Resistor circuits

Components are in series if they are joined end to end like this:

[[☃ image 6]]

We'll work with resistors to reveal the properties of the series connection.

Resistors in series

Resistors are in series when they are connected head-to-tail and there are no other wires branching off from the nodes between components.

In the following image, $\text{R1}$, $\text{R2}$, and $\text{R3}$ are in series:

[[☃ image 6]]

Resistors in series share the same current.

The resistors in the following image are not in series. There are extra branches leading away from the nodes between resistors. If these branches carry current (orange arrows), then $\text{R1}$, $\text{R2}$, and $\text{R3}$ do not share the same current.

[[☃ image 8]]

Properties of resistors in series

Here is a circuit with resistors in series:

[[☃ image 12]]

Voltage source $\text V_{\text{S}}$ is connected to the series resistor chain. Voltage $v_{\text S}$ is some constant value, but we don't yet know the current $i$ or how $v_{\text S}$ splits up between the three resistors.

Two things we do know are:

• The three resistor voltages have to add up to $v_{\text S}$.
• Current $i$ flows through all three resistors.

With this little bit of knowledge, and Ohm's Law, we can write these expressions:

$v_{\text{S}} = v_{\text{R1}} + v_{R2} + v_{R3}$

$v_{\text R1} = i\cdot\text R_1 \qquad v_{\text R2} = i\cdot\text R_2 \qquad v_{\text R3} = i\cdot\text R_3 \qquad$

This is enough to get going. Combining equations,

$v_{\text{S}} = i\cdot \text{R1} ,+, i\cdot \text{R2} ,+, i\cdot \text{R3}$

We can factor out the current and gather the resistors in one place:

$v_{\text{S}} = i,, (\text{R1} + \text{R2} + \text{R3})$

Since we know $v_{\text S}$, we solve for the unknown $i$,

$i = \dfrac{v_{\text{S}}} {(\text{R1} + \text{R2} + \text{R3})}$

This looks just like Ohm's Law for a single resistor, except the series resistors appear as a sum.

We conclude:

For resistors in series, the overall resistance is the sum of the individual resistors.

Equivalent series resistor

We can imagine a new bigger resistor equivalent to the sum of the series resistors. It is equivalent in the sense that, for a given $\text V_\text{S}$, the same current $\blueD i$ flows.

$\text R_{\text{series}} = \text{R1} + \text{R2} + \text{R3}$

[[☃ image 9]]

You may hear this slang expression for what is going on: From the "viewpoint" of the voltage source, the three series resistors "look like" one larger equivalent resistor. That means the current, $\blueD i$, provided by the voltage source is the same in both cases.

[[☃ explanation 1]]

If you have several resistors in series, the general form of the equivalent series resistance is,

$\text R_{\text{series}} = \text{R1} + \text{R2} + ... + \text{R}_\text N$

[[☃ image 11]]

Voltage distributes between resistors in series

We worked out the current $i$ through the series connection, so what's left is to figure out the voltage across the individual resistors.

Do this by applying Ohm's Law to the individual resistors,

$v_{\text R1} = i\cdot\text R_1 \qquad v_{\text R2} = i\cdot\text R_2 \qquad v_{\text R3} = i\cdot\text R_3 \qquad$

This becomes more interesting if you do an example, with real numbers. I encourage you to try this on your own before revealing the answer.

[[☃ image 4]]

$\tiny{\text{PROBLEM 1}}$
a. What is the current $\blueD i$ and what are the voltages across the three resistors?

b. Show the individual resistor voltages add up to $\text V_\text S$.

[[☃ explanation 1]]

Reflection

Based on the resistor voltages you just computed:
[[☃ graded-group 1]]

[[☃ graded-group 2]]

[[☃ graded-group 3]]

Summary

Resistors in series share the same current.

Resistors in series add,

$\text R_{\text{series}} = \text{R1} + \text{R2} + ... + \text{R}_\text N$

Voltage distributes amongst series resistors, with the largest voltage across the largest resistor.

[[☃ explanation 1]]

Components are in parallel if they share two nodes, like this:

[[☃ image 6]]

In this article we will work with resistors in parallel, to reveal the properties of the parallel connection. Later articles will cover capacitors and inductors in series and parallel.

Resistors in parallel

Resistors are in parallel when their two terminals connect to the same nodes.

In the following image, $\text{R1}$, $\text{R2}$, and $\text{R3}$ are in parallel. The two distributed nodes are represented by the two horizontal lines.
[[☃ explanation 1]]

[[☃ image 4]]

Resistors in parallel share the same voltage on their terminals.

The resistors in the following image are not in parallel. There are extra components (orange boxes) breaking up the common nodes between resistors. This circuit has four separate nodes, so $\text{R1}$, $\text{R2}$, and $\text{R3}$ do not share the same voltage.

