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.
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.
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 $-40 ~°\rm{C}$.
\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*}
\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.
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.
The following circuit with $R_1=200 ~\Omega$, $R_2=R_3=100 ~\Omega$ and the switch $S$ is given.
1. The switch shall now be open. Calculate the equivalent resistance $R_{\rm eq}$ between $\rm A$ and $\rm B$.
2. The switch shall now be closed. Calculate the equivalent resistance $R_{\rm eq}$ between $\rm A$ and $\rm B$.
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*}
The circuit in the following has to be simplified.
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!
The best thing is to re-think the wiring like rubber bands and adjust them:
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 $R_1$:
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:
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*}
The circuit shown in the following is used to control the brightness when turning on a small light bulb.
The circuit contains a voltage source $U=12 ~\rm{V}$, a switch $S_1$, a resistor of $R_1=20 ~\Omega$ and a capacitor of $C=100 ~\rm{µF}$.
The switch $S_2$ to an additional consumer $R_2$ will be considered to be open for the first tasks. At the moment $t_0=0 ~\rm{s}$ the switch $S_1$ is closed, the voltage across the capacitor is $u_c (t_0 )=0 ~\rm{V}$.
1. First do not consider the light bulb – it is not connected to the RC circuit.
Calculate the point of time $t_1$ when $u_c (t_1)=0.5\cdot U$.
So, here only R_1 and C gives the time constant: $\tau = R_1 \cdot C$
The following formula describes the time course of $u_C(t)$ which has to be $u_c (t_1)=0.5\cdot U$: \begin{align*} u_c (t) = U \cdot (1- e^{t/\tau}) = 0.5\cdot U \end{align*} It has to be rearranged to $t$ \begin{align*} (1- e^{t/\tau}) &= 0.5 \\ e^{t/\tau} &= 0.5 \\ t/\tau &= ln(0.5) \\ t &= \tau \cdot ln(0.5) \\ t &= R_1 \cdot C \cdot ln(0.5) \end{align*}
2. Calculate the overall energy dissipated by $R_1$ while charging the capacitor $0 ~\rm{V}$ to $12 ~\rm{V}$.
3. Now, consider the light bulb as a resistor of $R_\rm B=20 ~\Omega$, and ignore again the left side ($S_2$ is open).
The voltage across the capacitor is again $0 ~\rm{V}$ at the moment $t_0=0 ~\rm{s}$ when the switch $S_1$ is closed.
Calculate the voltage $u_c (t_2)$ across the capacitor at $t_2=1 ~\rm{ms}$ after closing the switch.
Hint: To solve this, first create an equivalent linear voltage source from $U$, $R_1$, and $R_\rm B$.
An equivalent linear voltage source can be given with $U$, $R_1$, and $R_\rm B$ as seen in yellow.
Therefore, the voltage of the equivalent linear voltage source is: $U_s = U \cdot {{R_\rm B}\over{R_1 + R_\rm B}} = 1/2 \cdot U$
The internal resistance is given by substituting the ideal voltage source with its resistance ($=0 ~\Omega$, short-circuit).
\begin{align*}
R_i &= R_1 || R_\rm B \\
&= 10 ~\Omega
\end{align*}
\begin{align*} u_c (t_2) &= U_s \cdot (1- e^{t_2/(R_i\cdot C)}) \\ &= {{1}\over{2}} \cdot U \cdot (1- e^{1~\rm{ms}/(10 ~\Omega \cdot 100 ~\rm{µF})}) \end{align*}
4. Explain (without calculation) how the situation in 3. would change once also $S_2$ is closed from the beginning on.
A circuit with an ideal voltage source ($U=50 ~\rm V$, $f=330 ~\rm Hz$) and two components ($R$ and $\underline{X}_1$) shall be given.
