# Example: Resistors in series

#### Prerequisites

## Understanding the situation

In this and the next two examples, we'll both show how Kirchhoff's principles are used to analyze electric networks and develop some useful results for analyzing more complex situations. We'll first consider the simplest non-trivial electric network: one battery and two resistors connected in series.

Two circuit elements are in ** series** means that they are connected so that whatever current flows into one of them flows out of that element and into the next.

Consider this simple example.

## Presenting a sample problem

Two resistors of resistance $R_1$ and $R_2$ are connected to each other and to a battery as shown in the diagram(s) at the right. If the battery maintains a voltage difference of $V_0$ across its terminals, find the current in and voltage drop across each resistor. Assume the battery and resistors are ideal and the connecting wires are resistanceless.

## Solving this problem

We've drawn the diagram in two forms: the semi-realistic one on the left and the more standard symbolic form on the right.

To answer the questions in the problem we have five tools to bring to bear: Kirchhoff's principles, the 0-resistance-wire heuristic, and our definition of what a battery is.

- The total amount of current flowing into any volume in an electrical network equals the amount flowing out. (K1)
- Across any single resistor, the potential drop across the resistor is proportional to the current through the resistor. The proportionality constant is the resistance: ΔV = IR. (K2)
- Following around any loop in an electrical network the potential has to come back to the same value (sum of drops = sum of rises). (K3)
- An element (e.g., a wire) having a 0 resistance is at a uniform potential: There is no potential drop since R = 0. (Heuristic)
- A battery is a device that maintains a fixed potential difference across its terminals. (Definition of a battery)

To approach this problem, let's first get a sense of what's happening ("tell the story of the problem"). The battery maintains a potential difference across its terminals by principle 5, so it's like it's trying to push current out of its high end. Once it's connected up into a loop, it can create a flow — an electric current. In steady state (which takes about a nanosecond to establish in a circuit like this that would fit on a table), since there is only one loop, taking any piece of wire as a "volume" in principle 1, we conclude that there must be a constant current running through the circuit. Whatever current enters a bit of wire, that same amount of current must leave it.

As the current runs around the circuit, as it passes through the wires, nothing happens to the potential (it doesn't change) by principle 4. When it passes through the resistors, there is a drop whose amount is governed by principle 2. The drops across the two resistors must add up to the rise in the battery by principle 3.

Let's play this out in detail in terms of the symbols. There are many ways to go about this analysis, but often a useful first step is to name all the currents. Since there are no branchings, the current must be the same in all parts of the circuit. Call it $I$.

A useful second step is often to "map the potential". An electric potential is like a "height" in a problem with gravitational potential energy. Any point in the circuit has a fixed "height" and we can either find it out or give it a name if we don't know it. The first step in doing this is choosing a 0 for our potential. Since we only have one battery (see the "Dangerous Bend" below), it's convenient to pick the low end of the battery as 0. Since the battery maintains a fixed potential difference across its terminals, we then know the high end. We can then follow wires in both directions, specifying the potential (since there is no change in a resistanceless wire). When we get to a resistor, there will be a change as we cross it. We may then have to give a name to potential values at points where we do not know what they are. Let's try it.

In the figure below on the left side, we have specified that the current is the same everywhere in the circuit and have given it the name I. In the figure at the right, we have put in the voltages at the places where we know it — at the bottom and top of the battery — and continued it along the circuit as far as we could using principle 4. We don't (yet) know the drop across the resistors so we don't know the potential between them. We just give it a name: $V_1$.

What else do we know? We've used principles 1, 4, and 5 in setting up our diagrams, and principle 3 is automatically satisfied. (That's what principle 4 does for us in a single loop network. It won't always do that for us.) Clearly, we have to satisfy principle 2: Ohm's law. Since this is ONLY true for a single resistor, we have to apply it once for each resistor using the drops we read off our map. It gives the following two equations:

$$(V_0 - V_1) = IR_1$$

$$V_1 = IR_2$$

Now what? What good does all this do us? What we've done is what we often do when we are trying to solve something in terms of symbols. Instead of trying to look for something we can calculate, we look to see what are our unknowns and what are the relationships between what we are given and what we know. It's only after we have set up as many relationships as we can (using every tool in our toolbox) do we stop and ask if we are done. Here, we are given the properties of the three elements: $V_0$ (the battery), $R_1$ and $R_2$ (the resistors). We want to find the currents and voltage drops and so have defined two new variables in order to be able to specify those everywhere: $I$ and $V_1$.

