Exam Summer Semester 2024
Additional permitted Aids
- non-programmable calculator,
- formulary (4 one-sided DIN A4 pages)
Hits
- The duration of the exam is 120 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 Electrostatics I
(written test, approx. 8 % of a 120-minute written test, SS2024)
Given is the arrangement of electric charges in the picture below. The values of the point charges are
- $q_0=-1 ~\rm nC$
- $q_1=-5 ~\rm nC$
- $q_2=q_3=+2 ~\rm nC$
In the beginning, the area charge is $q_4=0 ~\rm nC$.
The permittivity is $\varepsilon_{\rm r} \varepsilon_0 = 8.854 \cdot 10^{-12} ~\rm As/Vm$
1. Calculate the single forces $\vec{F}_{01}$, $\vec{F}_{02}$, $\vec{F}_{03}$, on the charge $q_0$!
First, calculate the magnitude of the forces, like $\vec{F}_{01}$.
The force $\vec{F}_{01}$ is purely on the $x$-axis and therefore equal to $F_{01,x}$.
\begin{align*}
\vec{F}_{01} = F_{01,x} &= {{1}\over{4\pi\varepsilon}}\cdot {{q_1\cdot q_0}\over{r^2_{01}}} \\
&= {{1}\over{4\pi\cdot 8.854 \cdot 10^{-12} ~\rm As/Vm}}\cdot {{1 \cdot 10^{-9} ~\rm C \cdot 5 \cdot 10^{-9} ~\rm C}\over{(7 \cdot 10^{-3} ~\rm m)^2}} \\
&= 917.\;.\!.\!.\! \cdot 10^{-6} {\rm {{(As)^2 \cdot Vm}\over{As \cdot m^2}}}
= 917.\;.\!.\!.\! \cdot 10^{-6} {\rm {{VAs}\over{m}}}
= 917.\;.\!.\!.\! \cdot 10^{-6} {\rm {{Ws}\over{m}}} \\
&= 917.\;.\!.\!.\! {\rm \mu N} \quad \text{(to the right)}
\end{align*}
Similarly, we get for $\vec{F}_{02}$ and $\vec{F}_{03}$ \begin{align*} \vec{F}_{02} = F_{02,x} &= -1997.\;.\!.\!.\! {\rm \mu N} \quad \text{(to the right)} \\ \vec{F}_{03} = F_{03,y} &= -1123.\;.\!.\!.\! {\rm \mu N} \quad \text{(to the top)} \\ \end{align*}
- $\vec{F}_{01} = \left( {\begin{array}{cccc} 917 {~\rm \mu N} \\ 0 \\ \end{array} } \right)$
- $\vec{F}_{02} = \left( {\begin{array}{cccc} 1997 {~\rm \mu N} \\ 0 \\ \end{array} } \right)$
- $\vec{F}_{03} = \left( {\begin{array}{cccc} 0 {~\rm \mu N} \\ -1123 {~\rm \mu N} \\ \end{array} } \right)$
2. What is the magnitude of the resulting force?
\begin{align*} F = |\vec{F}| &= \sqrt{\left( \sum_i F_{i,x} \right)^2 + \left( \sum_i F_{i,y} \right)^2} \\ &= \sqrt{\left( +917 {~\rm \mu N} - 1997 {~\rm \mu N} \right)^2 + \left( 1123 {~\rm \mu N} \right)^2} \\ &= 1560.\;.\!.\!.\! {~\rm \mu N} \\ \end{align*}
3. Now the charges $q_2=q_3$ are set to 0. The area charge $q_4$ generates a homogenous field of $E_4$. Which value needs $E_4$ to have to get a resulting force of $0 ~\rm N$ on $q_0$?
In the homogenous field the force is calculated by $F = E \cdot q$.
