Why does only the ratio of resistances matter?

Because the output is a fraction of the input determined by R_2/(R_1+R_2). Doubling both resistors leaves that fraction unchanged, though it halves the current drawn.

What is the loading effect?

Connecting a component across the output places it in parallel with R_2, lowering the effective resistance of that section and so reducing the output voltage below its unloaded value.

How do I avoid loading problems?

Ensure the load resistance is much larger than R_2 — a factor of ten or more usually suffices — or use smaller divider resistances so the load is comparatively large.

Should I use large or small resistor values?

It is a trade-off. Large values draw less current and waste less power but are more susceptible to loading; small values are robust against loading but dissipate more heat.

Back to course

42–44ElectricityStretch

Potential Dividers

Two resistors in series across a supply split the potential difference between them in proportion to their resistances. That simple arrangement — a $\textbf{potential divider}$ — is how a fixed supply is turned into any smaller voltage you need, and how sensors are made to control circuits.

35 min Topic 3 · Electric Circuits

What you'll be able to do

Derive and use the potential divider equation
Design a divider to produce a required output voltage
Analyse sensing circuits using thermistors and LDRs
Explain the loading effect and when it can be neglected

The divider equation

Two resistors $R_{1}$ and $R_{2}$ in series carry the same current $I = \dfrac{V_{\text{in}}}{R_{1}+R_{2}}$ . The p.d. across $R_{2}$ is then $IR_{2}$ , which gives the divider equation directly.

Read it as a ratio: the output is the same fraction of the input as $R_{2}$ is of the total resistance. A resistor taking a quarter of the total resistance takes a quarter of the voltage.

Two immediate consequences are worth noticing. Equal resistors halve the supply. And the output can never exceed the input, since the fraction is always less than one — an answer above $V_{\text{in}}$ means the resistors have been swapped.

V_{\text{out}} = V_{\text{in}} \times \dfrac{R_{2}}{R_{1}+R_{2}}

Output across

R_{2}

\dfrac{V_{1}}{V_{2}} = \dfrac{R_{1}}{R_{2}}

The p.d.s are in the same ratio as the resistances.

Example 1A 12 V supply is connected across a

3.0\ \mathrm{k}\Omega

and a

1.0\ \mathrm{k}\Omega

resistor in series. Find the p.d. across the

1.0\ \mathrm{k}\Omega

resistor.

1Total resistance:

4.0\ \mathrm{k}\Omega

2The

1.0\ \mathrm{k}\Omega

is one quarter of the total.

V_{\text{out}} = 12 \times \dfrac{1.0}{4.0}

Answer

3.0\ \mathrm{V}

Tip — Because it is a ratio, the units of resistance cancel — you can work in kilohms throughout without converting, provided both resistances use the same unit.

Designing a divider

To produce a required output, rearrange the equation for the ratio you need. Only the $\textbf{ratio}$ of the resistors is fixed by the voltage requirement, not their individual values.

That freedom is used to control the current. Large resistances draw little current and waste little power, but make the circuit more vulnerable to loading. Small resistances are robust against loading but waste more power as heat.

A potentiometer — a resistor with a sliding contact — is a divider whose ratio can be varied continuously, which is how a volume control works.

\dfrac{R_{2}}{R_{1}+R_{2}} = \dfrac{V_{\text{out}}}{V_{\text{in}}}

The design condition — a ratio.

Example 1Design a divider giving 5.0 V from a 9.0 V supply, using a

2.0\ \mathrm{k}\Omega

resistor as

R_{2}

1Required ratio:

\dfrac{R_{2}}{R_{1}+R_{2}} = \dfrac{5.0}{9.0}

2So

9.0 R_{2} = 5.0\left(R_{1}+R_{2}\right)

, giving

4.0R_{2} = 5.0R_{1}

R_{1} = \dfrac{4.0 \times 2.0}{5.0}

Answer

R_{1} = 1.6\ \mathrm{k}\Omega

Sensing circuits

Replace one resistor with a sensor and the output voltage becomes a measure of a physical quantity. A $\textbf{thermistor}$ (NTC) has resistance falling as temperature rises; an $\textbf{LDR}$ has resistance falling as light intensity rises.