[[☃ image 5]]

Properties of resistors in parallel

Figuring out parallel resistors is a little trickier than series resistors. Here is a circuit with resistors in parallel. (This circuit has a current source. We don't get to use those very often, so this should be fun.)

[[☃ image 9]]

Current source $\text I_\text s$ is driving current $i$ towards $\text{R1}$, $\text{R2}$, and $\text{R3}$. We know the value of current $i$ is some given constant, but we don't yet know the voltage $v$ or how $i$ splits up into the three resistor currents.

Two things we do know are:

• The three resistor currents have to add up to $i$.
• Voltage $v$ appears across all three resistors.

With just this little bit of knowledge, and Ohm's Law, we can write these expressions:

$i = i_{\text{R1}} + i_{\text{R2}} + i_{\text{R3}} $

$v = i_{\text{R1}} \cdot \text{R1} \qquad v = i_{\text{R2}} \cdot \text{R2} \qquad v = i_{\text{R3}} \cdot \text{R3}$

This is enough to get going. Rearrange the three Ohm's Law expressions to solve for current in terms of voltage and resistance:

$i_{\text{R1}} = \dfrac{v}{\text{R1}} \qquad i_{\text{R2}} = \dfrac{v}{\text{R2}} \qquad i_{\text{R3}} = \dfrac{v}{\text{R3}}$

Substitute these expressions into the sum of currents:

$i = \dfrac{v}{\text{R1}} +\dfrac{v}{\text{R2}} + \dfrac{v}{\text{R3}}$

Factor out the common term $v$,

$i = v \left (\dfrac{1}{\text{R1}} +\dfrac{1}{\text{R2}} + \dfrac{1}{\text{R3}} \right )$

Now remember that we already know $i$ (it is a property of the current source), so we can solve for $v$:

$v = i ,\dfrac{1}{\left (\dfrac{1}{\text{R1}} +\dfrac{1}{\text{R2}} + \dfrac{1}{\text{R3}} \right )}$

This expression looks just like Ohm's Law, $v = i,\text R$, but with the parallel resistors appearing in a double-reciprocal in place of of a single resistor.

We conclude:

For resistors in parallel, the overall resistance is the reciprocal of the sum of reciprocals of the individual resistors.

(This sounds complicated, but we will derive something simpler before we are done.)

Equivalent parallel resistor

The previous equation suggests we can define a new resistor, equivalent to the parallel resistors. The new resistor is equivalent in the sense that, for a given $i$, the same voltage $v$ appears.

$\text R_{\text{parallel}} = \dfrac{1}{\left (\dfrac{1}{\text{R1}} +\dfrac{1}{\text{R2}} + \dfrac{1}{\text{R3}} \right )}$

The equivalent parallel resistor is the reciprocal of the sum of reciprocals. We can write this equation another way by rearranging the giant reciprocal,

$\dfrac{1}{\text R_{\text{parallel}}} = \dfrac{1}{\text{R1}} +\dfrac{1}{\text{R2}} + \dfrac{1}{\text{R3}}$

Ohm's Law applied to parallel resistors,

$v = i ,\text{R}_{\text{parallel}}$

[[☃ image 6]]

From the "viewpoint" of the current source, the equivalent resistor $R_{\text{parallel}}$ is indistinguishable from the three parallel resistors, because in both circuits, $v$ is the same.

If you have multiple resistors in parallel, the general form of the equivalent parallel resistance is,

$\dfrac{1}{\text R_{\text{parallel}}} = \dfrac{1}{\text{R1}} +\dfrac{1}{\text{R2}} + ... + \dfrac{1}{\text{R}_\text N}$

[[☃ image 8]]

Current distributes between resistors in parallel

We worked out the voltage $v$ across the parallel connection, so what's left to figure out is the currents through the individual resistors.

Do this by applying Ohm's Law to the individual resistors.

$v = i_{\text R1}\cdot\text R_1 \qquad v = i_{\text R2}\cdot\text R_2 \qquad v = i_{\text R3}\cdot\text R_3 \qquad$

This becomes more informative if you do an example with real numbers.

[[☃ image 7]]

Find the voltage $v$ and the currents through the three resistors.
Show that the individual resistor currents add up to $i$.
[[☃ explanation 1]]

[[☃ explanation 4]]

Reflection

Based on the resistor currents you just solved:

[[☃ graded-group 1]]

[[☃ graded-group 2]]

[[☃ graded-group 4]]

[[☃ graded-group 5]]

[[☃ graded-group 3]]

Special case - two resistors in parallel

[[☃ image 3]]

Two resistors in parallel have an equivalent resistance of:

$\text R_{\text{parallel}} = \dfrac{1} {\left (\dfrac{1}{\text{R1}} +\dfrac{1}{\text{R2}} \right )}$

It's possible do a bit of manipulation to eliminate the reciprocals and come up with another expression with just one fraction. Rather than just telling you the answer, it is a rite of passage to work through the algebra the first time. The answer is tucked away so you can try this on your own before peeking.

