Physics 9702 · Paper 2

Sensing Circuits: Thermistor & LDR

A loses resistance when it gets hot, and an loses resistance in bright light.
Put either one into a potential divider and that resistance change becomes an output voltage that follows temperature or light.

A resistance change becomes a voltage change

Electronic systems cannot read “resistance” directly; they read voltages. The divider does the conversion: it turns a resistance change into a voltage change.
Put a fixed resistor on top and the sensor on the bottom, output across the sensor. The sensor's share of the input is $\dfrac{R_s}{R_s + R_{\text{fixed}}}$ : when its resistance falls, its share falls with it.

Worked example

A thermistor thermometer

A 6.0 V supply drives a 1.0 kΩ fixed resistor in series with a thermistor. The output is across the thermistor. Its resistance is 4.0 kΩ when cold and 250 Ω when hot. Find the output at each temperature.

Cold: $V_{\text{out}} = \dfrac{4000}{4000 + 1000} \times 6.0 = 4.8$ V.
Hot: $V_{\text{out}} = \dfrac{250}{250 + 1000} \times 6.0 = 1.2$ V.

Answer

4.8 \text{ V cold}, \; 1.2 \text{ V hot}

Look at the two shares. Cold: the sensor takes 4.8 V, so the fixed resistor takes 1.2 V. Hot: the sensor takes 1.2 V, so the fixed resistor takes 4.8 V.
The two shares always add up to the input. When one share falls, the other rises by the same amount.
Drag the temperature or light level below and watch the sensor's resistance fall sharply. Its share of the 6.0 V falls with it.

temperature20 °C

sensor R = 1.99 kΩ·Vout = 3.99 V of the 6.0 V

The explain chain when r can be ignored

When the cell has no internal resistance, the chain has four links. For an LDR divider with output across the LDR, as it gets dark: darker, so LDR resistance rises, so the LDR's share of the input rises, so the output p.d. rises. Four links, in order, every time.
To get the opposite behaviour, swap the two components. The two shares always add up to the full input, so the output across the fixed resistor moves the other way.

Common mistake

Vague answers like “the current changes so the voltage changes” score nothing. The mark scheme wants each link named: condition, sensor resistance, share of the input, output.

The full chain when the cell has internal resistance

Recent papers make this question harder: the cell now has internal resistance, and the chain must pass through the cell current and the lost volts.
The parallel pair is the whole outside circuit. So the p.d. across the pair is the terminal p.d. $V = E - Ir$ : when the lost volts rise, the p.d. across both R and T falls.
R itself never changes. What changes is the p.d. across R, and with it the current in R.

Worked example

A thermistor in parallel with a resistor

A cell of e.m.f. 1.50 V and internal resistance 0.12 Ω is connected to a 6.00 Ω resistor R, with a thermistor T in parallel with R. At one temperature the current in R is 0.200 A. Find the current in the cell and the resistance of T.

The p.d. across R (and so across T): $6.00 \times 0.200 = 1.20$ V.
The p.d. across r: $1.50 - 1.20 = 0.30$ V, so the cell current is $0.30/0.12 = 2.5$ A.
First law: the thermistor carries $2.5 - 0.200 = 2.3$ A.
$R_T = \dfrac{1.20}{2.3} = 0.52 \; \Omega$ .

Answer

I_{\text{cell}} = 2.5 \text{ A}, \; R_T = 0.52 \; \Omega

(9702/21/O/N/25 Q5)

The same paper asks: explain why the current in R falls as the thermistor gets hotter. The full chain, link by link: hotter, so the thermistor's resistance falls; so the total circuit resistance falls; so the cell current rises; so the lost volts $Ir$ rise; so the terminal p.d. falls; R itself is unchanged, so the current in R falls.

Your turn— tap to reveal the worked answer (9702/22/O/N/25 Q5)

In a similar circuit, an LDR is in the parallel pair. The light gets dimmer. State and explain what happens to the terminal p.d. of the battery.

