The potential divider circuit — Physics

Before this

This leads to

The transformer — stepping voltage up or down
Charging and discharging capacitors · coming soon

You have 9 volts, and you need 3

Here is a common problem in electronics. Your supply gives 9.0 V. One part of your circuit needs only 3.0 V. You could buy a different supply. But there is a cheaper way, and it uses only two resistors and the series rules you already have.

There is a second problem hiding behind the first. The lecture on Ohm's law met a thermistor whose resistance falls from 10 kΩ at 25 °C to about 3.6 kΩ at 50 °C. It said that a circuit can measure this change and use it as a thermometer. But a change in resistance is not something a display or a computer chip can read. Electronics reads potential differences. So how do you turn a resistance into a p.d.?

One circuit answers both questions. It is called the potential divider, and once you know it, you will find it inside almost every electronic device.

Two resistors share the supply

Connect two resistors, R₁ and R₂, in series across the supply. Take the output, $V_{out}$ , across R₂. That is the whole circuit.

A potential divider. The potentials, measured from the negative terminal, fall from 9.0 V to 3.0 V across R₁, and from 3.0 V to 0 V across R₂. The output is the p.d. across R₂.

Why is the output 3.0 V? Follow the reasoning one step at a time.

Step 1. The resistors are in series, so the same current $I$ flows through both. Charge has only one path, and it cannot pile up anywhere.
Step 2. Their total resistance is $R_{1} + R_{2} = 9.0 k Ω$ . So $I = 9.0 \div 9000 = 1.0 mA$ .
Step 3. The p.d. across R₂ is $I R_{2} = 0.0010 \times 3000 = 3.0$ V. The p.d. across R₁ is $0.0010 \times 6000 = 6.0$ V.
Step 4. Check: 6.0 V + 3.0 V = 9.0 V. The two p.d.s add up to the supply, as energy conservation demands.

Now do the same steps with letters. The current is $I = \frac{V _{in}}{R _{1} + R _{2}}$ , and the output is $V_{out} = I R_{2}$ . Put the first into the second:

V_{out} = \frac{V _{in}}{R _{1} + R _{2}} \times R_{2}

The potential divider equation

For two resistors in series across a supply $V_{in}$ , with the output taken across R₂:

V_{out} = V_{in} \times \frac{R _{2}}{R _{1} + R _{2}}

The p.d.s across the two resistors are in the same ratio as their resistances: $\frac{V _{1}}{V _{2}} = \frac{R _{1}}{R _{2}}$ .

Read the equation as a sentence. The output is a fraction of the input, and the fraction is "R₂'s share of the total resistance". A fraction of a whole is always between 0 and 1, so the output is always between 0 V and $V_{in}$ . The bigger resistor takes the bigger share of the p.d. This is because the same current flows through both, so the one with more resistance needs more p.d. to push that current through it.

That is it. That is the whole idea. Everything in this lecture is that one sentence, used in different ways.

Stop and think

In the figure, what happens to $V_{out}$ if both resistors are changed to ten times their values, 60 kΩ and 30 kΩ?

Show the answer

Nothing. $V_{out}$ is still 3.0 V. R₂'s share of the total is $\frac{30}{90} = \frac{1}{3}$ , exactly as before. The current is ten times smaller, 0.10 mA, but each resistor is ten times bigger, so each p.d. is unchanged. The output depends only on the ratio of the resistances, not their sizes. (The sizes do matter for something else: how much current the divider wastes, and how easily a load disturbs it. Both come later in this lecture.)

Test that idea before reading on.

Predict — double both resistors

A 6.0 V supply feeds R₁ = 1.0 kΩ and R₂ = 2.0 kΩ, so the output across R₂ is 4.0 V. Now R₁ is changed to 2.0 kΩ and R₂ to 4.0 kΩ. What is the new output?

This box works out what happens when both resistors in a divider are doubled.

Before: R₂'s share is 2.0 ÷ (1.0 + 2.0) = 2/3, so the output is 6.0 × 2/3 = 4.0 V. After: R₂'s share is 4.0 ÷ (2.0 + 4.0) = 2/3, so the output is still 4.0 V. The current halves, from 2.0 mA to 1.0 mA, but each resistance doubles, so each p.d. stays the same. Only the ratio of the resistances matters.