[[☃ explanation 1]]

Special special case - two equal resistors in parallel

If two resistors in parallel have the same value, what is the equivalent $R_{\text{parallel}}$?

Let $\text{R1} , \text{R2} = \text R$

$R_{\text{parallel}} = \dfrac{\text R \cdot \text R}{\text R + \text R} = \dfrac{\text R \cdot \text R}{2\text R}$

$\large R_{\text{parallel}} = \frac{1}{2} \text R$

Two identical resistors in parallel have an equivalent resistance half the value of either resistor. The current splits equally between the two.

Summary

Resistors in parallel share the same voltage.

The general form for three or more resistors in parallel is,

$\dfrac{1}{\text R_{\text{parallel}}} = \dfrac{1}{\text{R1}} +\dfrac{1}{\text{R2}} + ... + \dfrac{1}{\text{R}_\text N} $

For two parallel resistors it is usually easier to combine them as the product over the sum:

$\text R_{\text{parallel}} = \dfrac{\text{R1}\cdot\text{R2}}{\text R1+\text R2}$

$\text R_{\text{parallel}}$ is always smaller than the smallest parallel resistor.

Current distributes amongst parallel resistors, with the largest current flowing through the smallest resistor.

[[☃ explanation 1]]

In a previous article we studied parallel resistors.

[[☃ image 3]]

We derived this equation to combine parallel resistors into a single equivalent resistor,

$\text R_{\text{parallel}} = \dfrac{1}{\left (\dfrac{1}{\text{R1}} +\dfrac{1}{\text{R2}} + ... + \dfrac{1}{\text{R}_{\text N}} \right )}$

This is a fairly complex expression, with $1/\text R$ terms embedded inside another reciprocal. There is an alternate way to approach this problem, using the concept of conductance.

Conductance

Ohm's Law, $v = i,\text R$, defines resistance as the ratio of voltage over current,

$\text R = \dfrac{v}{i}$

The term conductance is the inverse of this expression. It is the ratio of current over voltage,

$\text G = \dfrac{i}{v}$

This gives us yet another way to write Ohm's Law,

$i = v,\text G$

The unit of conductance is the siemens, abbreviated $\text S$. It is named after Werner von Siemens, founder of the German industrial electronics and telecommunications company that bears his name. There is an s at the end of siemens even if it is singular, $1,\text{siemens}$. You may come across an older term, the mho, used as the unit of conductance. Mho is just "ohm" spelled backwards. That term isn't used anymore.

Using conductance instead of resistance for the same physical object simply emphasizes a different aspect of its behavior. Resistance reduces or impedes current flow, while conductance allows current to pass through. The terms are two aspects of the same idea. [[☃ explanation 2]]

A $100,\Omega$ resistor is the same as a conductance of $\dfrac{1}{100,\Omega}$ $= 0.01 ,\text S$.

[[☃ image 4]]

Parallel conductance

In this section, we'll repeat the analysis of parallel resistors, but this time, instead of calling each component a resistor, we will call it a conductance. The result for parallel conductance will have a strong resemblance to series resistors.
[[☃ explanation 1]]

Here is a circuit with conductances in parallel. We will analyze this circuit using the language of conductance, and the conductance form of Ohm's Law, $i = v,\text G.$

[[☃ image 3]]

The value of current $i$ is some given constant. We don't yet know $v$ or how $i$ splits up into three currents through the conductances.

Two things we do know are:

• The three conductance currents add up to $i$.
• Voltage $v$ appears across all three conductances.

With just this little bit of knowledge, and the conductance form of Ohm's Law, we can write these expressions:

$i = i_{\text{G1}} + i_{\text{G2}} + i_{\text{G3}} $

$i_{\text{G1}} = v \cdot \text{G1} \qquad i_{\text{G2}} = v \cdot \text{G2} \qquad i_{\text{G3}} = v \cdot \text{G3}$

This is enough to get going. Combining equations:

$i = v\cdot \text{G1} ,+, v\cdot \text{G2} ,+, v\cdot \text{G3}$

Factor out the voltage term and gather the conductance values in one place:

$i = v,, (\text{G1} + \text{G2} + \text{G3})$

This looks just like Ohm's Law for a single conductance, with the parallel conductances appearing as a sum.

We conclude:

For conductances in parallel, the overall conductance is the sum of the individual conductances.

Notice how much this looks like the formula for resistors in series. Conductances in parallel are like resistances in series, they add.

