Equivalent Resistance

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) 0.9Ω B) 1Ω C) 1.1Ω
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".