Which way the output moves depends on where the sensor is placed. With the sensor as $R_{2}$ — the one the output is taken across — its falling resistance reduces the output. With the sensor as $R_{1}$ instead, its falling resistance increases the output.

That choice is the design decision in these questions. Deciding what you want the output to do, then placing the sensor accordingly, is what is being examined.

Example 1A thermistor is used with a fixed resistor so that the output voltage

\textbf{rises}

as temperature increases. Where should the thermistor go, and why?

1An NTC thermistor’s resistance falls as temperature rises.

2For the output to rise, the resistance in the output branch must become a larger fraction of the total.

3So the thermistor must be the

\textit{other}

resistor,

R_{1}

AnswerPlace the thermistor as

R_{1}

, with the output taken across the fixed resistor. As temperature rises, the thermistor’s resistance falls, so the fixed resistor takes a larger share of the supply.

Tip — Reason it through with an extreme case. Ask what happens as the sensor’s resistance approaches zero: whichever resistor it is, it then takes almost none of the voltage.

The loading effect

Connecting a component across the output puts it in $\textbf{parallel}$ with $R_{2}$ , reducing the resistance of that section — and therefore reducing the output voltage below the value the unloaded divider would give.

The size of the effect depends on the ratio of resistances. If the load resistance is much greater than $R_{2}$ , the parallel combination is barely changed and the loading is negligible. If it is comparable to or smaller than $R_{2}$ , the output falls substantially.

This is why a voltmeter is designed with very high resistance: it must measure the output without appreciably changing it. It is also why a divider intended to feed a low-resistance load should use small resistor values.

R_{\text{eff}} = \dfrac{R_{2}R_{L}}{R_{2}+R_{L}}

The load sits in parallel with

R_{2}

R_{L} \gg R_{2} \Rightarrow \text{loading negligible}

The design condition to avoid the problem.

Example 1A divider of two

10\ \mathrm{k}\Omega

resistors across 12 V has a

10\ \mathrm{k}\Omega

load connected across the output. Find the output voltage with and without the load.

1Unloaded: equal resistors, so

V_{\text{out}} = 6.0

2Loaded:

R_{2}

in parallel with the load gives

\dfrac{10 \times 10}{20} = 5.0\ \mathrm{k}\Omega

3New divider:

V_{\text{out}} = 12 \times \dfrac{5.0}{10 + 5.0}

AnswerUnloaded

6.0

V; loaded

4.0

V — a drop of a third.

Equation recap

V_{\text{out}} = V_{\text{in}}\tfrac{R_{2}}{R_{1}+R_{2}}

The divider equation.

\tfrac{V_{1}}{V_{2}} = \tfrac{R_{1}}{R_{2}}

P.d.s in the ratio of resistances.

R_{\text{eff}} = \tfrac{R_{2}R_{L}}{R_{2}+R_{L}}

With a load connected.

R_{L} \gg R_{2}

Condition for negligible loading.

Common mistakes to avoid

Using

R_{1}

in the numerator when the output is across

R_{2}

The numerator is the resistance the output is measured across.

Ignoring a load connected across the output.

It sits in parallel with

R_{2}

and lowers the output voltage.

Assuming an NTC thermistor’s resistance rises with temperature.

It falls — which reverses the direction the output moves.

Converting kilohms to ohms unnecessarily.

The equation is a ratio, so consistent units cancel. Converting is harmless but wastes time.

Key takeaways

A potential divider splits the supply in the ratio of the resistances.
$V_{\text{out}} = V_{\text{in}}R_{2}/(R_{1}+R_{2})$ , with the numerator being the output resistor.
Only the ratio sets the voltage; the absolute values set the current.
Sensors make the output respond to temperature or light, with direction set by placement.
A load sits in parallel with $R_{2}$ and lowers the output unless it is much larger.

Test yourself

Answer a few questions, then review anything you missed.

Next step

Keep practising

Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.

Browse practice topics

Related lessons

All 6 lessons