Worked examples

Worked Example 1 — Three volts from nine

R₁ = 6.0 kΩ, R₂ = 3.0 kΩ and $V_{in} = 9.0$ V. Find $V_{out}$ across R₂ by two methods.

Think first. Which share does R₂ have? 3.0 kΩ out of 9.0 kΩ is one third.

By the equation: $V_{out} = 9.0 \times \frac{3.0}{6.0 + 3.0} = 9.0 \times \frac{1}{3} = 3.0$ V. The kilohms cancel in the fraction, so there is no need to convert to ohms.

By the ratio: the p.d.s are in the ratio 6.0 : 3.0 = 2 : 1. Split 9.0 V into 3 equal parts of 3.0 V. R₁ gets 2 parts (6.0 V) and R₂ gets 1 part (3.0 V).

Check. The smaller resistor has the smaller share: 3.0 V is less than 6.0 V. ✓

Worked Example 2 — Designing a divider

You need 2.0 V from a 12 V supply. You have a 1.0 kΩ resistor for R₂. What value should R₁ have?

Think first. R₂ must take 2.0 V, so R₁ must take the rest: 12 − 2.0 = 10 V.

The p.d.s are in the ratio of the resistances: $\frac{R _{1}}{R _{2}} = \frac{10}{2.0} = 5$ . So R₁ = 5 × 1.0 kΩ = 5.0 kΩ.

Check. $12 \times \frac{1.0}{5.0 + 1.0} = 2.0$ V. ✓

Turning a sensor into a signal

Now replace one fixed resistor with a component whose resistance changes with the world around it.

A thermistor (NTC type, as in the Ohm's law lecture) has a resistance that falls as its temperature rises.
A light-dependent resistor (LDR) has a resistance that falls as the light on it gets brighter.

The sensor's resistance changes, so its share of the total changes, so the output p.d. changes. A resistance has become a p.d. that electronics can read.

Which way does the output move? Ask one question: is the sensor the component the output is taken across? If it is, the output follows the sensor's resistance. If the sensor is the other component, the output moves the opposite way, because the fixed resistor's share rises when the sensor's share falls.

The explorer below uses the thermistor from the Ohm's law lecture with a fixed 10 kΩ resistor and a 6.0 V supply. (The thermistor's values at other temperatures come from a standard model of a typical thermistor, fitted to 10 kΩ at 25 °C and 3.6 kΩ at 50 °C.)

Explore — a thermometer from two components

Change the temperature, and watch the thermistor's resistance and the output. Then move the thermistor to the other position and try again.

Temperature: 25 °C

This table shows the output of a 6.0 V divider made from the thermistor and a fixed 10 kΩ resistor, with the thermistor in each position.

Temperature (°C)	Thermistor (kΩ)	Output, thermistor at bottom (V)	Output, thermistor at top (V)
0	≈ 33.5	≈ 4.62	≈ 1.38
25	10.0	3.00	3.00
50	≈ 3.60	≈ 1.59	≈ 4.41

With the thermistor at the bottom, the output falls as it gets hotter. With it at the top, the output rises. At 25 °C both resistances are 10 kΩ, so the output is half the supply in both positions.

You should have seen two things. First, the thermistor's resistance falls steeply at low temperatures and more slowly when warm, so the output is not a straight-line thermometer. Real circuits correct for this with a table or a formula. Second, swapping the positions turned a falling output into a rising one. The designer chooses the position to suit the job.

Worked Example 3 — A light sensor

An LDR has a resistance of about 10 kΩ in dim light and 500 Ω in bright light (typical values for a small LDR). It is in series with a fixed 2.0 kΩ resistor across a 5.0 V supply. The output is taken across the fixed resistor. Find the output in dim light and in bright light.

Think first. The output is across the fixed resistor, not the LDR. So the output should move the opposite way to the LDR's resistance: up in bright light.

Dim: $V_{out} = 5.0 \times \frac{2.0}{10 + 2.0} = 5.0 \times \frac{1}{6} \approx 0.83$ V.

