Exam Style Question - 2

QuestionA potential divider consists of a fixed 2.0 kΩ resistor in series with a thermistor, connected to a 6.0 V supply of negligible internal resistance. The output p.d. V_out is taken across the thermistor. At 20 °C the thermistor has a resistance of 4.0 kΩ.(a) Calculate V_out at 20 °C.
(b) The temperature rises and the thermistor's resistance falls to 1.0 kΩ. Calculate the new V_out.
(c) Explain, in terms of charge carriers, why the thermistor's resistance falls.(a)V_out = V_in × R₂/(R₁ + R₂) = 6.0 × 4.0/(2.0 + 4.0) = 4.0 V(b)V_out = 6.0 × 1.0/(2.0 + 1.0) = 2.0 VThe thermistor now has a smaller share of the total resistance, so it takes a smaller share of the supply p.d. The fixed resistor now has 4.0 V across it.(c) The thermistor is made of a semiconductor. As its temperature rises, more electrons gain enough energy to break free and become conduction electrons, so the number of charge carriers per unit volume, n, increases. From I = nqvA, more charge carriers give a larger current for the same p.d., so the resistance falls. (The lattice vibrates more too, but the increase in n has the much bigger effect.)Exam tip: in a potential divider you can leave the resistances in kΩ, because only their ratio matters.