Question

A thermistor has a resistance of \( 10\,\text{k}\Omega \) at \( 25^\circ\text{C} \) and \( 1\,\text{k}\Omega \) at \( 100^\circ\text{C} \). The range of operation is \( 0^\circ\text{C} \) to \( 150^\circ\text{C} \). The excitation voltage is 5 V and a series resistor of \( 1\,\text{k}\Omega \) is connected to the thermistor. The power dissipated in the thermistor is: 

 

• A. 4.0 mW
• B. 4.7 mW
• C. 5.4 mW
• D. 6.1 mW

Hint:

Use $P = I^2 R$ to calculate power dissipated in specific components, and consider worst-case resistance (minimum $R_T$ for max power).

Answer 1

The Correct Option is D

We are given:
- Excitation voltage \( V = 5\,\text{V} \)
- Series resistor \( R_1 = 1\,\text{k}\Omega \)
- Thermistor resistance \( R_T = 1\,\text{k}\Omega \) at \( 100^\circ\text{C} \) (use this as worst-case scenario since it gives maximum power dissipation)

Total resistance in the circuit:
\[ R_{\text{total}} = R_1 + R_T = 1\,\text{k}\Omega + 1\,\text{k}\Omega = 2\,\text{k}\Omega \]
Current through the circuit:
\[ I = \frac{V}{R_{\text{total}}} = \frac{5}{2000} = 2.5\,\text{mA} \]
Power dissipated in the thermistor:
\[ P = I^2 \cdot R_T = (2.5 \times 10^{-3})^2 \times 1000 = 6.25 \times 10^{-6} \times 1000 = 6.25 \times 10^{-3} = \textbf{6.25 mW} \]
Rounding down appropriately as per answer choices, the closest correct answer is: 6.1 mW.