Here, this field has to compensate for the force $\vec{F}_{01}$ from $q_1$ on $q_0$:
\begin{align*}
|\vec{F}_{01}| &= |E_4| \cdot |q_0| \\
\rightarrow E_4 &= {{|\vec{F}_{01}|}\over{|q_0|}} \\
&= {{917.\;.\!.\!.\! \cdot 10^{-6} {~\rm N}}\over{1 \cdot 10^{-9} ~\rm C }} \\
&= 917.\;.\!.\!.\! \cdot 10^{3}{{\rm N}\over{\rm C}} \\
&= 917.\;.\!.\!.\! \cdot 10^{3}{{\rm VAs/m}\over{\rm As}} \\
&= 917.\;.\!.\!.\! \cdot 10^{3}{{\rm V}\over{\rm m}} \\
\end{align*}
Exercise E2 Electrostatics II
(written test, approx. 10 % of a 120-minute written test, SS2024)
You must analyze an aluminum profile for usage in an environment critical for electrostatic discharge.
The figure on the right shows the cross-section of the aluminum element (hatched). During the application, it might get charged up. All areas in white consist of air (= dielectric).
Six designated areas are shown by dashed frames and numbers n, which are partly inside the object.
Arrange the designated areas clearly according to ascending field strengths $|\vec{E}_n|$ (absolute magnitude)! Indicate also, if designated areas have quantitatively the same field strength.
Exercise E3 Capacitor
(written test, approx. 12 % of a 120-minute written test, SS2024)
Older capacitive pressure sensors are based on a parallel plate capacitor setup (see left-side image).
In the following such a sensor is given with:
- Plate area : $A=25 ~\rm mm^2$
- Distance between both plates: $d=200 ~\rm \mu m$
- Air between the plates: $\varepsilon_{\rm r,air}=1$
- Supply voltage: $3.3 ~\rm V$
- Boundary effects on the end of the layers shall be ignored in the following calculations.
$\varepsilon_{0} =8.854 \cdot 10^{-12} ~\rm F/m $
1. Calculate the capacity $C$.
2. Calculate the value of the displacement field between the plates, when a voltage of $U=3.3 ~\rm V$ is applied.
The displacement field is given by: \begin{align*} D &= \varepsilon_0 \varepsilon_r E \\ &= \varepsilon_0 \varepsilon_r {{U}\over{d}} \\ &= 8.854 \cdot 10^{-12} ~\rm As/Vm \cdot 1 \cdot {{3.3 {~\rm V} }\over{200 \cdot 10^{-6} {~\rm m} }} \\ \end{align*}
3. Calculate the charge difference between both plates for a voltage of $U=3.3 ~\rm V$.
4. Due to a production problem, the right-side layer is covered with a contaminant, see the right-side image.
The contaminant has $\varepsilon_{\rm r,c}>\varepsilon_{\rm r,air}$, while the distance between the plates remains the same.
Give a generalized formula $C_2=f(A,d,x,\varepsilon_{\rm r,c}, \varepsilon_{\rm r,air})$.
Therefore: \begin{align*} C = {{1}\over{ {{1}\over{C_{\rm air} }} + {{1}\over{ C_{\rm c}}} }} \end{align*}
With \begin{align*} C_{\rm air} &= \varepsilon_0 \varepsilon_{\rm r, air} {{A}\over{d-x}} &&= \varepsilon_0 A {{\varepsilon_{\rm r, air}}\over{d-x}} \\ C_{\rm c} &= \varepsilon_0 \varepsilon_{\rm r, c} {{A}\over{x}} &&= \varepsilon_0 A {{\varepsilon_{\rm r, c} }\over{x}} \\ \end{align*}
This leads to: \begin{align*} C = \varepsilon_0 A {{1}\over{ {{d-x}\over{\varepsilon_{\rm r, air} }} + {{x}\over{ \varepsilon_{\rm r, c}}} }} \end{align*}
Exercise E4 Magnetic Field Lines
(written test, approx. 6 % of a 120-minute written test, SS2024)
The following setup shall be given:
- Four conductors are located perpendicular to the plane of the diagram
- All of them conduct a current with the same magnitude, but not in the same direction.
- A permanent magnet is located in between the conductors.
