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electrical_engineering_1:task_70jjg4yzznocarsq_with_calculation [2023/02/11 23:06] – mexleadminelectrical_engineering_1:task_70jjg4yzznocarsq_with_calculation [Unbekanntes Datum] (aktuell) – gelöscht - Externe Bearbeitung (Unbekanntes Datum) 127.0.0.1
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-{{tag>resistivity power exam_WS2022}} 
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-<panel type="info" > <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> 
-<fs x-large>**Exercise ~~#~~ : Temperature-dependent resistance of a wire** \\ (written test, approx. 6% of a 60-minute written test, WS2022) \\ \\ </fs> 
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-A heating element made of a Nichrome wire with a round cross-section is used in an electric oven. Nichrome is a common Nickel Chromium alloy for heating elements. \\ 
-The Nichrome wire has a resistivity of $1.10\cdot 10^{-6} \Omega m$. \\ 
-The heating element is $3 m$ long and has a diameter of $3.57 mm$. \\ \\ 
-1. Calculate the resistance R of the heating element. 
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-<button size="xs" type="link" collapse="RJ0R6J4APUMUKRJ6_1_path">{{icon>eye}} Solution</button><collapse id="RJ0R6J4APUMUKRJ6_1_path" collapsed="true"> 
-\begin{align*} 
-R &= \rho \cdot \frac{l}{A} && | \text{with } A = r^2 \cdot \pi = \frac{1}{4} d^2 \cdot \pi \\ 
-R &= \rho \cdot \frac{4 \cdot l}{d^2 \cdot \pi} && \\ 
-R &= 1.10\cdot 10^{-6} \Omega m \cdot \frac{4 \cdot 3m}{(3.57\cdot 10^{-3}m)^2 \cdot \pi} && \\ 
-\end{align*} 
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-</collapse> 
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-<button size="xs" type="link" collapse="RJ0R6J4APUMUKRJ6_1_solution">{{icon>eye}} Final result</button><collapse id="RJ0R6J4APUMUKRJ6_1_solution" collapsed="true"> 
-\begin{align*} 
-R &= 0.33 \Omega \\ 
-\end{align*} 
- \\ 
-</collapse> 
- 
-2. The heating element is used to heat the oven to a temperature of 180°C. For this, a power dissipation (= heat flow) of P=40 W is necessary. \\ Calculate the current I needed to operate it. 
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-<button size="xs" type="link" collapse="RJ0R6J4APUMUKRJ6_2_path">{{icon>eye}} Solution</button><collapse id="RJ0R6J4APUMUKRJ6_2_path" collapsed="true"> 
- 
-\begin{align*} 
-P = U \cdot I = R \cdot I^2  \quad \rightarrow \quad I= \sqrt{\frac{P}{R}} = \sqrt{\frac{40 W}{0.33 \Omega}}  
-\end{align*} 
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-</collapse> 
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-<button size="xs" type="link" collapse="RJ0R6J4APUMUKRJ6_2_solution">{{icon>eye}} Final result</button><collapse id="RJ0R6J4APUMUKRJ6_2_solution" collapsed="true"> 
-\begin{align*} 
-I = 11 A 
-\end{align*} 
- \\ 
-</collapse> 
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-</WRAP></WRAP></panel>