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The potential divider and its output

A potential divider is two resistors in series, and the output across one of them is the supply pd shared in the ratio of the resistances. A sliding contact moves that output between zero and the full supply, and along a uniform wire the pd is proportional to length. A load lowers it.

Potential dividers and output voltage

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Year 12AQA 3.5.1.5CIE 10.3.1OCR A 4.3.3(a), 4.3.3(c)(i)EDEXCEL 42, 43

Potential dividers, part 1 of 2. Part 2 is Sensing circuits and the potentiometer.

Builds on Circuit symbols and Kirchhoff's laws.

IN THIS TOPIC

  • Explain why series resistors share the supply pd in the ratio of their resistances.
  • Calculate a divider's output, and supply a variable pd with a sliding contact.
  • Describe how the potential varies along a uniform current-carrying wire, and take a pd off a length of it.
  • Predict what connecting a load across the output does to that output, and how to keep the effect small.

COMMON MISCONCEPTION

The battery decides the voltage across each component.

The battery fixes only the total pd across the series chain. The same current flows through each resistor, so each takes a share of the pd proportional to its resistance.

Sharing in the ratio of the resistances

Two resistors in series carry the same current, and each has a pd of V = IR across it, so the pds are proportional to the resistances. The supply fixes only the total pd; the ratio of the resistances fixes how it is shared. Taking the output across R2 gives

Vout=Vin×R2R1+R2V_{out} = V_{in} \times \frac{R_{2}}{R_{1} + R_{2}}NOT IN THE AQA DATA BOOKLET: LEARN IT
Two resistors in series across a twelve volt supply: four kilo-ohms then eight kilo-ohms. A cyan bracket across the eight kilo-ohm resistor reads eight volts out: two thirds of the resistance takes two thirds of the pd.
FIG. 1A 12 V supply across 4.0 kΩ and 8.0 kΩ in series. The 8.0 kΩ resistor holds two thirds of the resistance, so it takes two thirds of the pd: 8.0 V.

Check the formula at its limits. As R2 approaches zero the output falls to zero; as R2 becomes large the output approaches the full supply pd. Any value between those is available, provided the output supplies almost no current; a real load reduces the output, which is covered in the section below.

Variable dividers and the uniform wire

Make the split point movable and the divider becomes adjustable. A sliding contact on a resistance track divides it into two parts whose ratio changes as the contact moves, so the output follows the slider, continuously, from zero to the full supply.

The potential divider: a dial that behaves (animated figure)
Describe the drawing

A potential divider with a sliding tap. The wiper glides along the resistive track and the output p.d. is simply the share of the 6.0 V below the tap: the bar's height mirrors the wiper's position exactly, pixel for pixel. Slide high, nearly six volts; slide low, nearly none. It is a voltage dial, behaving like a dial.

FIG. 2A potential divider with a sliding tap. The output pd is the share of the 6.0 V across the part of the track below the tap, so moving the tap from one end to the other varies the output continuously between zero and 6.0 V.

Volume controls, dimmer switches and joystick axes are all built from one track and one wiper in this way.

If the track is a uniform wire, of one material at one cross-section along its whole length, the output can be read from a length. One current passes along the whole wire, and R = ρL/A gives every centimetre of it the same resistance, so every centimetre takes the same pd. The potential therefore falls at a steady rate along the wire, from the full supply at the end joined to the positive terminal down to zero at the far end, and the pd across any part of it is proportional to the length of that part. Writing V for the pd across the whole length L, the pd Vx across a length x measured from the far end is

Vx=VxLV_{x} = V\frac{x}{L}
A metre of uniform resistance wire between terminals A and B with a cell driving a steady current through it. Below the wire, a graph of potential against distance along it: a straight line falling from six volts at A to zero at B, with the halfway point P marked at three volts.
FIG. 3A metre of uniform wire carrying a steady current, and beneath it the potential plotted against distance along it: a straight line from 6.0 V down to zero, so the halfway point sits at 3.0 V.

A graph of potential against distance along the wire is a straight line whose gradient is the potential gradient, in volts per metre. Halfway along the wire the pd is half the total, and a fifth of the way along it is a fifth of the total. A uniform wire laid beside a metre rule is therefore a voltage scale, which is the basis of the potentiometer in the next lesson.

WORKED EXAMPLE

Designing for five volts

A circuit needs 5.0 V from a 9.0 V supply. The lower resistor of the divider is 10 kΩ and the output is taken across it. Find the upper resistance required.

The output takes its share of the supply in the ratio of the resistances. Put 5.0 of the 9.0 volts across the bottom resistor and 4.0 V is left for the top.

One current threads both, so the resistances sit in the same 4.0 to 5.0 ratio as their voltages. Rtop = 10 × 4.0/5.0 = 8.0 kΩ.

Check by running the divider forward: 9.0 × 10/(10 + 8.0) = 5.0 V.

What a load does to the output

The equation above assumes that the output supplies no current. A component connected across the output sits in parallel with whichever resistor the output is taken across. The parallel combination has a lower resistance than that resistor alone, so its share of the supply falls and the output is below the value the unloaded ratio predicts.

The size of the effect depends on the load. Two equal resistors give half the supply when unloaded. A load of the same resistance across the lower one reduces the output from half the supply to a third of it. A load ten times the divider resistance reduces the output by only a few per cent.

There are two ways to keep the effect small, and either is an acceptable answer. Build the divider from resistances well below the load resistance, or feed the output into a component that draws almost no current, such as a high-resistance voltmeter or an amplifier input. Low divider resistances have a cost: a current flows continuously through the chain, which shortens battery life.

Sensing circuits has moved to the next part of this lesson.

The potentiometer and null methods, a CIE extension has moved to the next part of this lesson.

ASSESSMENT FOCUS

  • Read the circuit before writing the formula. The numerator is the resistance the output is taken across, which is not always the lower resistor; the denominator is the total.
  • Every output lies between zero and the supply pd. A value outside that range means the ratio has been inverted.
  • Along a uniform wire the pd is proportional to length, so a ratio of two lengths answers the question without any resistance value. State that the wire is uniform, since that is the condition the step depends on.
  • If a load is connected across the output, state that it is in parallel with that resistor and that the output falls below the unloaded value. Combine the resistances before applying the ratio.

CHECK YOURSELF

A divider is built from a 4.0 kΩ resistor above an 8.0 kΩ resistor across a 12 V supply, output across the 8.0 kΩ. (a) Find the output. (b) A load of 8.0 kΩ is now connected across the output. Find the new output, and say why it is lower.

Show a hint

Part (b) needs the parallel combination before the ratio.

Show the answer

(a) Vout=12×8.04.0+8.0V_{out} = 12 \times \frac{8.0}{4.0 + 8.0} = 8.0 V.

(b) The load sits in parallel with the lower resistor, and 8.0 kΩ with 8.0 kΩ is 4.0 kΩ. So Vout=12×4.04.0+4.0V_{out} = 12 \times \frac{4.0}{4.0 + 4.0} = 6.0 V. The lower section now holds half the total resistance instead of two thirds, so it takes half the supply instead of two thirds. The load changed the resistance ratio, not the equation.

One current, so the shares follow the resistances.

WORKBOOK

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  • Explain why series resistors share the supply pd in the ratio of their resistances.
  • Calculate a divider's output, and supply a variable pd with a sliding contact.
  • Describe how the potential varies along a uniform current-carrying wire, and take a pd off a length of it.
  • Predict what connecting a load across the output does to that output, and how to keep the effect small.

Open the full revision checklist to track your progress across the whole unit.