1. Do not consider the permanent magnet at first. Draw at least 10 field lines of the H-field qualitatively. Give a a correct representation of their direction, and density for the shown area.
2. Discuss how the permanent magnet affects the H-field, based on the fundamental definition of the H-field.
- The H-field is defined by currents $\sum I = \int H {\rm d}s$ .
- In the permanent magnet, there are no free currents.
- The bound currents (of the permanent magnet) create also an H field.
- This exits on the north pole and enters the magnet on the south pole (similar to the B-field)_
- $H = B/\mu$
- The H-field from task 1 gets distracted
Exercise E5 Fields of an coax Cable
(written test, approx. 12 % of a 120-minute written test, SS2024)
A $0.5 ~\rm m$ long coax cable is used for signal transmission. The diagram shows the cross-section of the coax cable with the origin in the center of the coax cable. Due to the given load, the following situation appears:
- Inner conductor: $+3.3 ~\rm mA$, $+10 ~\rm nC$ (current into the plane of the diagram)
- Outer conductor: $-3.3 ~\rm mA$, $ 0 ~\rm nC$ (current out of the plane of diagram)
1. What is the magnitude of the magnetic field strength $H$ at $\rm (0.1 ~mm | 0)$ and $\rm (0.55 ~mm | 0)$?
The magnitude of the magnetic field strength $H$ can be calculated by: $H = {{I}\over{2 \pi \cdot r}} $
So, we get for $H_{\rm i}$ at $\rm (0.1 ~mm | 0)$, and $H_{\rm o}$ at $\rm (0.55 ~mm | 0)$:
\begin{align*} H_{\rm i} &= {{I}\over{2 \pi \cdot r_{\rm i}}} \\ &= {{+3.3 A}\over{2 \pi \cdot { 0.1 \cdot 10^{-3}~\rm m}}} \\ H_{\rm o} &= {{I}\over{2 \pi \cdot r_{\rm o}}} \\ &= {{+3.3 A}\over{2 \pi \cdot { 0.55 \cdot 10^{-3}~\rm m}}} \\ \end{align*}
Hint: For the direction, one has to consider the right-hand rule.
By this, we see that the $H$-field on the right side points downwards.
Therefore, the sign of the $H$-field is negative.
But here, only the magnitude was questioned!
- for $(0.1 ~/rm mm | 0)$ : $H_{\rm i} = 5.25... ~\rm A/m$
- for $(0.55 ~/rm mm| 0)$ : $H_{\rm o} = 0.955... ~\rm A/m$
2. Plot the graph of the magnitude of $H(x)$ with $x \in \rm [-0.6~mm, +0.6~mm]$ from $\rm (-0.6 ~mm | 0)$ to $\rm (0.6 ~mm | 0)$ in one diagram. Use proper dimensions and labels for the diagram!
- In general, the $H$-field is proportional to ${{1}\over{r}}$ for the situation between both conductors.
- For $H(x)$ with $x$ within the inner conductor, one has to consider how much current is conducted within a circle with the radius $x$.
This is proportional to the area within this radius. Therefore, Ther formula $H = {{I}\over{2 \pi \cdot r}} $ gets $H(x) = {{I}\over{2 \pi \cdot x}} \cdot {{\pi x^2}\over{\pi (0.1~\rm mm)^2}}$. This leads to a formula proportional to $x$. - For $x$ within the outer conductor one also gets a linear proportionality with a similar approach.
3. What is the magnitude of the electric displacement field $D$ at $\rm (-0.1 ~mm | 0)$ and $\rm (0.55 ~mm | 0)$?
Here, for any position radial to the center, the surrounding area is the surface of a cylindrical shape (here for simplicity without the round endings).
