A 15 V ideal source is connected across a series pair: a 11 kΩ fixed resistor above a thermistor. The output is measured across the thermistor with a voltmeter of effectively infinite resistance. The thermistor is 33 kΩ when cold and 11 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: 11.25 V cold and 7.5 V warm; it falls. Fixed-resistor output rises from 3.75 V to 7.5 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 = 15 × 33/(11 + 33) = 11.25 V. Matching kilo-ohm units cancel in the resistance ratio, so no separate conversion is needed here. When warm, Vout = 15 × 11/(11 + 11) = 7.5 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 15 V. Thus the fixed resistor's output rises from 3.75 V to 7.5 V. The infinite-resistance voltmeter avoids loading the divider and changing the intended resistance network.

A temperature-sensitive potential divider 10

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