Equivalent parallel conductances

We can imagine a new conductance equivalent to the sum of the parallel conductances. It is equivalent in the sense that the same voltage appears.

$\text G_{\text{parallel}} = \text{G1} + \text{G2} + \text{G3}$

[[☃ image 3]]

Conductance example

Let's solve the same circuit we did for parallel resistors, but using the new representation.

This is the circuit with conductances, $\text G = \dfrac{1}{\text R}$

[[☃ image 6]]

You can try to solve this yourself before looking at the answer. We want to find voltage $v$ and the individual currents, $i_{\text{G1}}$, $i_{\text{G2}}$, and $i_{\text{G3}}$, using the conductance form of Ohm's Law, $i = v,\text G$.

Find $v$ and the current through the three conductances.
Show that the individual currents add up to $i$.
[[☃ explanation 1]]

Summary

Conductances in parallel combine with a simple sum. The two ways to combine parallel resistors are:

$\text G_{\text{parallel}} = \text{G1} + \text{G2} + ... + \text G_\text N$

$\text R_{\text{parallel}} = \dfrac{1}{\left (\dfrac{1}{\text{R1}} +\dfrac{1}{\text{R2}} + ... + \dfrac{1}{\text{R}_{\text N}} \right )}$

The sum of conductances is simpler than the "reciprocal of reciprocals" we came up with for parallel resistors, and there are no special-case formulas to remember. This is the main reason to introduce the concept of conductance. The reciprocals did not go away, we just did them at the beginning when we derived $\text G$ values from the given $\text R$'s. Using conductance represents a rearrangement of the same computation.

How you choose to analyze parallel circuits, $\text G$ or $\text R$, is a matter of convenience and simplicity.

[[☃ explanation 1]]

Complicated resistor networks can be simplified by combining series and parallel resistors. Consider this example circuit:

[[☃ image 3]]

The diagram shows a voltage source connected to a resistor network. The two small circles at the left end represent the ports of the resistor network.

Imagine we want to calculate how much current flows from the voltage source. The answer is not immediately obvious, since there are many resistors and branches. However, by following a systematic process, we can combine the resistors until the resistor network is reduced to one equivalent resistor.

** Strategy for simplifying a resistor network **

• Begin as far away as possible from the circuit location in question (which in this case is the voltage source).

• Replace series or parallel resistors with their equivalent resistor.

• Continue one step at a time until a single equivalent resistor represents the entire resistor network.

##Example

The location in question is the input voltage source, so we start the simplification process on the far right and work our way toward the source.

Simplifying a circuit is a process of many small steps. Consider a chunk of circuit, simplify, then move to the next chunk.

Tip: Redraw the schematic after every step so you don't miss an opportunity to simplify.

Step 1. The shaded resistors, $2,\Omega$ and $8,\Omega$, are in series.

[[☃ image 4]]

[[☃ graded-group 1]]

The two resistors can be replaced by their equivalent resistance:

[[☃ image 6]]

Key insight: From outside the shaded box, the two series resistors and the equivalent resistor are indistinguishable from each other. The exact same current and voltage exist in both versions.

Step 2. We now find two $10,\Omega$ resistors in parallel at the new far right of the circuit.

[[☃ image 4]]

[[☃ graded-group 1]]

[[☃ image 6]]

Again, looking into the shaded box from the left, the current and voltage with the equivalent resistor is still indistinguishable from the original circuit.

Step 3. A pattern is emerging. We're working through the schematic from right to left, simplifying and redrawing as we go. Next we find two series resistors, $1,\Omega$ and $5,\Omega$.

[[☃ image 4]]

[[☃ graded-group 1]]

[[☃ image 5]]

Step 4. Now we have three resistors in parallel.

[[☃ image 4]]

[[☃ graded-group 1]]

[[☃ image 5]]

Step 5. We're down to the last two resistors, which are in series.

[[☃ image 4]]

You can do this one in your head:

[[☃ image 6]]

We're left with a single $3,\Omega$ resistor. It represents the entire network as far as the voltage source is concerned. The current demanded from the voltage source is:

$\text{I} = \dfrac{\text V}{3,\Omega}$

We started with $7$ resistors and simplified down to $1$, a significant reduction in complexity.

Just for fun, here's an animation of the circuit simplification:

[[☃ video 1]]

Sometimes when you are simplifying a resistor network, you get stuck. Some resistor networks cannot be simplified using the usual series and parallel combinations. This situation can often be handled by trying the $\Delta - \text Y$ transformation, or 'Delta-Wye' transformation.

The names Delta and Wye come from the shape of the schematics, which resemble letters. The transformation allows you to replace three resistors in a $\Delta$ configuration by three resistors in a $\text Y$ configuration, and the other way around.

