Exam Summer Semester 2023

Additional permitted Aids

  • non-programmable calculator,
  • formulary (2 DIN A4 pages)

Hits

  • The duration of the exam is 60 min.
  • Attempts to cheat will lead to exclusion and failure of the exam.
  • Withdrawal is no longer possible after these exam has been handed out.
  • Please write down intermediate calculations and results on the assignment sheet. (when more space is needed also on the reverse side. In this case: Mark it clearly).
  • Always use units in the calculation.
  • Use a document-proof, non-red pen.

Tasks

Exercise E1 Resistivity and temperature dependent Resistance
(written test, approx. 7 % of a 60-minute written test, SS2023)

The resistance of the dielectric material of a film capacitor has to be calculated.
The given film capacitor has an internal surface of $A=100 ~\rm dm^2$ and a distance between the plates of $d=0.8 ~\rm μm$.
The resistivity of the dielectric material is $\rho_{\rm PP}(20 ~\rm °C)=10^{17} ~\Omega m$.
For the given material the temperature coefficients in the range of $20 ~\rm °C$ and $55 ~\rm °C$ are given as $\alpha =-0.048 ~\rm 1/K$ and $\beta=+0.00057 ~\rm 1/K^2$.

electrical_engineering_1:kyt15w11e3sempb2circuit.svg

Calculate the resistance for the dielectric material for $20 ~\rm °C$.

Solution

\begin{align*} R(20 ~\rm °C) &= \rho\cdot {{d}\over{A}}\\ &= 10^{17} ~\Omega \rm m \cdot {{0.8\cdot 10^{-6} ~\rm m}\over{1 ~\rm m^2}} \end{align*}

Result

\begin{align*} R(20 ~\rm °C) &= 80 ~\rm G\Omega\\ \end{align*}

Calculate the resistance for the dielectric material for $55 ~\rm °C$.

Solution

\begin{align*} R(55 ~\rm °C) &= R(20 ~\rm °C) \cdot (1+\alpha\cdot\Delta T + \beta\cdot T^2 + ...)\\ &= 80 ~\rm G\Omega \cdot (1-0.048 ~\rm 1/K \cdot(35 ~{\rm K}) + 0.00057 ~\rm 1/K^2\cdot\Delta (35 ~{\rm K})^2 ) \end{align*}

Result

\begin{align*} R(55 ~\rm °C) &= 1.46 ~\rm G\Omega \end{align*}

(In reality, the relationship between $R$ and $T$ for Polypropylene is better described by the $B25$ value in an exponential formula. In this case, the best fit would be $B25 = 15’000$ for $T$ between $20 ~\rm °C$ and $100 ~\rm °C$)

Exercise E2 Analyzing a Scope Plot
(written test, approx. 12 % of a 60-minute written test, SS2023)

On an oscilloscope is the following plot visible.
The measured current curve shall be visible as a dashed line.
The continuous line shows the voltage.

electrical_engineering_1:exbtaqnp18vn0gricircuit.svg

Use the correct symbols and units in your answers!

1. Calculate the frequency $f$ of the periodic signals.

Solution

Frequency $f$ is given by the period $T$. The period can be measured in the imagine of the scope.

  1. The sine waves repeat after $6 ~\rm divisions$ (e.g. from falling turning point to falling turning point of one curve)
  2. The scale is $0.1 ~\rm ms/Div$

\begin{align*} f &= {{1} \over {T}} \\ T &= 6 ~\rm Div \cdot 0.1 ~ms/Div \\ \rightarrow f &= {{1} \over {6 ~\rm Div \cdot 0.1 ~ms/Div}} \end{align*}

Result

\begin{align*} f = 1.66... ~{\rm kHz} \rightarrow f &= 1.67 ~\rm kHz \end{align*}

2. Write down the amplitude values for the voltage and current.

Solution

Similar to 1. the amplitude can be derived. The amplitude is given by the voltage difference between maximum and turning point.

  1. The amplitude of the voltage is $3 ~\rm divisions$
  2. The scale is $5 ~\rm V/Div$

\begin{align*} \hat{U} &= 3 ~\rm Div \cdot 5 ~\rm V/Div \end{align*}

Similar for the current.

