Physics · 4 Electricity and magnetism

4.3.3 Action and use of circuit components

Before you start

$V = I R$ and its rearrangements (4.2.4).
In a series circuit the current is the same everywhere and the p.d.s add up (4.3.2).
That a thermistor's and an LDR's resistance fall as temperature or light rises (4.3.1).

Syllabus objectives covered (3)

Know that the p.d. across an electrical conductor increases as its resistance increases for a constant current
Describe the action of a variable potential divider
Recall and use the equation for two resistors used as a potential divider R1 / R2 = V1 / V2

Part 1 of 3

More resistance means more p.d.

Start from $V = I R$ . If the current $I$ is held constant, then $V$ is directly proportional to $R$ .
So for a fixed current, the component with the larger resistance has the larger p.d. across it. Double the resistance, double the p.d.
That is the whole of the Core objective, and it is the idea the rest of this subtopic is built on.
Picture it in energy terms: a coulomb passing through a high-resistance component gives up more of its energy there. P.d. is energy per coulomb, so a bigger p.d. means more energy given up in that component.
This is why a series circuit shares out the supply p.d. in proportion to the resistances — which is exactly what a potential divider exploits.

Part 2 of 3

The potential divider

A potential divider is simply two resistors in series across a supply, with the output taken across one of them.
Because they are in series, the same current flows through both. From $V = I R$ , each takes a share of the supply p.d. in proportion to its resistance.
That gives the equation you need to recall:
$\frac{R_{1}}{R_{2}} = \frac{V_{1}}{V_{2}}$
In words: the p.d.s are in the same ratio as the resistances. Double the resistance, double the share of the p.d.
Be careful to keep the subscripts matched. $R_{1}$ goes with $V_{1}$ on the same side of both fractions; crossing them over inverts the answer.
It is often quicker to reason with the ratio directly. If $R_{1}$ is three times $R_{2}$ , then $R_{1}$ takes three-quarters of the supply p.d. and $R_{2}$ takes one-quarter.
Output across R2. Its share of the supply p.d. is R2/(R1+R2).
The output is always a fraction of the supply, never more than it. If you calculate an output larger than the supply p.d., you have inverted the ratio.

Part 3 of 3

Making it variable, and making it sense

A variable potential divider replaces one of the fixed resistors with a variable resistor, so the output p.d. can be adjusted by hand.
As you increase the resistance of one, it takes a bigger share of the supply p.d., and the other takes correspondingly less. Sliding the contact sweeps the output smoothly from nearly zero up to nearly the full supply p.d.
This is how a volume control or a dimmer works: a continuously adjustable p.d. from a fixed supply.
Now replace one resistor with a thermistor or an LDR, and the divider becomes a sensor. Nothing is adjusted by hand — the component changes its own resistance as conditions change.
Take a thermistor at the top and a fixed resistor at the bottom, with the output across the fixed resistor. When it gets hot, the thermistor's resistance falls, so it takes a smaller share of the p.d. and the output across the fixed resistor rises.
Swap the two components round and the behaviour reverses: the output would then fall as it got hotter. Which arrangement you need depends on whether the circuit should switch on when hot or when cold.
So when a question asks you to explain a sensing circuit, always answer in this order: what happens to the sensor's resistance, what that does to its share of the p.d., and therefore what happens to the output.

Worked examples

Example 1. A $12$ V supply is connected across a $2.0 k Ω$ resistor in series with a $4.0 k Ω$ resistor. Calculate the p.d. across each resistor.

Show the working

The resistors are in series, so the same current flows through both and the p.d.s share out in the ratio of the resistances.
The ratio is $2.0 : 4.0$ , which simplifies to $1 : 2$ . So the total of $12$ V divides into three equal parts.
One part goes to the $2.0 k Ω$ resistor and two parts to the $4.0 k Ω$ resistor.
$V_{1} = \frac{1}{3} \times 12 = 4.0 V V_{2} = \frac{2}{3} \times 12 = 8.0 V$
Check with the equation: $R_{1} / R_{2} = 2 / 4 = 0.5$ and $V_{1} / V_{2} = 4 / 8 = 0.5$ . They match.
Check the total too: $4.0 + 8.0 = 12$ V, the supply p.d. Both checks pass, and note the kilohms cancelled out — the ratio is all that mattered.

Answer: $4.0$ V across the $2.0 k Ω$ resistor and $8.0$ V across the $4.0 k Ω$ resistor.

Example 2. In a potential divider across a $9.0$ V supply, the p.d. across $R_{1}$ is $6.0$ V and $R_{2}$ has a resistance of $150 Ω$ . Calculate $R_{1}$ .

Show the working

First find the p.d. across $R_{2}$ , since the two p.d.s must add to the supply p.d.
$V_{2} = 9.0 - 6.0 = 3.0 V$
Now apply the potential-divider equation, keeping the subscripts matched.
$\frac{R_{1}}{R_{2}} = \frac{V_{1}}{V_{2}} = \frac{6.0}{3.0} = 2.0$
So $R_{1}$ is twice $R_{2}$ .
$R_{1} = 2.0 \times 150 = 300 Ω$
Sense-check it: $R_{1}$ has the larger p.d. across it, so it should have the larger resistance. It does.

Answer: $R_{1} = 300 Ω$ .

Example 3. A potential divider consists of a thermistor in series with a fixed resistor across a battery. The output is taken across the fixed resistor. Explain what happens to the output p.d. as the temperature rises.