After analysis, the following formula for the impedance was extracted:
\begin{align*}
\underline{Z} = \left({{2}\over{3+4{\rm j}}}+5{\rm j} \right) \Omega
\end{align*}
1. Calculate the physical values of the two components.
\begin{align*} \underline{Z} &= \left({{2}\over{3+4{\rm j}}} + 5{\rm j} \right) ~\Omega \\ &= \left({{2}\over{3+4{\rm j}}} \cdot {{3-4{\rm j}}\over{3-4j}} + 5{\rm j} \right) ~\Omega \\ &= \left({{2}\over{9+16 }} \cdot (3-4{\rm j} ) + 5{\rm j} \right) ~\Omega \\ &= \left(0.24 - 0.32{\rm j} + 5{\rm j} \right) ~\Omega \\ &= 0.24 ~\Omega + {\rm j} \cdot 4.68 ~\Omega \\ &= R + {\rm j} X_L \\ \end{align*}
With the complex part comes the physical value: \begin{align*} X_L &= \omega L \\ L &= {{X_L}\over{2\pi \cdot f}} \\ &= {{4.68 ~\Omega}\over{2\pi \cdot 300 ~\rm{Hz}}} \\ \end{align*}
\begin{align*} R &= 0.24 ~\Omega \\ L &= 2.26 ~\rm{mH} \end{align*}
2. Calculate the phase and absolute value of complex current $\underline{I}$ through the circuit.
\begin{align*} \underline{I} &= {{\underline{U}}\over{\underline{Z}}} \\ &= {{50 ~\rm{V}}\over{ 0.24 ~\Omega + {\rm j} \cdot 4.68 ~\Omega }} \\ &= {{50 ~\rm{V}}\over{ 0.24 ~\Omega + {\rm j} \cdot 4.68 ~\Omega }} \cdot {{ 0.24 ~\Omega - {\rm j} \cdot 4.68 ~\Omega }\over{ 0.24 ~\Omega - {\rm j} \cdot 4.68 ~\Omega }} \\ &= {{50 ~\rm{V}}\over{(0.24 ~\Omega)^2 + (4.68 ~\Omega)^2 }} \cdot ( 0.24 ~\Omega - {\rm j} \cdot 4.68 ~\Omega ) \\ \end{align*}
The absolute value $|\underline{I}|$ can be calculated as: \begin{align*} |\underline{I}| &= {|{\underline{U}|}\over{|\underline{Z}|}} \\ &= {{50 ~\rm{V}}\over{| 0.24 ~\Omega + {\rm j} \cdot 4.68 ~\Omega |}} \\ &= {{50 ~\rm{V}}\over{\sqrt{ (0.24 ~\Omega)^2 + (4.68 ~\Omega)^2 }}} \end{align*}
The phase $\varphi_i$ can be calculated as \begin{align*} \varphi_i &= \arctan \left( {{\Im()}\over{\Re()}} \right) \\ &= \arctan \left( {{-4.68 ~\Omega}\over{0.24 ~\Omega}} \right) \\ \end{align*}
\begin{align*} |\underline{I}| &= 10.67 ~\rm{A} \\ \varphi_i &= -87.06° \end{align*}
3. Now an additional component $\underline{X}_2$ shall be added in series to the two components.
This component shall be dimensioned in such a way that the current and voltage are in phase. Calculate these component value!
Calculate the resistor values which have to be used in the following circuits.
1. A resistor $R_1$ shall have the same absolute value of the impedance as a capacitor $C_1=40 ~{\rm nF}$ at $f_1=4 ~{\rm MHz}$.
\begin{align*} R_1 &= |\underline{X}_{C1}| \\ &= {{1}\over{2\pi \cdot f \cdot C_1}} \\ &= {{1}\over{2\pi \cdot 4 ~{\rm MHz} \cdot 40 ~\rm nF}} \\ \end{align*}
2. A $RL$ series circuit with $L_2=4.7 ~\rm µH$, where an AC voltage source of $U_2=1.0 ~\rm V$ with $f_2=450 ~\rm kHz$ generates a current $I_2=60 ~\rm mA$.
This can be rearranged to get $R_2$: \begin{align*} R_2 &= \sqrt{ \left( {{U }\over{I }} \right)^2 - X_{L2}^2 } \\ &= \sqrt{ \left( {{1~{\rm V}}\over{60~\rm mA}} \right)^2 - (2\pi \cdot 450~{\rm kHz} \cdot 4.7 ~ {\rm µH})^2 } \\ \end{align*}
\begin{align*} R_2 &= 10.0 ~\Omega \\ \end{align*}
3. A $RC$ parallel circuit with $C_3=4.7 ~\rm nF$ on an AC current source ($I_{3S}=1.3 ~\rm A$,$f_3=200 ~\rm kHz$), which generates a current of $I_{3R}=1.0 ~\rm A$ through $R_3$.
