{{tag>efficiency charges power exam_ee1_SS2023}}
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#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~  Efficiency       \\ <fs medium>(written test, approx. 14 % of a 60-minute written test, SS2023)</fs>
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A lithium-ion battery cell can be considered a linear voltage source with an internal resistance $R_\rm i$ and a source voltage $U_\rm s=3.5 ~\rm V$. 
The battery shall provide energy for a mobile device with a load resistance of $R_\rm L=2 ~\Omega$ 
The following values are from the lithium-ion battery datasheet: 
  * Internal impedance: 	$R_\rm i	=50 ~\rm m\Omega$ 
  * Maximum discharge current: 	$I_{\rm Dis max}	=3 ~\rm A$ 
  * Typical capacity: 		$2'600 ~\rm mAh$

1. Draw an **equivalent circuit diagram** with the internal resistance and an external load. Label all voltages and currents.

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{{drawio>electrical_engineering_1:w3wf215v2u98ty07Circuit.svg}}

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2. Calculate the **efficiency of the battery** in this case. 
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\begin{align*}
\eta &= {{R_\rm L}\over{R_\rm L + R_\rm i}} \\
    &= {{2 ~\Omega}\over{2 ~\Omega + 0.05 ~\Omega}} 
\end{align*}

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#@ResultBegin_HTML~2~@#
\begin{align*}
\eta = 97.56... \% \rightarrow  \eta = 97.6 \% 
\end{align*}
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3. (**HARD**) Once the load resistance is changed, the efficiency for discharging also changes. What would be the **lowest possible efficiency**?  
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Lowest efficiency for highest current, so for $I_{\rm Dis max}.
In this case, the efficiency is: 

\begin{align*}
\eta &= {{U_\rm S - R_\rm i \cdot I_{\rm Dis max}}\over{U_\rm S}} \\
     &= 1 - R_\rm i \cdot {{I_{\rm Dis max}}\over{U_\rm S}} \\
     &= 1 - 0.05 {~\rm \Omega} \cdot {{3~\rm A}\over{3.5 ~\rm V}} \\     
\end{align*}

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#@ResultBegin_HTML~3~@#
\begin{align*}
\eta = 95.71... ~\% \rightarrow  \eta = 95.7 ~\% 
\end{align*}
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4. Calculate the **voltage drop on the load resistance** $R_\rm L=2 ~\Omega$.
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Voltage divider with $R_\rm i$ and $R_\rm L$:

\begin{align*}
U_\rm L= U_\rm S \cdot {{R_\rm L}\over{R_\rm L + R_\rm i}}
\end{align*}

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\begin{align*}
U_\rm L= 3.414... ~\rm V \rightarrow U_\rm L= 3.4 ~\rm V 
\end{align*}
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5. How much **charged $\rm Li$ ions** have **to be moved** in the battery to charge it from $0~ \%$ to $100~\%$? \\
Lithium is monovalent – so, there are only $\rm Li^+$ ions. The elementary charge is $q_\rm e=1.602 \cdot 10^{-19} ~\rm C$.

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\begin{align*}
n_{\rm Li^+}={{Q}\over{q_\rm e}} \\ \\

Q &= 2.6 {~\rm Ah} \\
  &= 2.6 \cdot 3600 {~\rm As} \\
  &= 9360 {~\rm As} = 9360 {~\rm C}
\end{align*}

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#@ResultBegin_HTML~5~@#
\begin{align*}
n_{\rm Li^+}=5.842... \cdot 10^{22} \rightarrow n_{\rm Li^+}=5.84 \cdot 10^{22} \\
\end{align*}
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