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.

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
$$R_{eq}=R_1+R_2+R_3+\dots$$

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.

A) 0.9Ω
B) 1Ω
C) 1.1Ω

D) 5.5Ω E) 10Ω F) 11Ω

D) 5.5Ω E) 10Ω F) 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".