[[☃ image 8]]

The $\Delta - \text Y$ drawing style emphasizes these are 3-terminal configurations. Something to notice is the different number of nodes in the two configurations. $\Delta$ has three nodes, while $\text Y$ has four nodes (one extra in the center).

The configurations can be redrawn to square up the resistors. This is called a $\pi - \text T$ configuration,

[[☃ image 10]]

The $\pi - \text T$ style is a more conventional drawing you would find in a typical schematic. The transformation equations developed next apply to $\pi - \text T$ as well.

$\Delta - \text Y$ transformation

For the transformation to be equivalent, the resistance between each pair of terminals must be the same before and after. It is possible to write three simultaneous equations to capture this constraint.

[[☃ image 3]]

Consider terminals $x$ and $y$ (and for the moment assume terminal $z$ isn't connected to anything, so the current in $\text R3$ is $0$ ). In the $\Delta$ configuration, the resistance between $x$ and $y$ is $Rc$ in parallel with $Ra +Rb$ .

On the $\text Y$ side, the resistance between $x$ and $y$ is the series combination $R1+R2$ (again, assume terminal $z$ isn't connected to anything, so $\text R1$ and $\text R2$ carry the same current and can be considered in series). We set these equal to each other to get the first of three simultaneous equations,

$R1+R2 = \dfrac{Rc,(Ra+Rb)}{Rc+(Ra+Rb)}$

We can write two similar expressions for the other two pairs of terminals. Notice the $\Delta$ resistors have letter names, $(Ra$, etc.$)$ and the $\text Y$ resistors have number names, $(R1$, etc.$)$.

After solving the simultaneous equations (not shown), we get the equations to transform either network into the other.

$\Delta \rightarrow \text Y$ transformation

Equations for transforming a $\Delta$ network into a $\text Y$ network:

$R1 = \dfrac{Rb,Rc}{Ra + Rb + Rc}$

$R2 = \dfrac{Ra,Rc}{Ra + Rb + Rc}$

$R3 = \dfrac{Ra,Rb}{Ra + Rb + Rc}$

Transforming from $\Delta$ to $\text Y$ introduces one additional node.

$\text Y \rightarrow\Delta$ transformation

Equations for transforming a $\text Y$ network into a $\Delta$ network:

$Ra = \dfrac{R1,R2 + R2,R3 + R3,R1}{R1}$

$Rb = \dfrac{R1,R2 + R2,R3 + R3,R1}{R2}$

$Rc = \dfrac{R1,R2 + R2,R3 + R3,R1}{R3}$

Transforming from $\text Y$ to $\Delta$ removes one node.

Example

Let's do a symmetric example. Assume we have a $\Delta$ circuit with $3,\Omega$ resistors. Derive the $\text Y$ equivalent by using the $\Delta \rightarrow \text Y$ equations.

$R1 = \dfrac{Rb,Rc}{Ra + Rb + Rc} = \dfrac{3 \cdot 3}{3 + 3 + 3} = 1,\Omega$

$R2 = \dfrac{Ra,Rc}{Ra + Rb + Rc} = \dfrac{3 \cdot 3}{3 + 3 + 3} = 1,\Omega$

$R3 = \dfrac{Ra,Rb}{Ra + Rb + Rc} = \dfrac{3 \cdot 3}{3 + 3 + 3} = 1,\Omega$

[[☃ image 3]]

Going in the other direction, from $\text Y \rightarrow\Delta$, looks like this,

$Ra = \dfrac{R1,R2 + R2,R3 + R3,R1}{R1} = \dfrac{1\cdot1 + 1\cdot1 + 1\cdot1}{1} = 3,\Omega$

$Rb = \dfrac{R1,R2 + R2,R3 + R3,R1}{R2} = \dfrac{1\cdot1 + 1\cdot1 + 1\cdot1}{1} = 3,\Omega$

$Rc = \dfrac{R1,R2 + R2,R3 + R3,R1}{R3} = \dfrac{1\cdot1 + 1\cdot1 + 1\cdot1}{1} = 3,\Omega$

Example

Now for an example that's a little less tidy. We want to find the equivalent resistance between the top and bottom terminals.

[[☃ image 6]]

Try as we might, there are no resistors in series or in parallel. But we are not stuck. First, let's redraw the schematic to emphasize we have two $\Delta$ connections stacked one on the other.

[[☃ image 8]]

Now select one of the $\Delta$'s to convert to a $\text Y$. We will perform a $\Delta \rightarrow \text Y$ transformation and see if it breaks the logjam, opening up other opportunities for simplification.

We go to work on the bottom $\Delta$ (an arbitrary choice). Very carefully label the resistors and nodes. To get the right answers from the transformation equations, it is critical to keep the resistor names and node names straight. $Rc$ must connect between nodes $x$ and $y$, and so on for the other resistors. Refer to Diagram 1 above for the labeling convention.

