A 8 V ideal source is connected across a series pair: a 4 kΩ fixed resistor above a thermistor. The output is measured across the thermistor with a voltmeter of effectively infinite resistance. The thermistor is 12 kΩ when cold and 4 kΩ when warm. (a) Calculate both output voltages. (b) Explain their change. (c) State the trend if output were instead measured across the fixed resistor.

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Accepted Answer

Thermistor output: 6 V cold and 4 V warm; it falls. Fixed-resistor output rises from 2 V to 4 V. The two components carry the same current because they are in series. Their voltage drops are proportional to resistance, so the thermistor output is Vout = Vs × Rthermistor/(Rfixed + Rthermistor). When cold, Vout = 8 × 12/(4 + 12) = 6 V. Matching kilo-ohm units cancel in the resistance ratio, so no separate conversion is needed here. When warm, Vout = 8 × 4/(4 + 4) = 4 V. Although total current increases as total resistance falls, the thermistor receives a smaller share of the supply voltage. The two voltage drops must still add to 8 V. Thus the fixed resistor's output rises from 2 V to 4 V. The infinite-resistance voltmeter avoids loading the divider and changing the intended resistance network.

A temperature-sensitive potential divider 3

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