Show the working

Start with the sensor. As the temperature rises, the resistance of the NTC thermistor decreases.
The two components are in series, so the p.d. shares out in proportion to their resistances.
The thermistor's share of the supply p.d. therefore falls, because its resistance has fallen relative to the fixed resistor.
The two p.d.s must still add to the supply p.d., so whatever the thermistor gives up, the fixed resistor gains.
The output, taken across the fixed resistor, therefore increases as the temperature rises — which could be used to switch on a cooling fan.

Answer: The thermistor's resistance decreases as the temperature rises, so its share of the supply p.d. falls. Since the p.d.s must still sum to the supply p.d., the p.d. across the fixed resistor — the output — increases.

Where people go wrong

Mistake: Crossing over the subscripts in $R_{1} / R_{2} = V_{1} / V_{2}$ .

Instead: $R_{1}$ pairs with $V_{1}$ on the same side of both fractions.

Crossing them inverts the ratio, giving an answer that is wrong by exactly the factor you were trying to find.

Mistake: Calculating an output p.d. larger than the supply p.d.

Instead: The output is always a fraction of the supply, so it can never exceed it.

The two resistors share one supply p.d. between them. An answer above the supply is a sure sign the ratio was inverted.

Mistake: Saying the thermistor's resistance rises when it gets hot.

Instead: An NTC thermistor's resistance falls as the temperature rises.

It is the only type in this syllabus, and every sensing-circuit answer depends on getting this direction right.

Mistake: Answering a sensing-circuit question by naming the outcome only.

Instead: Give the chain: sensor resistance changes, its share of the p.d. changes, so the output changes.

The marks are on the reasoning steps, not the conclusion. Each link is a separate marking point.

Mistake: Assuming the output always rises when the sensor's resistance falls.

Instead: It depends which component the output is taken across.

Swapping the two components reverses the behaviour, which is how the same parts make both a 'switch on when hot' and a 'switch on when cold' circuit.

Check yourself

Two components carry the same current. One has twice the resistance of the other. How do their p.d.s compare?
Answer
The one with twice the resistance has twice the p.d. across it, since $V = I R$ with $I$ constant.
A $10 Ω$ and a $30 Ω$ resistor form a potential divider across $8.0$ V. What is the p.d. across each?
Answer
The ratio is $1 : 3$ , so $2.0$ V across the $10 Ω$ and $6.0$ V across the $30 Ω$ .
In a potential divider the p.d.s are $3.0$ V and $9.0$ V. If the smaller resistance is $200 Ω$ , what is the larger?
Answer
The p.d.s are in the ratio $1 : 3$ , so the resistances are too: $600 Ω$ .
Describe what happens to the output of a variable potential divider as the variable resistor's resistance is increased, with the output taken across the fixed resistor.
Answer
The variable resistor takes a larger share of the supply p.d., so the p.d. across the fixed resistor — the output — decreases.
An LDR is in series with a fixed resistor, with the output across the LDR. Explain what happens to the output as it gets darker.
Answer
In the dark the LDR's resistance increases, so it takes a larger share of the supply p.d. and the output across it increases.

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. A potential divider is made from a $400 Ω$ resistor in series with a $600 Ω$ resistor, connected across a $5.0$ V supply.

(a) Calculate the p.d. across the $600 Ω$ resistor. 3 marks

Mark scheme

C1 recognises the p.d.s are in the ratio of the resistances, or finds the current 5.0/1000
C1 correct fraction 600/1000 applied to 5.0 V
A1 3.0 V; unit required

(b) Explain why this resistor has the larger p.d. across it. 2 marks No calculator

Mark scheme

B1 the same current flows through both, as they are in series
B1 V = IR, so for a constant current the larger resistance has the larger p.d.

Question 2. A circuit is designed to switch on a heater automatically when a greenhouse becomes cold. It uses a thermistor and a fixed resistor as a potential divider, with the output taken across the thermistor.

(a) Describe the action of this potential divider as the temperature falls. 3 marks No calculator

Mark scheme

B1 the resistance of the thermistor increases as the temperature falls
B1 the thermistor therefore takes a larger share of the supply p.d.
B1 so the output p.d. across the thermistor increases

(b) Predict how the behaviour of the circuit would change if the thermistor and the fixed resistor were interchanged. 2 marks No calculator

Mark scheme

B1 the output would now be taken across the fixed resistor
B1 so the output p.d. would decrease as the temperature falls, i.e. the behaviour is reversed, ORA

Question 3. In a potential divider across a $6.0$ V supply, the output p.d. is $4.5$ V across resistor $R_{2}$ . Resistor $R_{1}$ has a resistance of $500 Ω$ .

(a) Determine the p.d. across $R_{1}$ . 1 mark

Mark scheme

B1 1.5 V

(b) Calculate the resistance of $R_{2}$ . 3 marks

Mark scheme

C1 R1 / R2 = V1 / V2 used
C1 correct substitution 500 / R2 = 1.5 / 4.5; ecf from (a)
A1 1500 Ω; unit required

Words to know

Potential divider: Two resistors in series across a supply, with an output taken across one of them, giving a chosen fraction of the supply p.d.
Variable potential divider: A potential divider in which one resistance can be adjusted, so the output p.d. can be varied continuously.
Output p.d.: The p.d. taken from a potential divider, always a fraction of the supply p.d.

All Physics lessons Next: 4.4 Electrical safety