Result

\begin{align*} \hat{U} &= 15 ~\rm V \\ \hat{I} &= 0.5 ~\rm A \end{align*}

3. What are the RMS values of voltage and current?

Solution

The RMS values for sine waves are given as \begin{align*} \square_{\rm RMS} ={{1} \over {2} } \sqrt{2} \cdot \hat{\square} \end{align*}

\begin{align*} U_{\rm RMS} & ={{1} \over {2} } \sqrt{2} \cdot \hat{U} ={{1} \over {2} } \sqrt{2} \cdot 15 ~\rm V \\ I_{\rm RMS} & ={{1} \over {2} } \sqrt{2} \cdot \hat{I} ={{1} \over {2} } \sqrt{2} \cdot 0.5 ~\rm A \end{align*}

Result

\begin{align*} U_{\rm RMS} & = 10.606... ~\rm V \rightarrow U_{\rm RMS} = 10.6 ~\rm V \\ I_{\rm RMS} &= 0.3535... ~\rm A \rightarrow I_{\rm RMS} = 0.35 ~\rm A \end{align*}

4. What is the phase shift on the system under test (in radiant and degree)?

Solution

The phase shift $\varphi$ is given as the angle between the current and voltage phasor. The phase shift is negative when the voltage lags the current.
A full period has an angle of $360°$ or $2\pi$ in $\rm radian$.
One has to find out the phase shift as a fraction of a period to get the value of the phase shift:

\begin{align*} \varphi &= \rm -{{0.5 ~Div} \over { 6 ~Div}} \cdot 2\pi \end{align*}

Result

\begin{align*} \varphi &= -{{1} \over { 6}} \pi \\ \varphi &= - 30° \end{align*}

Exercise E3 Complex voltage dividers
(written test, approx. 16 % of a 60-minute written test, SS2023)

The circuit below shall be given with the following values:

  • $R = 1.1 ~\rm k\Omega$
  • $L = 3.5 ~\rm mH$
  • $\underline{U}_\rm I = 5 ~\rm V$
  • $f_0 = 150 ~\rm kHz$

electrical_engineering_1:c9fj1si7l797equscircuit.svg

1. Calculate the impedance $\underline{Z}_L$.

Solution

\begin{align*} \underline{Z}_L &= {\rm j} \cdot \omega \cdot L \\ &= {\rm j} \cdot 2\pi \cdot 150 ~\rm kHz \cdot 3.5 ~\rm mH \end{align*}

Result

\begin{align*} \underline{Z}_L &= {\rm j}\cdot 3.3 ~\rm k\Omega \end{align*}

2. Draw the two impedance phasors and the resulting phasor for the overall impedance in a diagram. Choose an appropriate scaling factor and write it down.

Result

electrical_engineering_1:c9fj1si7l797equsphasors.svg

3. Calculate the output voltage $|\underline{U}_\rm O|$ and the phase shift between $\underline{U}_\rm O$ and $\underline{U}_\rm I$.

Solution

The formula to start with is the (complex) voltage divider: \begin{align*} \underline{U}_{\rm O} &= \underline{U}_{\rm I} \cdot {{R} \over {R+{\rm j}\cdot Z_L }} \\ &= \underline{U}_{\rm I} \cdot {{R} \over {R+{\rm j}\cdot Z_L }} \cdot {{R-{\rm j}\cdot Z_L} \over {R-{\rm j}\cdot Z_L }}\\ \end{align*}

Result

\begin{align*} \underline{U}_{\rm O} &= 0.5 ~\rm V - j \cdot 1.5 ~V \end{align*}

4. Calculate the cut-off frequency of this setup.

Solution

At cut off frequency the absolute values of impedances $\underline{Z}_L$ is equal to $\underline{Z}_R=R$. This leads to: \begin{align*} f_{\rm cutoff} &= {{R} \over {2\pi \cdot L} } \\ &= {{1.1 \cdot 10^3 {\rm {V} \over{A} } } \over {2\pi \cdot 3.5 \cdot 10^{-3} {\rm {As} \over{V} } }} \end{align*}

Result

\begin{align*} f_{\rm cutoff} &= 50 ~\rm kHz \end{align*}

Exercise E4 Pure Resistor Network Simplification
(written test, approx. 12 % of a 60-minute written test, SS2023)

The circuit below shall be given.

The values in the circuit are

  • $R_1 = 60 ~\Omega$
  • $R_2 = 40 ~\Omega$
  • $R_3 = 40 ~\Omega$
  • $R_4 = 100 ~\Omega$
  • $U_{\rm AB} = 10 ~\rm V$

electrical_engineering_1:cgeyprm6oboukccicircuit.svg

1. Calculate the voltage at node $\rm K$, when switch $\rm S$ is open.
It might be beneficial to redraw the circuit first.