This leads to: \begin{align*} D(x) &= {{Q}\over{A}} \\ &= {{Q}\over{l \cdot 2\pi \cdot x}} \\ \end{align*}
So, we get for $D_{\rm i}$ at $\rm (0.1 ~mm | 0)$, and $D_{\rm o}$ at $\rm (0.55 ~mm | 0)$:
\begin{align*} D_{\rm i} &= {{Q }\over{2 \pi \cdot r_{\rm i} \cdot l}} \\ &= {{10 \cdot 10^{-9} C}\over{2 \pi \cdot { 0.1 \cdot 10^{-3}~\rm m} \cdot 0.5 ~\rm m}} \\ D_{\rm o} &= {{Q }\over{2 \pi \cdot r_{\rm o} \cdot l}} \\ &= {{10 \cdot 10^{-9} C}\over{2 \pi \cdot { 0.55 \cdot 10^{-3}~\rm m}\cdot 0.5 ~\rm m}} \\ \end{align*}
Hint: For the direction, one has to consider the sign of the enclosed charge.
By this, we see that the $D$-field is positive.
But here, again only the magnitude was questioned!
- for $(0.1 ~\rm mm | 0)$ : $D_{\rm i} = 31.8... ~\rm uC/m^2$
- for $(0.55 ~\rm mm| 0)$ : $D_{\rm o} = 5.78... ~\rm uC/m^2$
4. Plot the graph of the magnitude of $D(x)$ with $x \in \rm [-0.6~mm, +0.6~mm]$ from $\rm (-0.6 ~mm | 0)$ to $\rm (0.6 ~mm | 0)$ in one diagram. Use proper dimensions and labels for the diagram!
- In general, the $D$-field is proportional to ${{1}\over{r}}$ for the situation between both conductors.
- Since the charges are on the surface of the conductor, there is no $D$-field within the conductor.
Exercise E6 Lorentz Force
(written test, approx. 8 % of a 120-minute written test, SS2024)
A robotic shuttle system uses magnets to lift the mobile shuttle over the fixed floor. To do so, coils in the floor repel the permanent magnets of the mobile shuttle (see image).
A single coil shall have the following properties:
- $N=500$ windings
- radius $r=40 ~\rm mm$
- current $I = 1.6 ~\rm A$
- magnetic field of the shuttle of is homogeneous with $B=0.5 ~\rm T$
1. Calculate the magnitude of the resulting force on one coil!
2. For one winding of the left coil, the cross-sections are marked in bold in the image. Draw the resulting force vectors into the image for each side of the winding.
- $B$-field downwards
- conductors inwards and outwards
The resulting force has to be perpendicular to $B$-field and conductor.
The right-hand rule leads to forces pointing radially into the coil
3. Does the Lorentz force lift the shuttle for a homogeneous $B$-field of the shuttle? Explain.
The Lorentz force can only have a lifting effect in an inhomogeneous field.
In this case, there sum of the forces results in a repulsing force, see image.
Beside boundary effects, The field gets also inhomogeneous, by the additional field of the coils.
Exercise E7 Magnetic Potential
(written test, approx. 8 % of a 120-minute written test, SS2024)
Calculate the magnetic potential difference V_m for the following paths as shown by the solid lines.
Dotted lines are only for there for symmetry aspects!
The wires conduct the following currents:
- $|I_1|=2 ~\rm A$
- $|I_2|=5 ~\rm A$
- $|I_3|=11 ~\rm A$
- $|I_4|=7 ~\rm A$
Pay attention to the signs of the currents (given by the diagrams) and of the results!
- Task: $+I_1 - I_2 = -3 ~\rm A$
- Task: $+{{1}\over{4}} I3 = 11/4 ~\rm A$ (it does not matter which way the path goes from the startpoint to the endpoint, as long as it has the same direction and number of revolutions)
- Task: $-{{1}\over{4}} I1 = -0.5 ~\rm A$
- Task: $+2 \cdot I_2 - 1 \cdot I_3 = -1 ~\rm A$
Exercise E8 Self-Induction
(written test, approx. 8 % of a 120-minute written test, SS2024)
A coil with a length of $30 ~\rm cm$ and a radius of $2 ~\rm cm$ has $500$ turns.
The current through the coil changes linearly from $0$ to $3 ~\rm A$ in $0.02 ~\rm ms$.