Bright: 500 Ω = 0.50 kΩ, so $V_{out} = 5.0 \times \frac{2.0}{0.50 + 2.0} = 5.0 \times 0.8 = 4.0$ V.

Check. The output rose in bright light, as predicted. ✓ Both answers are between 0 and 5.0 V. ✓ This circuit could switch on a fan when sunlight falls on a window. For a street light that switches on in the dark, swap the two components.

The hidden assumption: nothing draws current from the output

Every answer so far assumed that the same current flows through R₁ and R₂. That is true only if nothing connected to the output takes current. As soon as you connect a load, such as a lamp, a motor or a meter, it sits in parallel with R₂. The bottom part of the divider now has a lower resistance, so it takes a smaller share, and the output drops.

Worked Example 4 — A load pulls the output down

A divider with R₁ = R₂ = 10 kΩ is connected to 12 V. (a) Find the output with nothing connected. (b) A 10 kΩ load is connected across R₂. Find the new output. (c) Repeat with a voltmeter of resistance 1.0 MΩ instead.

(a) Equal resistors share equally: $V_{out} = 6.0$ V.

(b) R₂ and the load are in parallel: $\frac{10 \times 10}{10 + 10} = 5.0 k Ω$ . Now $V_{out} = 12 \times \frac{5.0}{10 + 5.0} = 4.0$ V. The load has cut the output by a third.

(c) In parallel: $\frac{10 \times 1000}{10 + 1000} \approx 9.90 k Ω$ . So $V_{out} = 12 \times \frac{9.90}{10 + 9.90} \approx 5.97$ V. The meter changes the reading by only 0.03 V.

Check. A load can only lower the bottom resistance, so the output can only fall. ✓ The larger the load's resistance, the smaller the fall. ✓

So the divider equation is exact only when the output feeds something with a very high resistance, such as a good voltmeter or the input of a chip. A sensor divider feeding a microchip works well. A divider used to run a motor does not: the motor's low resistance would drag the output down. That is the deeper reason why a divider is a signal circuit, not a power supply. It also wastes energy as heat in R₁ all the time it is on.

Three traps

Trap 1 — The wrong resistor on top of the fraction

The resistor on top of the fraction is the one the output is taken across. If the output is across R₁, the fraction is $\frac{R _{1}}{R _{1} + R _{2}}$ . Always look at the circuit diagram, not at which letter is called R₂.

Trap 2 — Forgetting the load

The equation assumes that no current leaves through the output. If a load with a resistance similar to R₂ is connected, combine it with R₂ in parallel first, then use the equation.

Trap 3 — "Bright light means a higher output"

That depends on where the sensor is. Across the sensor, the output follows its resistance. Across the fixed resistor, it moves the other way. Write the reason as a chain: light up, LDR resistance down, LDR's share down, output (across which component?) up or down.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

A 12 V supply is connected across a 2.0 kΩ resistor and a 4.0 kΩ resistor in series. Find the p.d. across each, by the ratio method.

Hint 1
The p.d.s are in the ratio 2.0 : 4.0 = 1 : 2.

Hint 2
Split 12 V into 3 equal parts.

Full answer
Three parts of 4.0 V each. The 2.0 kΩ resistor takes 1 part, 4.0 V. The 4.0 kΩ resistor takes 2 parts, 8.0 V. Check with the equation: $12 \times \frac{4.0}{2.0 + 4.0} = 8.0$ V. ✓ And 4.0 + 8.0 = 12 V. ✓
The thermistor from this lecture (10 kΩ at 25 °C, 3.6 kΩ at 50 °C) is placed at the bottom of a divider. A fixed 10 kΩ resistor is at the top, and the supply is 6.0 V. The output is across the thermistor. Find the output at 25 °C and at 50 °C. Does the output rise or fall as the temperature rises?

Hint 1
At 25 °C the two resistances are equal.

Hint 2
At 50 °C: $6.0 \times \frac{3.6}{10 + 3.6}$ .