Solution

Rearranging the circuit one can get: electrical_engineering_1:cgeyprm6oboukccicircuitsolve1.svg

Once the switch $\rm S$ is opened, the upper part is a parallel circuit. Therefore, $R_{\rm eq}$ is given as: \begin{align*} R_{\rm eq} &= (R_1+R_2)||(R_1+R_2)+R_4 \\ &= {{1}\over{2}}\cdot(R_1+R_2)+R_4 \\ &= {{1}\over{2}}\cdot(60~\Omega + 40~\Omega) + 100~\Omega \\ \end{align*}

Result

\begin{align*} R_{\rm eq} &= 150~\Omega \end{align*}

2. Calculate the voltage at node $\rm K$, when switch $\rm S$ is closed.

Solution

The voltage divider for node $\rm K$ has the same proportionality as the voltage divider for node $\rm K'$. Therefore, the potential of $\rm K$ is the same as for $\rm K'$. There will be no current flow through $R_3$. The resistance does not create a voltage drop and therefore does not interfere with the circuit.

Result

The equivalent resistance is similar to the circuit with opened switch. \begin{align*} R_{\rm eq} &= 150~\Omega \end{align*}

Exercise E5 Pure Resistor Network Simplification I
(written test, approx. 14 % of a 60-minute written test, SS2023)

The circuit below shall be given.

electrical_engineering_1:erlctd760zmvox0tcircuit.svg

1. What is the equivalent resistance $R_{\rm eq}$?

Solution

Part of the circuit is shorted. Here the resistors (marked in red) are shorted by the connections marked in blue: electrical_engineering_1:erlctd760zmvox0tcircuitsolve1.svg

The circuit can then be rearranged for better interpretation: electrical_engineering_1:erlctd760zmvox0tcircuitsolve2.svg

Therefore, $R_{\rm eq}$ is given as: \begin{align*} R_{\rm eq} &= (2R||2R + R + R)||6R &&+ 6R ||(2R + 2R + 4R||4R) \\ &= (R + R + R)||6R &&+ 6R ||(2R + 2R + 2R) \\ &= 3R||6R &&+ 6R ||6R \\ &= {{3R\cdot 6R}\over{3R+6R}} &&+3R \\ \end{align*}

Result

\begin{align*} R_{\rm eq} &= 5R \end{align*}

2. Now the input voltage shall be given as $U_{\rm AB} = 60 ~\rm V$. What is the value for $U$ in the circuit?

Solution

The current through the circuit is given as $I_{\rm AB} = U_{\rm AB} \cdot R_{\rm eq} $.
This current has to flow in summary through parallel branches. The voltage $U$ in question in the upper right branch given by $(4R||4R)+2R+2R$. Its resistance is just the same as the upper left branch $6R$.
Therefore, half of the current flows to the left half to the right side.
The voltage $U$ is consequently: \begin{align*} U &= {{I_{\rm AB} }\over{ 2\cdot R_{\rm eq} }} \\ &= {{U_{\rm AB} \cdot 2R }\over{ 2\cdot R_{\rm eq} }} \\ &= {{60 ~\rm V }\over{ 5 }} \end{align*}

Result

\begin{align*} U &= 12 ~\rm V \\ \end{align*}

Exercise E6 Equivalent Linear Source
(written test, approx. 10 % of a 60-minute written test, SS2023)

The circuit below has to be simplified. Use equivalent linear sources for simplification.
Calculate the internal resistance $R_\rm i$ and the source voltage $U_\rm s$ of an equivalent linear voltage source.

  • $R_1 = 5 ~\Omega$
  • $U_0 = 10 ~\rm V$
  • $R_2 = 7.5 ~\Omega$
  • $I_3 = 0.5 ~\rm A$
  • $R_4 = 10 ~\Omega$
  • $U_5 = 4 ~\rm V$

electrical_engineering_1:lefxcuaxiu8ewcr9circuit.svg

Solution

The principle idea here is to find parts of the circuit which are already a linear (voltage or current) source. Then this can be transformed into the equivalent other source, as shown in the next picture.

electrical_engineering_1:lefxcuaxiu8ewcr9circuitsolve1.svg

In order to get the currents one has to calculate it by $I_x = {{U_x}\over{R_x}}$ \begin{align*} I_0 &= {{U_0}\over{R_1}} = {{10~\rm V}\over{5 ~\Omega}} = 2 ~\rm A\\ I_5 &= {{U_5}\over{R_4}} = {{4 ~\rm V}\over{10~\Omega}} = 0.4 ~\rm A \\ \end{align*}

$I_3$ and $I_0$ can be combined to $I_{03}=I_0 - I_3$ facing upwards: \begin{align*} I_{03}=1.5~\rm A \end{align*}

Then, the linear current source $I_{03}$ with $R_1$ gets transformed into a linear voltage source with $U_{03}=R_1\cdot I_{03}$ facing down. \begin{align*} U_{03}=7.5~\rm V \end{align*}

Then, the resistors $R_1$ and $R_2$ can be combined to $R_{12}= R_1 + R_2$.