The arrangement is located in air ($\mu_{\rm r}=1$).
$\mu_0= 4\pi \cdot 10^{-7} ~\rm Vs/Am$
1. Calculate the (self-)inductance of the coil.
2. Determine the induced voltage in the coil during the change in current.
Exercise E9 Magnetic Circuit
(written test, approx. 9 % of a 120-minute written test, SS2024)
A toroidal core (ferrite, $μ_{\rm r}=900$) has a cross-sectional area of $A=300 ~\rm mm^2$ and an average circumference of $l=3 ~\rm dm$.
On the core, there are three coils with:
- Coil 1: $N_1 = 1200$, $I_1=100 ~\rm mA$
- Coil 2: $N_2 = 33 $, $I_2= 3 ~\rm A$
- Coil 3: $N_3 = 270 $, $I_3=0.3 ~\rm A$
Refer to the drawing for the direction of the windings, current, and flux!
1. Draw the equivalent magnetic circuit that fully represents the setup.
Name all the necessary magnetic resistances, fluxes, and voltages.
- Since the material, and diameter of the core is constant, one can directly simplify the magnetic resistor into a single $R_\rm m$.
- For the orientation of the magnetic voltages $\theta_1$, $\theta_2$, and $\theta_3$, the orientation of the coils and the direction of the current has to be taken into account by the right-hand rule.
- There is only one flux $\Phi$
- The magnetic voltages are antiparallel to the flux for sources and parallel for the load.
2. Calculate the magnetic resistance $R_\rm m$.
3. Calculate the resulting magnetic flux in the core.
To get the flux $\Phi$, the Hopkinson's Law can be applied - similar to the Ohm's Law: \begin{align*} \Phi &= {{\theta_{\rm R} }\over {R_{\rm m}}} \\ &= {{-60~\rm A }\over { 0.884 \cdot 10^{6} \rm {{1}\over{H}} }} \\ &= 67.8 ... \cdot 10^{-6} { \rm A \cdot H} \\ &= 67.8 ... ~\rm \mu Wb \\ &= 67.8 ... ~\rm \mu Vs \\ \end{align*}
Exercise E10 Magnetic Circuit
(written test, approx. 10 % of a 120-minute written test, SS2024)
A real capacitor behaves like an $RLC$ resonant circuit, with an equivalent series resistance $R$ and an equivalent series inductance $L$.
A given capacitor shall have the following values:
- $C=10 ~\rm nF$
- $R=20 ~\rm m\Omega$
- $L=1.6 ~\rm nH$
1. What is the impedance $Z_{RLC}$ of this real capacitor for $f_0=44 ~\rm MHz$? (Phase and magnitude)
The impedance is based on the resistance $R$ and the reactance $X_{LC}= {\rm j}\cdot (X_L - X_C)$: \begin{align*} \underline{Z}_{RLC} &= R + {\rm j}\cdot (X_L - X_C) \\ &= R + {\rm j}\cdot (\omega L - {{1}\over{\omega C}}) \\ &= R + {\rm j}\cdot (2\pi f \cdot L - {{1}\over{2\pi f \cdot C}}) \\ \end{align*}
The reactive part is \begin{align*} X_{LC} &= 2\pi f \cdot L - {{1}\over{2\pi f \cdot C}} \\ &= 2\pi 44 \cdot 10^{6} {~\rm MHz} \cdot 1.6 \cdot 10^{-9} {~\rm H} - {{1}\over{2\pi \cdot 10^{6} {~\rm MHz} \cdot 10 \cdot 10^{-9} {~\rm F}}} \\ &= +0.08062... ~\Omega \\ \end{align*}
To get the magnitude of the impedance $|\underline{Z}_{RLC}|$ one can use the Pythagorean Theorem: \begin{align*} |\underline{Z}_{RLC}| &= \sqrt{R^2 + X_{LC}^2} \\ &= \sqrt{(0.020~\Omega)^2 + ( 0.08062... ~\Omega )^2} \\ &= 0.0830 ... ~\Omega \\ \end{align*}
For the phase $\varphi$ the $\arctan$ can be applied: \begin{align*} \varphi &= \arctan \left( {{X_{LC}}\over{R}} \right) \\ &= \arctan \left( {{0.08062... ~\Omega}\over{0.020 ~\Omega}} \right) \\ &= 1.3276 ... \hat{=} +76° \\ \end{align*}
