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electrical_engineering_1:task_5ztn80yw2uibcsxr_with_calculation [2023/04/03 12:14] – angelegt mexleadminelectrical_engineering_1:task_5ztn80yw2uibcsxr_with_calculation [Unbekanntes Datum] (aktuell) – gelöscht - Externe Bearbeitung (Unbekanntes Datum) 127.0.0.1
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-{{tag>conversions battery chapter1_1}} 
-{{include_n>100}} 
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-#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~  Conversions: Battery                   #@TaskText_HTML@#    
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-1. How many minutes could an ideal battery with $10~{\rm kWh}$ operate a consumer with $3~{\rm W}$? 
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-#@PathBegin_HTML~1~@# 
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-\begin{align*} 
-W &= 10~{\rm kWh} &&= 10'000~{\rm Wh}\\ \\ 
-t &= {{W             }\over{P        }}  
- &&= {{10'000~{\rm Wh}}\over{3~{\rm W}}} \\ 
-&&&= 3'333~{\rm h}= 200'000~{\rm min} \quad (= 139~{\rm days}) 
-\end{align*} 
-#@PathEnd_HTML@# 
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-#@ResultBegin_HTML~1~@# 
-\begin{align*} 
-t = 200'000~{\rm min} 
-\end{align*} 
-#@ResultEnd_HTML@# 
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-2. Hard: Why is a real Lithium-ion battery with $10~{\rm kWh}$ not able to provide the given power for the calculated time? 
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-#@ResultBegin_HTML~2~@# 
-There are additional losses: 
-  * The battery has an internal resistance. Depending on the current provided by the battery, this leads to internal losses.  
-  * The internal resistance of the battery is depending on the state of charge (SoC) of the battery. 
-  * The wires also add additional losses to the system. 
-#@ResultEnd_HTML@# 
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-#@TaskEnd_HTML@#