no matter how difficult, will
always increase the conductance. Conductances add in parallel.
What is the equivalent resistance of these two resistors in
parallel?
A) 0.9ΩB) 1ΩC) 1.1Ω D) 5.5ΩE) 10ΩF) 11Ω
Because the two resistors are in parallel, their equivalent resistance
must be less than either individual resistance, just as it's
easier to get into the castle if there are two roads. That rules
out every answer except for A.
Numerically, conductances add in parallel. The conductance of
the 1Ω resistor is 1℧, and the conductance of the 10Ω
resistor is 0.1℧, so the equivalent conductance is
$$G_{eq}=G_1+G_2=1\mho+0.1\mho=1.1\mho,$$
for an equivalent resistance of 0.9Ω.
Most textbooks don't talk about conductance, but instead use the
formula
$$\frc{R_{eq}}=\frc{R_1}+\frc{R_2}$$
for resistors in parallel, which is basically the same thing. A
common mistake here is to flip the two resistances, add them
together to get \(\frc{R_{eq}}\), but then forget to flip the
result to get \(R_{eq}\). Thinking in terms of conductance might
help you avoid this error; the ℧ symbol serves as a reminder
that your answer is "upside-down".