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The values of the material constant p for thermistors P and Q are 4000 K and 3000 K, respectively. The resistance of each thermistor at 298 K is 2 kΩ. At 373 K, the ratio of the resistance of thermistor P to that of thermistor Q will be closest to (2009)
  • a)
    1.33
  • b)
    1.00
  • c)
    0.75
  • d)
    0.50
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
The values of the material constant p for thermistors P and Q are 400...
For thermistor P,
= 0.1345
For thermistor Q,
= 0.264
Then, the ratio of the resistance of thermistor P to thermistor Q
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Most Upvoted Answer
The values of the material constant p for thermistors P and Q are 400...
For thermistor P,
= 0.1345
For thermistor Q,
= 0.264
Then, the ratio of the resistance of thermistor P to thermistor Q
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Community Answer
The values of the material constant p for thermistors P and Q are 400...
Understanding Thermistors
Thermistors are resistors whose resistance varies significantly with temperature. The relationship between resistance and temperature can be described using the material constant (p) and the resistance at a reference temperature.
Given Data
- Thermistor P:
- Material constant (p) = 4000 K
- Resistance at 298 K = 2 kΩ
- Thermistor Q:
- Material constant (p) = 3000 K
- Resistance at 298 K = 2 kΩ
Resistance Calculation at 373 K
The resistance (R) of a thermistor at temperature T can be expressed as:
R(T) = R0 * exp(p * (1/T - 1/T0))
Where:
- R0 = resistance at reference temperature T0 (298 K)
- T0 = 298 K
- T = 373 K
Now, we will calculate R_P (for thermistor P) and R_Q (for thermistor Q) at T = 373 K.
Calculating R_P
- For Thermistor P:
R_P = 2 kΩ * exp(4000 * (1/373 - 1/298))
Calculating R_Q
- For Thermistor Q:
R_Q = 2 kΩ * exp(3000 * (1/373 - 1/298))
Ratio of Resistances
To find the ratio of the resistances at 373 K:
R_P / R_Q = (R_P) / (R_Q) = (exp(4000 * (1/373 - 1/298))) / (exp(3000 * (1/373 - 1/298)))
This simplifies to:
R_P / R_Q = exp((1000) * (1/373 - 1/298))
Calculating this gives a ratio of approximately 0.50.
Conclusion
Thus, the ratio of the resistance of thermistor P to that of thermistor Q at 373 K is closest to 0.50, making the correct answer option 'D'.
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The values of the material constant p for thermistors P and Q are 4000 K and 3000 K, respectively. The resistance of each thermistor at 298 K is 2 kΩ. At 373 K, the ratio of the resistance of thermistor P to that of thermistor Q will be closest to (2009)a)1.33b)1.00c)0.75d)0.50Correct answer is option 'D'. Can you explain this answer?
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The values of the material constant p for thermistors P and Q are 4000 K and 3000 K, respectively. The resistance of each thermistor at 298 K is 2 kΩ. At 373 K, the ratio of the resistance of thermistor P to that of thermistor Q will be closest to (2009)a)1.33b)1.00c)0.75d)0.50Correct answer is option 'D'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about The values of the material constant p for thermistors P and Q are 4000 K and 3000 K, respectively. The resistance of each thermistor at 298 K is 2 kΩ. At 373 K, the ratio of the resistance of thermistor P to that of thermistor Q will be closest to (2009)a)1.33b)1.00c)0.75d)0.50Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The values of the material constant p for thermistors P and Q are 4000 K and 3000 K, respectively. The resistance of each thermistor at 298 K is 2 kΩ. At 373 K, the ratio of the resistance of thermistor P to that of thermistor Q will be closest to (2009)a)1.33b)1.00c)0.75d)0.50Correct answer is option 'D'. Can you explain this answer?.
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