[[☃ image 10]]

When we perform the transform on the lower $\Delta$, the black $\Delta$ resistors will be replaced by the new gray $\text Y$ resistors, like this:

[[☃ image 11]]

Perform the transform yourself before looking at the answer. Check that you select the right set of equations.
Compute three new resistor values to convert the $\Delta$ to a $\text Y$, and draw the complete circuit.

[[☃ explanation 1]]

And voilà! Check out our circuit. It now has series and parallel resistors where it had none before. Continue simplification with series and parallel combinations until we get down to a single resistor between the terminals. Redraw the schematic again to square up the symbols into a familiar style.

[[☃ image 7]]

We proceed through the remaining simplification steps just as we did before in the article on Resistor Network Simplification.

On the left branch, $3.125 + 1.25 = 4.375 ,\Omega$

On the right branch, $4 + 1 = 5,\Omega$

[[☃ image 9]]

The two parallel resistors combine as $4.375,||, 5 = \dfrac{4.375 \cdot 5}{4.375 + 5} = 2.33,\Omega$

And we finish by adding the last two series resistors together,

$R_{equivalent} = 2.33 + 1.66 = 4,\Omega$

[[☃ image 10]]

Summary

$\Delta - \text Y$ transformations are another tool in your bag of tricks for simplifying circuits prior to detailed analysis.

Don't memorize the transformation equations. If the need arises, you can look them up.

[[☃ explanation 1]]

A very common and useful series resistor circuit goes by the nickname voltage divider. We will work out how this circuit operates, and you will see where the nickname comes from.

A voltage divider looks like this:

[[☃ image 7]]

Our goal is to come up with an expression relating output $v_{out}$ to input $v_{in}$. A good place to start is finding the current through $\text{R1}$ and $\text{R2}$.

Assumption: Assume $0$ current is flowing out of the divider. (Before we are done we will check to see what happens if this zero-current assumption doesn't hold).

[[☃ image 9]]

With this assumption, $\text{R1}$ and $\text{R2}$ have the same current, and we can consider them to be in series.

$i_{\text{1}} = i_{\text{2}}\qquad$ and for now let's just call this $i$.

To find the current, we apply Ohm's law and what we know about resistors in series, (reminder: resistors in series add),

$v = i,\text{R}\qquad$ Ohm's Law

$v_{in} = i ,(\text{R1} + \text{R2})$

Rearranging to solve for $i$,

$i = v_{in},\dfrac{1}{\text{R1} + \text{R2}}$

We've solved for current $i$ in terms of $v_{in}$ and both resistors.

Next, we write an expression for $v_{out}$ using Ohm's Law,

$v_{out} = i , \text{R2}$

We can substitute for $i$ in the previous equation to get,

$v_{out} = \left (v_{in},\dfrac{1}{\text{R1} + \text{R2}} \right ) \text{R2}$

and we have derived the voltage divider equation:

[[☃ image 8]]

The output voltage equals the input voltage scaled by a ratio of resistors: the bottom resistor divided by the sum of the resistors.

The ratio of resistors is always less than $1$ for any values of $\text{R1}$ and $\text{R2}$. This means $v_{out}$ is always less than $v_{in}$. Input voltage $v_{in}$ is scaled down to $v_{out}$ by a fixed ratio determined by the resistor values. This is where the circuit gets its nickname: voltage divider.

Example - use the voltage divider equation to find $v_{out}$

We want to find $v_{out}$ using the voltage divider relationship.

$v_{out} = v_{in},\dfrac{\text{R2}}{\text{R1} + \text{R2}}$

[[☃ image 7]]

We insert the actual input voltage and resistor values into the equation, remembering the equation tells us the bottom resistor, $\text{R2}$, goes in the numerator.

$v_{out} = 12,\text V\cdot\dfrac{3,\text k\Omega}{1,\text k\Omega + 3,\text k\Omega}$

$v_{out} = 12,\text V\cdot\dfrac{3,\text k\Omega}{4,\text k\Omega}$

$v_{out} = 12,\text V\cdot\dfrac{3}{4} = 9 ,\text V$

Let's do an optional step to check the current.

$i = \dfrac{v_{in}}{\text R1 + \text R2} = \dfrac{12,\text V}{1,\text k\Omega + 3,\text k\Omega} = \dfrac{12,\text V}{4,\text k\Omega} = 3,\text{mA}$

Now we know the current, so we can compute how much power is being dissipated by our voltage divider,

$p = i,v = 3,\text{mA} \cdot 12,\text V = 36,\text{mW}$

Summary: Our voltage divider takes an input voltage (in this case $12,\text V$, but it could be any value) and scales it down to create an output voltage that's $3/4$ of the input voltage. The $3/4$ ratio is determined by our choice of the two resistors. As long as $v_{in}$ is turned on, a current of $3,\text{mA}$ flows down through the voltage divider, so it dissipates $12,\text V \times 3,\text{mA} = 36,\text{mW}$.

Voltage divider practice problems

All of these problems use this circuit diagram,

[[☃ image 1]]

[[☃ graded-group 1]]

[[☃ explanation 1]]

[[☃ graded-group 2]]

[[☃ explanation 2]]

[[☃ graded-group 3]]

[[☃ explanation 3]]

[[☃ graded-group 4]]

[[☃ explanation 4]]

Review the assumption (advanced)

A voltage divider doesn't do anything useful unless its output is connected to something. You should be aware of what happens when a divider is connected to a load.
Remember we made an assumption at the beginning? We assumed the current flowing out of the output was $0$. That let us treat $\text{R1}$ and $\text{R2}$ as if they were in series, and we developed the voltage divider equation. Let's check what happens if the assumption is not true.

Operating the voltage divider near its mid-range

To start this discussion, we let $\text{R1} = \text{R2} $. With matched resistors, the expected $v_{out}$ of the voltage divider is the mid-point of the divider's range, $0.5,v_{in}$. To cause some current to flow out of the divider, we connect a resistor $\text R_{\text L}$. Does the divider still work? Does our voltage divider story collapse?

[[☃ image 4]]

Resistor $\text R_\text L$ acts as a load on the output of the voltage divider, meaning that it causes a current $i_\text L$ to flow. The presence of $\text R_\text L$ means $\text{R1}$ and $\text{R2}$ are no longer strictly in series. Let's make $\text{R}{\text L}$ fairly big, to make $i{\text L}$ fairly small relative to $i_2$. Let $\text R_\text L$ be ten times bigger than $\text{R2}$,

$\text{R}_{\text L} = 10,\text{R2}$

$\text{R2}$ and $\text{R}{\text L}$ are in parallel with each other. Combine the two parallel resistors using the parallel resistor formula to get $\text{R2} ,||, \text{R}{\text L}$,

$\text{R2} ,||, \text{R}{\text L} = \dfrac{\text{R2}\cdot\text{R}{\text L}}{\text{R2}+\text{R}_{\text L}} = \dfrac{\text{R2}\cdot10,\text{R2}}{\text{R2}+10,\text{R2}} = \dfrac{10}{11},\text{R2} = 0.91,\text{R2}$

[[☃ explanation 5]]
[[☃ explanation 3]]

This is our loaded voltage divider circuit, redrawn to show the equivalent resistance of $\text R2$ in parallel with $\text R_\text L$,

[[☃ image 6]]

The $10{\times}$ load resistor has the effect of reducing the resistance at the bottom of the voltage divider by roughly $9%$. What is the impact of this additional load on the divider's output voltage? Without the load, the expected output is $0.5,v_{in}$. Now we figure out the output voltage in the presence of a load resistor.

$v_{out} = v_{in}, \dfrac{0.91, \text R2}{\text R1 + 0.91, \text R2} $

We designed our divider with $\text R1 = \text R2$, so they cancel out,

$v_{out} = v_{in}, \dfrac{0.91}{1 + 0.91}$

$v_{out} = v_{in}, \dfrac{0.91}{1 .91} = 0.48,v_{in}$

The output voltage drops to $48%$ of the input voltage. How big an error is this?

$\dfrac{0.48}{0.50} = 0.96 = 96%$

The actual output of the voltage divider is low by $4%$ compared to the expected voltage. (Note the voltage error of $4%$ is significantly less than the $9%$ resistance change.) Does a few $%$ error matter? That's for you alone to decide. It depends on how accurate the voltage divider needs to be for your application.

The nugget to tuck away from this analysis: If the effective load resistance is $10{\times}$ greater than the bottom resistor in the voltage divider, you get roughly "one hand" of $%$ error $(4-5%)$ in the output voltage. This holds when the output voltage is near the center of its range (in the neighborhood of $v_{\text{in}}/2$).

Operating the voltage divider near its extremes

If you design the voltage divider to operate near its extremes, with the output voltage close to $0$ or $v_{\text{in}}$, the percentage error in output voltage will be different. We repeat the analysis with the output voltage set to $90%$ and $10%$ of the divider range. We keep the load resistor ten times the bottom resistor, so the parallel combination of $\text R2$ and $\text R_{\text L}$ is still $0.91,\text R2$.

Case 1: $v_{out} = 90%$ of $v_{in}$

Let $v_{out} = 90%$ of $v_{in}$. The expected output is $0.90 ,v_{in}$.

