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electrical_engineering_1:task_70jjg4yzznocarsq_with_calculation [2023/03/29 18:01] – mexleadminelectrical_engineering_1:task_70jjg4yzznocarsq_with_calculation [Unbekanntes Datum] (aktuell) – gelöscht - Externe Bearbeitung (Unbekanntes Datum) 127.0.0.1
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-{{tag>temperature_dependent_resistance power heat exam_ee1_WS2022}} 
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-#@TaskTitle_HTML@#~~#@ee1_taskctr.#~~  Temperature-dependent Resistance \\ <fs medium>(written test, approx. 6 % of a 60-minute written test, WS2022)</fs> #@TaskText_HTML@#     
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-A thermistor is used as a temperature sensor in a refrigeration system. The thermistor has a resistance of $10 ~\rm{k}\Omega$ at $+25 ~°\rm{C}$. \\ 
-Its temperature coefficients are: $\alpha=0.01 ~{{1}\over{\rm{K}}}$ and $\beta=71 \cdot 10^{-6}~{{1}\over{\rm{K}^2}}$  \\  
-The temperature inside the refrigeration system can reach down to $-40 ~°\rm{C}$. 
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-1. Calculate the resistance of the thermistor at $-40 ~°\rm{C}$.  
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-\begin{align*} 
-R &= R_0 \cdot (1 + \alpha \cdot \Delta T + \beta \cdot \Delta T^2) && | \text{with  } \Delta T = T_{\rm end} - T_{\rm start}\\ 
-R &= 10 ~\rm{k}\Omega \cdot \left(1 + 0.01 ~{{1}\over{\rm{K}}} \cdot (-40~°\rm{C} - 25~°\rm{C} ) + 71 \cdot 10^{-6}~{{1}\over{\rm{K}^2}} \cdot (-40~°\rm{C} - 25~°\rm{C})^2 \right) \\ 
-\end{align*} 
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-\begin{align*} 
-R &= 6.5 ~\rm{k}\Omega \\ 
-\end{align*} 
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-2. Additionally, explain which effect a resistive temperature sensor can have on the refrigeration system.  
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-Resistors transfer electrical energy out of the circuit and generate heat. Therefore, a resistive sensor might heat up the refrigeration system. \\ \\ 
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-3. Regarding question 2.: Given a constant sensor voltage, would a sensor with tenfold resistance be better or worse? Give an explanation for your answer.  
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-The power of the resistor $P = U \cdot I = R \cdot I^2 = {{U^2}\over{R}}$ is equivalent to the heat flow. \\  
-Therefore, with constant $U$ and increasing $R$ the power decreases. Ten times more resistance decreases the heat flow to one-tenth. 
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