Answer: dimmer light, so the LDR resistance rises; total resistance rises; the cell current falls; the lost volts fall; so the terminal p.d. rises. (9702/22/O/N/25 Q5)

One question, two versions. No internal resistance: think in shares of the input. With internal resistance: walk the chain through the cell current and the lost volts. Look for r in the circuit before you start writing.

PreviousAdjusting the Divider NextThe Potentiometer

Sensing Circuits: Thermistor & LDR

A loses resistance when it gets hot, and an loses resistance in bright light.
Put either one into a potential divider and that resistance change becomes an output voltage that follows temperature or light.

A resistance change becomes a voltage change

Electronic systems cannot read “resistance” directly; they read voltages. The divider does the conversion: it turns a resistance change into a voltage change.
Put a fixed resistor on top and the sensor on the bottom, output across the sensor. The sensor's share of the input is $\dfrac{R_s}{R_s + R_{\text{fixed}}}$ : when its resistance falls, its share falls with it.

Worked example

A thermistor thermometer

Cold: $V_{\text{out}} = \dfrac{4000}{4000 + 1000} \times 6.0 = 4.8$ V.
Hot: $V_{\text{out}} = \dfrac{250}{250 + 1000} \times 6.0 = 1.2$ V.

Answer

4.8 \text{ V cold}, \; 1.2 \text{ V hot}

Look at the two shares. Cold: the sensor takes 4.8 V, so the fixed resistor takes 1.2 V. Hot: the sensor takes 1.2 V, so the fixed resistor takes 4.8 V.
The two shares always add up to the input. When one share falls, the other rises by the same amount.
Drag the temperature or light level below and watch the sensor's resistance fall sharply. Its share of the 6.0 V falls with it.

temperature20 °C

sensor R = 1.99 kΩ·Vout = 3.99 V of the 6.0 V

The explain chain when r can be ignored

When the cell has no internal resistance, the chain has four links. For an LDR divider with output across the LDR, as it gets dark: darker, so LDR resistance rises, so the LDR's share of the input rises, so the output p.d. rises. Four links, in order, every time.
To get the opposite behaviour, swap the two components. The two shares always add up to the full input, so the output across the fixed resistor moves the other way.

Common mistake

Vague answers like “the current changes so the voltage changes” score nothing. The mark scheme wants each link named: condition, sensor resistance, share of the input, output.

The full chain when the cell has internal resistance

Recent papers make this question harder: the cell now has internal resistance, and the chain must pass through the cell current and the lost volts.
The parallel pair is the whole outside circuit. So the p.d. across the pair is the terminal p.d. $V = E - Ir$ : when the lost volts rise, the p.d. across both R and T falls.
R itself never changes. What changes is the p.d. across R, and with it the current in R.

Worked example

A thermistor in parallel with a resistor

The p.d. across R (and so across T): $6.00 \times 0.200 = 1.20$ V.
The p.d. across r: $1.50 - 1.20 = 0.30$ V, so the cell current is $0.30/0.12 = 2.5$ A.
First law: the thermistor carries $2.5 - 0.200 = 2.3$ A.
$R_T = \dfrac{1.20}{2.3} = 0.52 \; \Omega$ .

Answer

I_{\text{cell}} = 2.5 \text{ A}, \; R_T = 0.52 \; \Omega

(9702/21/O/N/25 Q5)

The same paper asks: explain why the current in R falls as the thermistor gets hotter. The full chain, link by link: hotter, so the thermistor's resistance falls; so the total circuit resistance falls; so the cell current rises; so the lost volts $Ir$ rise; so the terminal p.d. falls; R itself is unchanged, so the current in R falls.

Your turn— tap to reveal the worked answer (9702/22/O/N/25 Q5)

In a similar circuit, an LDR is in the parallel pair. The light gets dimmer. State and explain what happens to the terminal p.d. of the battery.

Answer: dimmer light, so the LDR resistance rises; total resistance rises; the cell current falls; the lost volts fall; so the terminal p.d. rises. (9702/22/O/N/25 Q5)