First we design a voltage divider that gives us the desired output. Figure out $\text R2$ in terms of $\text R1$ for a $90%$ voltage divider,

$\dfrac{v_{out}}{v_{in}} = 0.90 = \dfrac{\text R2}{\text R1 + \text R2}$

$0.90 ,(\text R1 + \text R2) = \text R2$

$0.90 ,\text R1 = \text R2 - 0.90,\text R2$

$0.90 ,\text R1 = 0.10 ,\text R2$

$\text R2 = \dfrac{0.90,\text R1}{0.10} = 9,\text R1$

$\text R2$ is $9$ times bigger than $\text R1$.

Now we load the circuit with $\text R_\text L$ and see how the output voltage changes. The expression we derived above for the loaded voltage divider is,

$\dfrac{v_{out}}{v_{in}} = \dfrac{0.91, \text R2}{\text R1 + 0.91, \text R2} $

We replace $\text R2$ with $9,\text R1$,

$\dfrac{v_{out}}{v_{in}} = \dfrac{0.91 ,(9,\text R1)}{\text R1 + \text 0.91,(9,\text R1)}$

All the $\text R1$'s cancel out, leaving,

$\dfrac{v_{out}}{v_{in}} = \dfrac{0.91 ,(9)}{1 + 0.91,(9)} = \dfrac{8.19}{9.19} = 0.89$

The actual output voltage is $89%$ of $v_{in}$ instead of $90%$.

The actual output voltage divided by the expected output is,

$\dfrac{0.89}{0.90} = 0.99$

So the actual voltage is lower than the expected by only $1%$.

Case 2: $v_{out} = 10%$ of $v_{in}$

Let $v_{out} = 10%$ of $v_{in}$. The expected output is $0.10 ,v_{in}$.

Express $\text R1$ in terms of $\text R2$ for a $10%$ voltage divider.

$\dfrac{v_{out}}{v_{in}} = 0.10 = \dfrac{\text R2}{\text R1 + \text R2}$

$0.10 ,(\text R1 + \text R2) = \text R2$

$0.10 ,\text R1 = \text R2 - 0.10,\text R2$

$0.10 ,\text R1 = 0.90 ,\text R2$

$\text R1 = \dfrac{0.90,\text R2}{0.10} = 9,\text R2$

$\text R1$ is $9$ times bigger than $\text R2$.

Now we load the circuit with $\text R_\text L$ and evaluate the change in output voltage. The expression we derived above for the loaded voltage divider is,

$\dfrac{v_{out}}{v_{in}} = \dfrac{0.91, \text R2}{\text R1 + 0.91, \text R2} $

We replace $\text R1$ with $9,\text R2$,

$\dfrac{v_{out}}{v_{in}} = \dfrac{0.91 ,\text R2}{9,\text R2 + \text 0.91,\text R2}$

All the $\text R2$'s cancel out,

$\dfrac{v_{out}}{v_{in}} = \dfrac{0.91}{9 + 0.91} = \dfrac{0.91}{9.91} = 0.092$

The actual output voltage is $9.2%$ of $v_{in}$ instead of the expected $10%$.

The actual output voltage divided by the expected output is,

$\dfrac{0.092}{0.10} = 0.92$

So the actual voltage differs by $8%$ from the expected. This is nearly twice the error compared to the mid-range divider.

Lessons for a loaded voltage divider

With a $10\times\text R2$ load resistor connected to a voltage divider:

• Near mid-range, the output voltage is reduced by $5%$.

• Near the top of its range, the error goes down substantially, to around $1%$.

• Near the bottom of its range, the error roughly doubles compared to mid-range. The output voltage is $8%$ lower than expected.

Controlling error in a loaded voltage divider

If your design requires the error to be significantly smaller, the load needs to be much larger than $10\times\text{R2}$, like an additional $10{\times}$ or more. You can get an additional $10{\times}$ two ways. Increase the load resistance. Or, redesign the voltage divider to have smaller $\text{R1}$ and $\text{R2}$, (at the cost of more power dissipated by the voltage divider).

Real-world resistor tolerance also impacts accuracy

Real-world resistors always have a $\pm$ tolerance on their value. If the accuracy of the voltage divider is critical to your application, use resistors with tight tolerances, and check for acceptable performance by analyzing the voltage divider at the anticipated tolerance extremes.

What's in a nickname

We mentioned at the start that the nickname of this circuit is a voltage divider. In many situations, that is exactly what it does. However, we showed that under certain conditions when there is a load on the divider, the actual output voltage is slightly lower than the value predicted by the voltage divider equation. The lesson: Call a circuit by its nickname; just remember that it's only a nickname.

Summary

[[☃ image 3]]

Voltage divider:

$v_{out} = v_{in},\dfrac{\text{R2}}{\text{R1} + \text{R2}}$

where $\text{R2}$ is the resistor on the bottom of the divider.

[[☃ explanation 1]]