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Exam Winter Semester 2022

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.

Only EEE1-relevant Part

This part is only for about 25 minutes !

1 Resistance of a Wire by Resistivity
(written test, approx. 6 % of a 60-minute written test, WS2022)

A heating element made of a Nichrome wire with a round cross-section is used in an electric oven.
Nichrome is a common Nickel Chromium alloy for heating elements.
The Nichrome wire has a resistivity of $1.10\cdot 10^{-6} ~\Omega \rm{m}$.
The heating element is $3 ~\rm{m}$ long and has a diameter of $3.57 ~\rm{mm}$.
1. Calculate the resistance $R$ of the heating element.

Solution steps

\begin{align*} R &= \rho \cdot \frac{l}{A} && | \text{with } A = r^2 \cdot \pi = \frac{1}{4} d^2 \cdot \pi \\ R &= \rho \cdot \frac{4 \cdot l}{d^2 \cdot \pi} && \\ R &= 1.10\cdot 10^{-6} ~\Omega \rm{m} \cdot \frac{4 \cdot 3~\rm{m}}{(3.57\cdot 10^{-3}~\rm{m})^2 \cdot \pi} && \\ \end{align*}

Result

\begin{align*} R &= 0.33 ~\Omega \\ \end{align*}

2. The heating element is used to heat the oven to a temperature of $180~°\rm{C}$. For this, a power dissipation (= heat flow) of $P=40 ~\rm{W}$ is necessary.
Calculate the current $I$ needed to operate it.

Solution steps

\begin{align*} P = U \cdot I = R \cdot I^2 \quad \rightarrow \quad I= \sqrt{\frac{P}{R}} = \sqrt{\frac{40 ~\rm{W}}{0.33 ~\Omega}} \end{align*}

Result

\begin{align*} I = 11 ~\rm{A} \end{align*}

1 Temperature-dependent Resistance
(written test, approx. 6 % of a 60-minute written test, WS2022)

A thermistor is used as a temperature sensor in a refrigeration system. The thermistor has a resistance of $10 ~\rm{k}\Omega$ at $+25 ~°\rm{C}$.
Its temperature coefficients are: $\alpha=0.01 ~{{1}\over{\rm{K}}}$ and $\beta=71 \cdot 10^{-6}~{{1}\over{\rm{K}^2}}$
The temperature inside the refrigeration system can reach down to $-40 ~°\rm{C}$.

1. Calculate the resistance of the thermistor at $-40 ~°\rm{C}$.

Solution steps

\begin{align*} R &= R_0 \cdot (1 + \alpha \cdot \Delta T + \beta \cdot \Delta T^2) && | \text{with } \Delta T = T_{\rm end} - T_{\rm start}\\ R &= 10 ~\rm{k}\Omega \cdot \left(1 + 0.01 ~{{1}\over{\rm{K}}} \cdot (-40~°\rm{C} - 25~°\rm{C} ) + 71 \cdot 10^{-6}~{{1}\over{\rm{K}^2}} \cdot (-40~°\rm{C} - 25~°\rm{C})^2 \right) \\ \end{align*}

Result

\begin{align*} R &= 6.5 ~\rm{k}\Omega \\ \end{align*}

2. Additionally, explain which effect a resistive temperature sensor can have on the refrigeration system.

Result

Resistors transfer electrical energy out of the circuit and generate heat. Therefore, a resistive sensor might heat up the refrigeration system.

3. Regarding question 2.: Given a constant sensor voltage, would a sensor with tenfold resistance be better or worse? Give an explanation for your answer.

Result

The power of the resistor $P = U \cdot I = R \cdot I^2 = {{U^2}\over{R}}$ is equivalent to the heat flow.
Therefore, with constant $U$ and increasing $R$ the power decreases. Ten times more resistance decreases the heat flow to one-tenth.

1 Pure Resistor Network Simplification
(written test, approx. 13 % of a 60-minute written test, WS2022)

The following circuit with $R_1=200 ~\Omega$, $R_2=R_3=100 ~\Omega$ and the switch $S$ is given.
electrical_engineering_1:70jjg4yzwctqarsqcircuit.svg

1. The switch shall now be open. Calculate the equivalent resistance $R_{\rm eq}$ between $\rm A$ and $\rm B$.

Solution steps

electrical_engineering_1:70jjg4yzwctqarsqcircuitsolution1.svg The equivalent resistor is given by a parallel configuration of resistors in series: \begin{align*} R_{\rm eq} &= (R_2 + R_1 + R_1)||(R_2 + R_2)\\ R_{\rm eq} &= (100 ~\Omega + 200 ~\Omega + 200 ~\Omega )&&||(100 ~\Omega + 100 ~\Omega ) &&\\ R_{\rm eq} &= (500 ~\Omega )&&||(200 ~\Omega )&&\\ R_{\rm eq} &= {{500 ~\Omega \cdot 200 ~\Omega }\over{500 ~\Omega + 200 ~\Omega}}&&\\ \end{align*}

