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Question #42127
The Question
A thermistor has a temperature-dependent resistance given by , where is the absolute temperature in Kelvin. It is given that and . The thermistor is connected in series with an inductor of inductance and an AC voltage source . The temperature is measured using a faulty linear scale , on which the freezing and boiling points of water are and , respectively. In steady state, the thermistor reaches a constant operating temperature measured as , while the surrounding environment is at . Assuming Newton's law of cooling holds, what is the thermal dissipation constant of the thermistor?
Options
- A.
- B.
- C.
- D.
Check Your Answer
Correct Answer
Step-by-Step Solution
Explanation
How the Examiner Thinks
Intuition_Breaker
Train your brain to resist "obvious" answers that are designed to deceive your intuition.
Concepts Behind This Question
Celsius-Fahrenheit Calibration Errors|Dynamic Resistance (AC Resistance)
Why This Question Is Tricky
Common Mistakes
This question tests the synthesis of thermal calibration, AC circuit impedance, and steady-state energy conservation.
It employs a 'Dynamic Resistance' trap where the resistance is coupled to the operating temperature, which in turn depends on the power dissipated.
A common JEE Advanced pitfall tested here is confusing apparent power () with true active power () in circuits with reactive components, leading to Option C.
Another trap is using peak current instead of RMS current for power calculations, leading to Option B.
Finally, it forces students to correctly map a faulty linear scale to absolute Kelvin before evaluating the non-linear exponential resistance function.
Quick Facts About This Question
Question Information
| Subject | Physics |
| Chapter | Thermal Properties of Matter - Class 11 Part 2 |
| Concept | Celsius-Fahrenheit Calibration Errors |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |