A Level Physics 9702: Potential Divider with a Thermistor

A potential divider circuit consists of a fixed resistor of resistance $2.0\text{ k}\Omega$ connected in series with a negative temperature coefficient (NTC) thermistor. This series combination is connected across a battery of e.m.f. $6.0\text{ V}$ and negligible internal resistance. The output potential difference $V_{\text{out}}$ is taken across the thermistor.

At room temperature, the resistance of the thermistor is $8.0\text{ k}\Omega$ .

(a) State how the resistance of an NTC thermistor changes as its temperature increases. [1]

(b) Calculate $V_{\text{out}}$ at room temperature. [2]

(c) The temperature of the thermistor increases until its resistance falls to $2.0\text{ k}\Omega$ . Calculate the new value of $V_{\text{out}}$ . [2]

(d) State and explain how $V_{\text{out}}$ changes as the temperature of the thermistor increases, and suggest one practical device in which this potential divider circuit could be used. [3]

Part (a): How NTC thermistor resistance changes with temperature

For a negative temperature coefficient (NTC) thermistor, resistance decreases as temperature increases (more charge carriers become available as the semiconductor material is heated).

Part (b): Output p.d. at room temperature

In a potential divider, the p.d. across a component is in proportion to its share of the total resistance: $V_{\text{out}} = V_{\text{supply}}\times\frac{R_{\text{thermistor}}}{R_{\text{fixed}}+R_{\text{thermistor}}}$

Substituting $V_{\text{supply}}=6.0\text{ V}$ , $R_{\text{fixed}}=2.0\text{ k}\Omega$ , $R_{\text{thermistor}}=8.0\text{ k}\Omega$ : $V_{\text{out}} = 6.0\times\frac{8.0}{2.0+8.0} = 6.0\times\frac{8.0}{10.0} = 6.0\times0.80 = 4.8\text{ V}$

Check using the other component instead: the p.d. across the fixed resistor is $6.0\times\frac{2.0}{10.0}=6.0\times0.20=1.2\text{ V}$ , and $4.8+1.2=6.0\text{ V}$ , matching the supply e.m.f. Confirmed.

Part (c): Output p.d. after the temperature rise

The thermistor’s resistance has fallen to $2.0\text{ k}\Omega$ , so the total resistance is now $2.0+2.0=4.0\text{ k}\Omega$ : $V_{\text{out}} = 6.0\times\frac{2.0}{2.0+2.0} = 6.0\times\frac{2.0}{4.0} = 6.0\times0.50 = 3.0\text{ V}$

Check using decimals for the fraction: $\frac{2.0}{4.0}=0.50$ exactly, so $V_{\text{out}}=6.0\times0.50=3.0\text{ V}$ . Same result both ways.

Part (d): Trend and application

As the thermistor’s temperature increases, its resistance falls (part (a)), so it takes a smaller share of the fixed $6.0\text{ V}$ supply. Since $V_{\text{out}}$ is taken across the thermistor, $V_{\text{out}}$ decreases as temperature increases, consistent with the fall from $4.8\text{ V}$ to $3.0\text{ V}$ found in parts (b) and (c).

This kind of circuit, where the output p.d. varies with temperature, is used in devices such as a thermostat or overheat/fire alarm: the changing $V_{\text{out}}$ can be compared (e.g. via an operational amplifier or comparator circuit) against a fixed reference voltage to trigger a switch when a temperature threshold is crossed.

Final answers

(a) NTC thermistor resistance $\boxed{\text{decreases}}$ as temperature increases
(b) $V_{\text{out}} = \boxed{4.8}\text{ V}$ at room temperature
(c) $V_{\text{out}} = \boxed{3.0}\text{ V}$ after the temperature rise
(d) $V_{\text{out}}$ decreases as temperature increases; usable in a temperature-controlled switch or overheat/fire alarm