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Exercise E6 Efficiency
(written test, approx. 14 % of a 60-minute written test, SS2023)

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.

Result

electrical_engineering_1:w3wf215v2u98ty07circuit.svg

2. Calculate the efficiency of the battery in this case.

Solution

\begin{align*} \eta &= {{R_\rm L}\over{R_\rm L + R_\rm i}} \\ &= {{2 ~\Omega}\over{2 ~\Omega + 0.05 ~\Omega}} \end{align*}

Result

\begin{align*} \eta = 97.56... \% \rightarrow \eta = 97.6 \% \end{align*}

3. (HARD) Once the load resistance is changed, the efficiency for discharging also changes. What would be the lowest possible efficiency?

Solution

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*}

Result

\begin{align*} \eta = 95.71... ~\% \rightarrow \eta = 95.7 ~\% \end{align*}

4. Calculate the voltage drop on the load resistance $R_\rm L=2 ~\Omega$.

Solution

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*}

Result

\begin{align*} U_\rm L= 3.414... ~\rm V \rightarrow U_\rm L= 3.4 ~\rm V \end{align*}

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 $q_\rm e=1.602 \cdot 10^{-19} ~\rm C$.

Solution

\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*}

Result

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