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 $\beta=+0.00057 ~\rm 1/K^2$.
Calculate the resistance for the dielectric material for $20 ~\rm °C$.
Calculate the resistance for the dielectric material for $55 ~\rm °C$.
(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 $100 ~\rm °C$)
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.
Use the correct symbols and units in your answers!
1. Calculate the frequency $f$ of the periodic signals.
Frequency $f$ is given by the period $T$. The period can be measured in the imagine of the scope.
\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*}
2. Write down the amplitude values for the voltage and current.
Similar to 1. the amplitude can be derived. The amplitude is given by the voltage difference between maximum and turning point.
\begin{align*} \hat{U} &= 3 ~\rm Div \cdot 5 ~\rm V/Div \end{align*}
Similar for the current.
3. What are the RMS values of voltage and current?
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*}
\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)?
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*}
\begin{align*} \varphi &= -{{1} \over { 6}} \pi \\ \varphi &= - 30° \end{align*}
The circuit below shall be given with the following values:
1. Calculate the impedance $\underline{Z}_L$.
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.
3. Calculate the output voltage $|\underline{U}_\rm O|$ and the phase shift between $\underline{U}_\rm O$ and $\underline{U}_\rm I$.
\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.
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*}
The circuit below shall be given.
The values in the circuit are
1. Calculate the voltage at node $\rm K$, when switch $\rm S$ is open.
It might be beneficial to redraw the circuit first.
Rearranging the circuit one can get:
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*}
2. Calculate the voltage at node $\rm K$, when switch $\rm S$ is closed.
The circuit below shall be given.
1. What is the equivalent resistance $R_{\rm eq}$?
Part of the circuit is shorted. Here the resistors (marked in red) are shorted by the connections marked in blue:
The circuit can then be rearranged for better interpretation:
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*}
2. Now the input voltage shall be given as $U_{\rm AB} = 60 ~\rm V$. What is the value for $U$ in the circuit?
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.
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.
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*}
The circuit below has to be analyzed. The component values are:
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.
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*}
2. What is the voltage $u_C$ at $t_1=t_0+4 ~\rm ms$?
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?
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*}
4. Draw the overall course of the voltage $u_C(t)$ over time in the diagram below.
Use an appropriate $x$-axis scale.
5. The switch $S_2$ is then closed, with $S_1$ still as shown in the drawing. What will be the maximum voltage of $u_C$?
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.
\begin{align*} U_C &= 50 ~\rm V \end{align*}
At which frequencies show the following components the given values?
1. An inductor with $X_{L1} = 60 ~\rm m\Omega$ and $L_1 = 15.9 ~\rm \mu H$.
\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*}
\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 $I_2 = 1 ~\rm mA$.
\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*}
\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 $C_3 = 5.6 ~\rm nF$.
\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*}
\begin{align*} f &= 300'774.5 ... ~\rm Hz \rightarrow f &= 300 ~\rm kHz \end{align*}
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:
1. Draw an equivalent circuit diagram with the internal resistance and an external load. Label all voltages and currents.
2. Calculate the efficiency of the battery in this case.
3. (HARD) Once the load resistance is changed, the efficiency for discharging also changes. What would be the lowest possible efficiency?
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*}
4. Calculate the voltage drop on the load resistance $R_\rm L=2 ~\Omega$.
\begin{align*} U_\rm L= U_\rm S \cdot {{R_\rm L}\over{R_\rm L + R_\rm i}} \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 $q_\rm e=1.602 \cdot 10^{-19} ~\rm C$.