Parallel circuit means that the voltage is the same on $R_3$ and $C_3$:
\begin{align*}
\underline{U}_3 = R_3 \cdot \underline{I}_{3R} = -{\rm j}\cdot {X}_{3C} \cdot \underline{I}_{3C}
\end{align*}
So it gets clear, that $\underline{I}_{3R}$ is perpendicular to $\underline{I}_{3C}$ (It has to, since $R_3$ is perpendicular to $-{\rm j}\cdot {X}_{3C}$, too).
Therefore, the resulting current of the parallel circuit is given as:
\begin{align*}
\underline{I}_{3} &= \underline{I}_{3R} + \underline{I}_{3C} \\
|\underline{I}_{3}|^2 &= |\underline{I}_{3R}|^2 + |\underline{I}_{3C}|^2 \\
{I}_{3C} &= \sqrt{|{I}_{3}|^2 - |{I}_{3R}|^2}
\end{align*}
Back to the first formula: \begin{align*} R_3 \cdot {I}_{3R} &= {X}_{3C} \cdot {I}_{3C} \\ R_3 &= {X}_{3C} \cdot {{{I}_{3C}}\over{{I}_{3R}}} \\ &= {{1}\over{2\pi \cdot f \cdot C_3}} \cdot {{\sqrt{|{I}_{3}|^2 - |{I}_{3R}|^2}}\over{{I}_{3R}}} \\ \end{align*}
\begin{align*} R_3 &= 70.0 ~\Omega \\ \end{align*}
A circuit designed to filter the noise from a signal shall be analyzed.
The input is given by a voltage source $u(t) = 3.0 ~{\rm V} \cdot \sin(2\pi \cdot 15 ~{\rm kHz} \cdot t)$ with an internal resistance of $10 ~\Omega$.
This linear source is connected with an inductor of $330 ~ {\rm µH}$ and a capacitor of $0.22 ~{\rm µF}$, all in series.
1. Draw the circuit diagram of the given circuit.
Label all components, voltages, and currents.
2. Calculate the single impedance $|\underline{Z}_C|$, $|\underline{Z}_L|$ such as $|\underline{Z}|$ of the overall circuit.
\begin{align*} Z_C &= {{1}\over{2\pi \cdot f \cdot C}}\\ &= {{1}\over{2\pi \cdot 15 ~{\rm kHz} \cdot 0.22 ~{\rm µF}}}\\ \end{align*}
\begin{align*} Z_L &= 2\pi \cdot f \cdot L\\ &= 2\pi \cdot 15 ~{\rm kHz} \cdot 0.22 ~{\rm µF}\\ \end{align*}
\begin{align*} Z_C &= {{1}\over{2\pi \cdot f \cdot C}}\\ &= {{1}\over{2\pi \cdot 15 ~{\rm kHz} \cdot 330 ~{\rm µH}}}\\ \end{align*}
\begin{align*} \underline{Z} &= R + \underline{Z}_L + \underline{Z}_C \\ &= R + j \cdot {Z}_L - j \cdot {Z}_C \\ &= R + j \cdot ({Z}_L - {Z}_C) \\ |\underline{Z}| &= \sqrt{R^2 + (\underline{Z}_L - \underline{Z}_C)^2 }\\ \end{align*}
\begin{align*} Z_L &= 31.1 ~\Omega \\ Z_C &= 48.2 ~\Omega \\ Z &= 19.8 ~\Omega \end{align*}
3. Draw the three impedance phasors $|\underline{Z}_C|$, $|\underline{Z}_L|$ and $|\underline{Z}_R|$ in a diagram.
Choose an appropriate scaling factor and write it down.
4. Calculate the current $|\underline{I}|$.
\begin{align*} Z &= {{\hat{U}}\over{\hat{I}}} \\ \hat{I} &= {{\hat{U}}\over{Z}} \\ \end{align*}
With $I = {{1}\over{\sqrt{2}}}\cdot \hat{I}$: \begin{align*} I &= {{1}\over{\sqrt{2}}}\cdot {{\hat{U}}\over{Z}} \\ &= {{1}\over{\sqrt{2}}}\cdot {{3.0 ~{\rm V}}\over{19.28 ~\Omega}} \\ \end{align*}
\begin{align*} I = 107 ~{\rm mA} \end{align*}