A student investigated the electrical properties of a negative temperature coefficient (NT… — ACT

A student investigated the electrical properties of a negative temperature coefficient (NTC) thermistor. The thermistor was placed in a temperature-controlled water bath, and its electrical resistance, $R$ (in kilohms, $\text{k}\Omega$ ), was measured at various temperatures, $T$ (in degrees Celsius, $^\circ\text{C}$ ). The data from this experiment are presented in the table below.

Temperature ( $T$ , $^\circ\text{C}$ )	Resistance ( $R$ , $\text{k}\Omega$ )
10	48.0
30	24.0
50	12.0
70	6.0

Based on the trend shown in the table, what is the predicted electrical resistance, in $\text{k}\Omega$ , of the thermistor at a temperature of $110^\circ\text{C}$ ?

Answer: 1.5 k\Omega

Answer

The predicted electrical resistance of the thermistor at

110^\circ\text{C}

1.5\text{ k}\Omega

The correct calculation identifies that the resistance decreases by a factor of 2 for every

20^\circ\text{C}

increase in temperature. Following this exponential trend, the resistance at

90^\circ\text{C}

3.0\text{ k}\Omega

, and at

110^\circ\text{C}

it is half of that, which equals

1.5\text{ k}\Omega

Step-by-Step Solution

Analyze the pattern of temperature changes in the table.

The temperature increments are constant at

\Delta T = 20^\circ\text{C}

(e.g.,

30 - 10 = 20^\circ\text{C}

50 - 30 = 20^\circ\text{C}

70 - 50 = 20^\circ\text{C}

Establishing a constant independent variable interval simplifies trend extrapolation.

Analyze the corresponding ratio of resistance values at each temperature step.

At each interval, the resistance value is divided by 2:

\frac{24.0}{48.0} = 0.5

\frac{12.0}{24.0} = 0.5

, and

\frac{6.0}{12.0} = 0.5

. This indicates a non-linear, exponential decay trend.

Determining the mathematical relationship allows for precise calculation of values outside the dataset range.

Extrapolate the trend to

90^\circ\text{C}

by applying the factor of

0.5

to the resistance at

70^\circ\text{C}

R(90^\circ\text{C}) = R(70^\circ\text{C}) \times 0.5 = 6.0 \times 0.5 = 3.0\text{ k}\Omega

Since

90^\circ\text{C}

is exactly

20^\circ\text{C}

above

70^\circ\text{C}

, the pattern dictates that the resistance halves.

Extrapolate the trend further to

110^\circ\text{C}

by applying the factor of

0.5

to the resistance at

90^\circ\text{C}

R(110^\circ\text{C}) = R(90^\circ\text{C}) \times 0.5 = 3.0 \times 0.5 = 1.5\text{ k}\Omega

Since

110^\circ\text{C}

is exactly

20^\circ\text{C}

above

90^\circ\text{C}

, the resistance halves once more.

Key Concept

Extrapolating non-linear (exponential) relationships by identifying constant ratios over equal intervals of the independent variable.

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