The current in this simple circuit is *I *everywhere, so if we could find *I* in terms of the givens, we would have our answer about the current for every element. For the voltage drops, we know it across the battery (that's a property of an ideal battery) and we have assumed that it's given. Reading off our voltage map, the voltage drop across $R_1$ is $V_0-V_1$ and the voltage drop across $R_2$ is $V_1 - 0 = V_1$. So if we can find $I$ and $V_1$ in terms of the givens, $V_0$, $R_1$, and $R_2$, we are done.

Looking at our two Ohm's law equations, we see we have two equations in two unknowns so we should be able to solve them. Since everything is symbols, it's a little tricky keeping track of what's known and what's unknown, but you'll get used to it once you start thinking about the symbols not as arbitrary math symbols but as representing real physical things in the circuit.

The second equation gives us $V_1$ in terms of $I$ so we can put that into the first equation and get one equation in one unknown: $I$.

$$V_0 - IR_2 = IR_1$$

$$IR_1 + IR_2 = V_0$$

$$I = \frac{V_0}{R_1 + R_2}$$

We can now use the second equation to find the drop across resistor $R_2$ which is $\Delta V_{R2} = V_1 - 0 = V_1$:

$$\Delta V_{R2} = V_1 = IR_2 = \frac{V_0R_2}{R_1 + R_2}$$

That's the drop across $R_2$.

The drop across $R_1$ is $\Delta V_{R1} = V_0 - V_1$.

$$V_0 -V_1 = V_0 - \frac{V_0R_2}{R_1 + R_2} = V_0 \bigg(1 - \frac{R_2}{R_1 + R_2}\bigg) = V_0 \frac{R_1 + R_2 - R_2}{R_1 + R_2}$$

$$V_0 -V_1 = V_0\frac{R_1}{R_1 + R_2}.$$

Although these look like a mess, they actually make a lot of sense.

## Making sense of the result

In general, it's not enough to solve a problem; we need to look at our answer and see that it makes sense. Here, that's especially important since resistors in series are going to be components of more complex problems we will analyze. Let's step back and see what's happened. Here are the results:

$$I = \frac{V_0}{R_1 + R_2}$$

$$V_{R1} = V_0\frac{R_1}{R_1 + R_2}$$

$$V_{R2} = V_0\frac{R_2}{R_1 + R_2}$$

Since *V*_{0} is the drop across the battery, we see that the current through our pair of resistors looks just the same as if we have a single resistor connected across the battery with resistance $R_1 + R_2$. This suggests the useful result:

*To the outside circuit, a pair of resistors in series behaves the same as a single resistor with a resistance equal to the sum of the resistances:*$$R^{series}_{effective} = R_1 + R_2$$.

Inside the resistors, they share the voltage drop in proportion to their fraction of the total resistance, with the large resistance getting the large drop. This seems reasonable since the current has to go through both and will do the most work (lost the most energy) going through the bigger resistance.

There are a couple of traps — "dangerous bends" — in this procedure. It's a bit tricky at first separating the current and voltage. The current is describing the motion of the charges, the voltage (potential) is describing the energy of a charge *when it passes a particular point*. Since the high potential of one end of the battery is what's causing current to flow, there is a temptation to say that "the potential moves around the circuit". This is like saying, since a hill's height is what causes a ball to roll down the hill, "the height of the hill moves down into the valley." We know from our everyday experience that this makes no sense, but since we don't have everyday experience with voltage, we can easily get confused. Keeping the gravitational analogy in mind can help.

A second is to assume that the low potential end of every battery is at 0 potential. We can only choose one 0. This doesn't matter here, but it will when we consider problems with multiple batteries. (See Example: Batteries in series and parallel.) This will be important since our circuit models of neuronal membranes include multiple ion pumps that play the role of batteries.

Joe Redish 2/21/16

#### Follow-ons

Last Modified: May 22, 2019