- $|\underline{Z}_{RLC}| = 83.0 ~\rm m \Omega$
- $\varphi = +76°$
2. What is the resonance frequency $f_r$ for the given capacitor? What is the impedance in this case?
The impedance at resonance is purely the resistance.
- $f_r = 39.79 ~\rm MHz$
- $|\underline{Z}_{RLC}(f_r)| = 20.0 ~\rm m \Omega$
3. For an application, the component shall be used in resonance on a supply of $5 ~\rm V$. What is the voltage on the ideal capacity $C$ in the shown circuit?
The voltage on the ideal capacity is the input voltage by the $Q$-factor increased: \begin{align*} U_C &= U_{\rm s} \cdot Q \\ &= U_{\rm s} \cdot \sqrt{ {{L}\over{C}} } \cdot {{1}\over{R}}\\ &= 5 {~\rm V} \cdot \sqrt{ {{ 1.6 \cdot 10^{-9} {~\rm H} }\over{ 10 \cdot 10^{-9} {~\rm F} }} } \cdot {{1}\over{0.020~\Omega}}\\ &= 100 ~\rm V \end{align*}
Exercise E11 Magnetic Circuit
(written test, approx. 10 % of a 120-minute written test, SS2024)
A symmetric and balanced three-phase motor is driven with a $230 ~\rm V$ / $400 ~\rm V$ / $50 ~\rm Hz$ three-phase power net. Each single string has a resistor $R=5 ~\Omega$ and an inductance of $L=10 ~\rm mH$.
1. Calculate the $\cos \varphi$, and the magnitude of the impedance $|Z|$ for a single string.
The phase $\varphi$ is given by: \begin{align*} \varphi &= \arctan \left( {{X_{L}}\over{R}} \right) \\ &= \arctan \left( {{2\pi \cdot f \cdot L}\over{R}} \right) \\ &= \arctan \left( {{2\pi \cdot 50 {~\rm Hz} \cdot 10 \cdot 10^{-3} {~\rm H}}\over{5 ~\Omega}} \right) \\ &= 0.5609 ... \hat{=} +32° \\ \end{align*}
With this, the $\cos \varphi$ becomes \begin{align*} \cos \varphi &= \cos(0.5609 ...) \\ &= 0.84673...\\ \end{align*}
The impedance is given by: \begin{align*} |\underline{Z}_{RL}| &= \sqrt{X_{L}^2 + R^2} \\ &= \sqrt{( 2\pi \cdot f \cdot L )^2 + R^2} \\ &= \sqrt{( 2\pi \cdot 50 {~\rm Hz} \cdot 10 \cdot 10^{-3} {~\rm H})^2 +(5 ~\Omega)^2} \\ &= 5.905... ~\Omega\\ \end{align*}
- $|\underline{Z}_{RL}| = 5.90 ~\Omega$
- $\cos\varphi = 0.84673$
2. Calculate the true power, the apparent power, and the reactive power of the motor.
The active power is \begin{align*} P &= S \cdot \cos \varphi \\ &= 26.898... \cdot 0.84673 ~\rm kW \\ &= 22.775... ~\rm kW \\ \end{align*}
The reactive power is \begin{align*} Q &= \sqrt{S^2 - P^2} \\ &= \sqrt{ (26.898... ~\rm kVA)^2 - (22.775... ~\rm kW)^2} \\ &= 14.310...~\rm kVAr \\ \end{align*}
- active power:
$P=22.775~\rm kW$
- reactive power:
$Q=14.310~\rm kVAr$
- apparent power:
$S=26.898~\rm kVA$