After this, the next step is to make a linear current source out of $U_{03}$ and $R_{12}$. The current will be $I_{0123}={{U_{03}}\over{R_{12}}}$, facing up again. \begin{align*} I_{0123}=0.6~\rm A \end{align*}

The second-last step is the sum up of the current sources $I_{0123}$ and $I_5$ as $I_{01235}=I_{0123}-I_5$ and the resistors as $R_{124}=R_{12}||R_4$. \begin{align*} I_{01235} &=0.2~\rm A \\ R_{124} &=5.55... ~\Omega \end{align*}

The final step is the back-transformation to a linear voltage source, with $U_{\rm AB} = R_{124} \cdot I_{01235}$.

The simplest and fastest (= for exams) is to work with interim results in the calculation.
Here, there there is also a full final formula given:

\begin{align*} U_\rm s &= U_{\rm AB} = I_{01235} \cdot R_{124} \\ &= (I_{0123}-I_5) \cdot (R_{12}||R_4) \\ &= \left({{U_{03}}\over{R_{12}}}-I_5\right) \cdot \left((R_1 + R_2)||R_4\right) \\ &= \left({{R_1\cdot I_{03}}\over{R_1+R_2}}-I_5\right) \cdot \left((R_1 + R_2)||R_4\right) \\ &= \left({{R_1\cdot \left({{U_0}\over{R_1}} - I_3\right)}\over{R_1+R_2}}-I_5\right) \cdot \left((R_1 + R_2)||R_4\right) \\ \end{align*}

Result

\begin{align*} U_\rm s &= U_{\rm AB} =1.11...~\rm V \\ R_{\rm i} &= 5.55... ~\Omega \end{align*}

Exercise E7 (Dis)Charging Capacities
(written test, approx. 14 % of a 60-minute written test, SS2023)

The circuit below has to be analyzed. The component values are:

  • $U = 10 ~\rm V$
  • $C_1 = 200 ~\rm nF$
  • $R_1 = 8.0 ~\rm k\Omega$
  • $R_2 = 17 ~\rm k\Omega$
  • $R_3 = 7.0 ~\rm k\Omega$
  • $I = 2.0 ~\rm mA$

electrical_engineering_1:p8yrdjr60k6bvc4ncircuit.svg

Before $t_0$ all switches are switched as shown and the capacitor is fully discharged.
At $t_0=0 ~\rm s$ the switch $S_1$ shall switch to the voltage source.

1. Calculate the time constant for charging the capacitor.

Solution

The time constant is generally given as: \begin{align*} \tau &= R\cdot C \end{align*}

Once $S_1$ is closed and $S_2$ is open at $t_0$, the source $U$ drives the current through the series circuit given by $S_1$, $C$, $R_1$ and $R_3$.
Therefore, $R= R_1 + R_3$ \begin{align*} \tau_1 &= (R_1+R_3)\cdot C \\ &= (8~\rm k\Omega + 7~\rm k\Omega) \cdot 0.2 ~\rm \mu F \\ &= 15\cdot 10^{3} \cdot 0.2 \cdot 10^{-6} ~\rm {{V}\over{A}}{{As}\over{V}}\\ \end{align*}

Result

\begin{align*} \tau_1 = 3.0 ~\rm ms \end{align*}

2. What is the voltage $u_C$ at $t_1=t_0+4 ~\rm ms$?

Solution

\begin{align*} u_C(t_1) &= U_0 \cdot(1-e^{-t/\tau} ) \\ &= 10~\rm V \cdot(1-e^{-4~ms/3~ ms} ) \end{align*}

Result

\begin{align*} u_C(t_1) &= 7.36~\rm V \end{align*}

3. The capacitors shall be charged to $U=10 ~\rm V$ at the time $t_2$.
At this point in time, the switch $S_1$ switches to the situation shown in the drawing. What is the new time constant?

