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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 $100 ~\rm °C$)

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Exercise E1 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

\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

\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*}

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