Result

\begin{align*} R_{\rm eq} &= 142.8 ~\Omega \\ \end{align*}

2. The switch shall now be closed. Calculate the equivalent resistance $R_{\rm eq}$ between $\rm A$ and $\rm B$.

Solution steps

Now a wye-delta transformation is necessary. electrical_engineering_1:70jjg4yzwctqarsqcircuitsolution2.svg Since $R_2=R_3$ and based on the equations for the transformation, the transformed $R_Y$ is given as: \begin{align*} R_{Y} &= {{R_2 \cdot R_2}\over{R_2 + R_2 + R_2}} \\ &= {{(100 ~\Omega)^2}\over{3 \cdot 100 ~\Omega}} \\ &= {{1}\over{3}} \cdot 100 ~\Omega = 33.33 ~\Omega \end{align*}

The equivalent resistor is given by a parallel configuration of resistors in series: \begin{align*} R_{\rm eq} &= R_Y + (R_Y + R_1 + R_1)||(R_Y + R_2)\\ R_{\rm eq} &= 33.33 ~\Omega + (33.33 ~\Omega + 400 ~\Omega)||(33.33 ~\Omega + 100 ~\Omega)\\ \end{align*}

Result

\begin{align*} R_{\rm eq} &= 135.3 ~\Omega \\ \end{align*}

1 Equivalent linear Source
(written test, approx. 14 % of a 60-minute written test, WS2022)

The circuit in the following has to be simplified.

electrical_engineering_1:6tqttque1e2nf2c7circuit.svg

Calculated the internal resistance $R_\rm i$ and the source voltage $U_\rm s$ of an equivalent linear voltage source on the connectors $\rm A$ and $\rm B$. \begin{align*} R_1=5.0 ~\Omega, && U_2=6.0 ~\rm{V}, && R_3= 10 ~\Omega, \\ I_4=4.2 ~\rm{A}, && R_5=10 ~\Omega , && R_6=7.5 ~\Omega, \\ R_7=15 ~\Omega && && \end{align*} Use equivalent sources in order to simplify the circuit!

Solution steps

The best thing is to re-think the wiring like rubber bands and adjust them: electrical_engineering_1:6tqttque1e2nf2c7circuitsolution1.svg

The linear voltage source of $U_2$ and $R_1$ can be transformed into a current source $I_2={{U_2}\over{R_1}}$ and $R_1$: electrical_engineering_1:6tqttque1e2nf2c7circuitsolution2.svg

Now a lot of them can be combined. The resistors $R_1$, $R_3$, $R_5$ are in parallel, like also $I_2$ and $I_4$: \begin{align*} R_{135} &= R_1||R_3||R_5\\ I_{24} &= I_2 - I_4 = {{U_2}\over{R_1}} - I_4 \end{align*} The resulting circuit can again be transformed: electrical_engineering_1:6tqttque1e2nf2c7circuitsolution3.svg
Here, the $U_{24}$ is calculated by $I_{24}$ as the following: \begin{align*} U_{24} &= R_{135} \cdot I_{24} \\ &= ({{U_2}\over{R_1}} - I_4) \cdot R_1||R_3||R_5 \end{align*}

On the right side of the last circuit, there is a voltage divider given by $R_{135}$, $R_6$, and $R_7$.
Therefore the voltage between $A$ and $B$ is given as: \begin{align*} U_{\rm AB} &= U_{24} \cdot {{R_7}\over{R_6 + R_7 + R_1||R_3||R_5}} \\ &= ({{U_2}\over{R_1}} - I_4) \cdot {{R_7 \cdot R_1||R_3||R_5}\over{R_6 + R_7 + R_1||R_3||R_5}} \\ \end{align*}

For the internal resistance $R_i$ the ideal voltage source is substituted by its resistance ($=0\Omega$, so a short-circuit): \begin{align*} R_{\rm AB} &= R_7 || ( R_6 + R_1||R_3||R_5) \\ \end{align*}

with $R_1||R_3||R_5 = 5 ~\Omega || 10 ~\Omega || 10 ~\Omega = 5 ~\Omega || 5 ~\Omega = 2.5 ~\Omega$:

\begin{align*} U_{\rm AB} &= ({{6.0 ~\rm{V}}\over{5.0 ~\Omega}} - 4.2 ~\Omega) \cdot {{15 ~\Omega \cdot 2.5 ~\Omega}\over{7.5 ~\Omega + 15 ~\Omega + 2.5 ~\Omega}} \\ R_{\rm AB} &= 15 ~\Omega|| ( 7.5 ~\Omega + 2.5 ~\Omega) \\ \end{align*}

Result

\begin{align*} U_{\rm s} &= U_{\rm AB} &&= 4.5 ~\rm{V}\\ R_{\rm i} &= R_{\rm AB} &&= 6 ~\Omega \end{align*}