Question
Some digital thermometers measure the current through a semiconductor to determine a patient's temperature. If a thermometer uses a germanium wire that has a resistance of $R$ at $37.0^{\circ} \mathrm{C}$ (normal body temperature), what is its resistance at $40.0^{\circ} \mathrm{C} ?$
Step 1
Step 1: We know that the resistance of a device is given by $R = \rho \frac{L}{A}$, where $\rho$ is the resistivity, $L$ is the length, and $A$ is the cross-sectional area. Show more…
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A digital thermometer employs a thermistor as the temperature-sensing element. A thermistor is a kind of semiconductor and has a large negative temperature coefficient of resistivity $\alpha$. Suppose $\alpha=-0.060\left(\mathrm{C}^{\circ}\right)^{-1}$ for the thermistor in a digital thermometer used to measure the temperature of a patient. The resistance of the thermistor decreases to $85 \%$ of its value at the normal body temperature of $37.0^{\circ} \mathrm{C}$. What is the patient's temperature?
A digital thermometer employs a thermistor as the temperature-sensing element. A thermistor is a kind of semiconductor and has a large negative temperature coefficient of resistivity α. Suppose α = -0.035 (°C)^-1 for the thermistor in a digital thermometer used to measure the temperature of a patient. The resistance of the thermistor decreases to 75% of its value at the normal body temperature of 37.0°C. What is the patient's temperature? °C = ?
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