Full answer
At 25 °C: equal resistances share equally, so 3.0 V. At 50 °C: $6.0 \times \frac{3.6}{13.6} \approx 1.59$ V. The output falls as the temperature rises, because the output is across the thermistor, and its resistance falls. Check: this matches the explorer with the thermistor at the bottom. ✓
A street light must switch on when the output of a divider rises above 3.0 V. You have the LDR from Worked Example 3 (10 kΩ dim, 500 Ω bright), a fixed 2.0 kΩ resistor and a 5.0 V supply. Should the output be taken across the LDR or across the fixed resistor? Calculate the output in dim and bright light to show your design works.

Hint 1
The output must be high in the dark. In the dark, which component has the larger resistance?

Hint 2
Try the output across the LDR: $5.0 \times \frac{R _{LDR}}{R _{LDR} + 2.0}$ .

Full answer
Take the output across the LDR. In the dark its resistance is large, so its share is large. Dim: $5.0 \times \frac{10}{10 + 2.0} \approx 4.17$ V, above 3.0 V, so the light switches on. Bright: $5.0 \times \frac{0.50}{0.50 + 2.0} = 1.0$ V, below 3.0 V, so it stays off. Check: these are the reverse of Worked Example 3 (0.83 V and 4.0 V), and each pair adds to 5.0 V. ✓
A stretch. A divider of two 20 kΩ resistors is connected to 6.0 V. (a) What should the output be? (b) A student measures it with a cheap voltmeter whose resistance is 20 kΩ. What does the meter read? (c) Roughly how large must a voltmeter's resistance be for the reading to be within 1% of the true value?

Hint 1
(b) The meter is in parallel with the bottom resistor. Combine them first.

Hint 2
(c) Within 1% means a reading of at least 2.97 V. Find the bottom resistance that gives 2.97 V, then find the meter that makes that resistance with 20 kΩ in parallel.

Full answer
(a) Equal resistors: 3.0 V. (b) 20 kΩ in parallel with 20 kΩ is 10 kΩ. So the reading is $6.0 \times \frac{10}{20 + 10} = 2.0$ V, a third lower than the truth. (c) For 2.97 V, the bottom part needs a resistance $R_{p}$ with $\frac{R _{p}}{20 + R _{p}} = \frac{2.97}{6.0}$ , which gives $R_{p} \approx 19.6$ kΩ. A meter $R_{L}$ in parallel with 20 kΩ gives 19.6 kΩ when $R_{L} \approx 990$ kΩ, so about 1 MΩ: fifty times the resistor it is measuring. Check: Worked Example 4 found that a 1.0 MΩ meter on a 10 kΩ divider changed the reading by only 0.03 V, about 0.5%. ✓

The bigger idea

You have met a potential divider before, without the name. In the lecture on internal resistance, a battery was drawn as a perfect e.m.f. ε with an internal resistance r inside it. Connect an outside resistor R, and the terminal p.d. is $V = ε \times \frac{R}{R + r}$ . That is the divider equation, with r as the top resistor and R as the bottom one. A battery with ε = 12 V and r = 2.0 Ω gives 10 V to a 10 Ω resistor: the "lost volts" are simply r's share.

The same idea runs through modern electronics. Sensors for temperature, light, pressure and moisture usually change a resistance. A divider turns that change into a p.d. A converter inside a microchip turns the p.d. into a number. From two components to a digital thermometer, the chain is short. The divider cannot supply power efficiently, as you saw. For changing the p.d. of a power supply itself, a later lecture builds the transformer, which steps an alternating p.d. up or down without wasting most of the energy as heat.

Key concepts in this lesson

A potential divider is two resistors in series across a supply, with the output taken across one of them.
The same current flows through both resistors, so the p.d.s are in the ratio of the resistances, and $V_{out} = V_{in} \times \frac{R _{2}}{R _{1} + R _{2}}$ when the output is across R₂.
The output depends only on the ratio of the resistances, so it always lies between 0 V and the supply p.d.
Replacing a resistor with a thermistor or an LDR turns a change in temperature or light into a change in p.d.
Across the sensor, the output follows the sensor's resistance; across the fixed resistor, it moves the opposite way.
A load connected to the output lowers it, so the equation is exact only when the load's resistance is much larger than R₂.