Solution

Again, the time constant is generally given as: \begin{align*} \tau &= R\cdot C \end{align*}

Now, $\rm S_1$ is opened and $\rm S_2$ is closed. Then, the source $U$ drives the current through the series circuit given by $\rm S_1$, $C$, $R_1$ and $R_2$.
Therefore, $R= R_1 + R_2$ \begin{align*} \tau_2 &= (R_1+R_2)\cdot C \\ &= (8~\rm k\Omega + 17~\rm k\Omega) \cdot 0.2 ~\rm \mu F \\ &= 25\cdot 10^{3} \cdot 0.2 \cdot 10^{-6} ~\rm {{V}\over{A}}{{As}\over{V}}\\ \end{align*}

Result

\begin{align*} \tau_2 = 5.0 ~\rm ms \end{align*}

4. Draw the overall course of the voltage $u_C(t)$ over time in the diagram below.
Use an appropriate $x$-axis scale.

electrical_engineering_1:p8yrdjr60k6bvc4nutdiagramquestion.svg

Solution

Both courses of the voltage for charging and discharging are described with an exponential function. However, the curve for charging increases first steep and flattens out for longer time scales ($\propto (1-e^{-x})$).

Result

electrical_engineering_1:p8yrdjr60k6bvc4nutdiagramsolution.svg

5. The switch $S_2$ is then closed, with $S_1$ still as shown in the drawing. What will be the maximum voltage of $u_C$?

Solution

The current of the source flows through the circuit consisting of $C$ in parallel with $R_1+R_2$. Without the parallel resistors, the current source would charge the capacitor „to infinity“ ($u_C \rightarrow \infty$) . This is here limited by the parallel resistors $R_1+R_2$.

The maximum voltage on the branch with the resisors $R_1+R_2$ is

\begin{align*} U_{12} &= R \cdot I \\ &= (R_1+R_2) \cdot I \end{align*}

This is also the maximum voltage on the capacitor, since it is in parallel with the resisors.

Result

\begin{align*} U_C &= 50 ~\rm V \end{align*}

Exercise E8 Impedances at Frequencies
(written test, approx. 14 % of a 60-minute written test, SS2023)

At which frequencies show the following components the given values?

1. An inductor with $X_{L1} = 60 ~\rm m\Omega$ and $L_1 = 15.9 ~\rm \mu H$.

Solution

\begin{align*} X_{L1} &= \omega_1 \cdot L = 2\pi f_1 \cdot L_1 \\ \rightarrow f_1 &= {{X_{L1}} \over {2\pi \cdot L_1}} \\ &= {{60 ~\rm m\Omega} \over {2\pi \cdot 15.9 ~\rm \mu H}} \\ &= {{60 \cdot 10^{-3}} \over {2\pi \cdot 15.9 \cdot 10^{-6}}} \rm {{ {{V}\over{A}} } \over { {{Vs}\over{A}}}} \\ \end{align*}

Result

\begin{align*} f_1 = 600.58...~{\rm Hz} \rightarrow f_1 = 600~{\rm Hz} \end{align*}

2. A capacitor with $C_2 = 5.2 ~\rm nF$, where an AC voltage of $U_2 = 6.8 ~\rm V$ generates a current $I_2 = 1 ~\rm mA$.

Solution

\begin{align*} X_{C2} &= {{1} \over{\omega_2 \cdot C_2} } = {{1} \over{2\pi f_2 \cdot C_2}}={{U_2} \over {I_2}} \\ \rightarrow f_2 &= {{I_2} \over{2\pi \cdot U_2 \cdot C_2}} \\ &= {{1 \cdot 10^{-3}} \over{2\pi \cdot6.8 \cdot 5.2 \cdot 10^{-9}}} \rm {{A} \over{V \cdot {{As} \over{V}} }} \end{align*}

Result

\begin{align*} f_2 = 4'500.9...~{\rm Hz} \rightarrow f_2 = 45.0 ~{\rm kHz}\end{align*}

3. An inductor with $L_3 = 50 ~\rm \mu H$, which shows the same absolute value of the impedance as a capacitor with $C_3 = 5.6 ~\rm nF$.

Solution

\begin{align*} X_{L3} &=X_{C3} \\ \omega_3 \cdot L_3 &= {{1} \over{\omega_3 \cdot C_3} } \\ 2\pi f_3 \cdot L_3 &= {{1} \over{2\pi f_3 \cdot C_3} } \\ \rightarrow f_3 &= {{1} \over{2\pi}} \sqrt{{{1} \over {C_3 \cdot L_3}}} \\ f_3 &= {{1} \over{2\pi}} \sqrt{{{1} \over {5.6\cdot 10^{-9} \cdot 50\cdot 10^{-6}}}. {{1}\over\rm {V/As \cdot A/Vs}} } \\ \end{align*}

Result

\begin{align*} f &= 300'774.5 ... ~\rm Hz \rightarrow f &= 300 ~\rm kHz \end{align*}

Exercise E9 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*}