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--- electrical_engineering_and_electronics_2:block06 [2026/04/11 12:05] – mexleadmin
+++ electrical_engineering_and_electronics_2:block06 [2026/04/11 12:22] (current) – mexleadmin
@@ Line 648: / Line 648: @@
 </WRAP></WRAP></panel>
+#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~  Mains Input of a Control Cabinet: Resistor-Capacitor Parallel Branch                    #@TaskText_HTML@#
+<WRAP right>{{drawio>electrical_engineering_and_electronics_2:EMIcheckExerciseV01.svg}}</WRAP>
+At the input of an industrial control cabinet, an RC parallel branch is connected to the AC mains.
+The resistor represents a damping and discharge path, while the capacitor models an EMI suppression capacitor.
+For thermal design and power-quality assessment, the effective currents and powers of this branch shall be determined.
+Data:
+\begin{align*}
+u_1 &= \hat U_1 \cos(\omega t) \\
+\hat U_1 &= 325~{\rm V} \\
+f &= 50~{\rm Hz} \\
+R &= 220~{\rm \Omega} \\
+C &= 4.7~{\rm \mu F}
+\end{align*}
+. Determine the RMS values of the voltage $U_1$ and of the currents $I$, $I_R$, and $I_C$.
+<WRAP group>
+<WRAP half column rightalign>
+#@PathBegin_HTML~501~@#
+<WRAP leftalign>
+First convert the voltage amplitude into the RMS value:
+\begin{align*}
+U_1 &= \frac{\hat U_1}{\sqrt{2}}
+    = \frac{325~{\rm V}}{\sqrt{2}}
+    = 229.8~{\rm V}
+\end{align*}
+The angular frequency is
+\begin{align*}
+\omega &= 2\pi f
+       = 2\pi \cdot 50~{\rm s^{-1}}
+       = 314.16~{\rm s^{-1}}
+\end{align*}
+The resistor current is
+\begin{align*}
+I_R &= \frac{U_1}{R}
+    = \frac{229.8~{\rm V}}{220~{\rm \Omega}}
+    = 1.045~{\rm A}
+\end{align*}
+The capacitor current is
+\begin{align*}
+I_C &= U_1 \omega C \\
+    &= 229.8~{\rm V}\cdot 314.16~{\rm s^{-1}}\cdot 4.7\cdot 10^{-6}~{\rm F} \\
+    &= 0.339~{\rm A}
+\end{align*}
+Because $I_R$ and $I_C$ are perpendicular in the phasor diagram, the total current is
+\begin{align*}
+I &= \sqrt{I_R^2+I_C^2} \\
+  &= \sqrt{(1.045~{\rm A})^2 + (0.339~{\rm A})^2} \\
+  &= 1.098~{\rm A}
+\end{align*}
+</WRAP>
+#@PathEnd_HTML@#
+</WRAP>
+<WRAP half column>
+#@ResultBegin_HTML~501~@#
+\begin{align*}
+U_1 &= 229.8~{\rm V} \\
+I_R &= 1.045~{\rm A} \\
+I_C &= 0.339~{\rm A} \\
+I   &= 1.098~{\rm A}
+\end{align*}
+#@ResultEnd_HTML@#
+</WRAP>
+</WRAP>
+. What active, reactive, and apparent power does the circuit absorb?
+<WRAP group>
+<WRAP half column rightalign>
+#@PathBegin_HTML~502~@#
+<WRAP leftalign>
+The resistor absorbs only active power:
+\begin{align*}
+P &= \frac{U_1^2}{R}
+   = \frac{(229.8~{\rm V})^2}{220~{\rm \Omega}} \\
+  &= 240.1~{\rm W}
+\end{align*}
+The capacitor absorbs only reactive power. For a capacitor, the reactive power is negative:
+\begin{align*}
+Q &= -U_1^2 \omega C \\
+  &= -(229.8~{\rm V})^2 \cdot 314.16~{\rm s^{-1}} \cdot 4.7\cdot 10^{-6}~{\rm F} \\
+  &= -78.0~{\rm var}
+\end{align*}
+The apparent power is
+\begin{align*}
+S &= U_1 I \\
+  &= 229.8~{\rm V}\cdot 1.098~{\rm A} \\
+  &= 252.4~{\rm VA}
+\end{align*}
+Check with the power triangle:
+\begin{align*}
+S &= \sqrt{P^2+Q^2}
+   = \sqrt{(240.1~{\rm W})^2+(-78.0~{\rm var})^2} \\
+  &= 252.4~{\rm VA}
+\end{align*}
+</WRAP>
+#@PathEnd_HTML@#
+</WRAP>
+<WRAP half column>
+#@ResultBegin_HTML~502~@#
+\begin{align*}
+P &= 240.1~{\rm W} \\
+Q &= -78.0~{\rm var} \\
+S &= 252.4~{\rm VA}
+\end{align*}
+#@ResultEnd_HTML@#
+</WRAP>
+</WRAP>
+. Determine the maximum and minimum value of the instantaneous power.
+<WRAP group>
+<WRAP half column rightalign>
+#@PathBegin_HTML~503~@#
+<WRAP leftalign>
+For a sinusoidal AC circuit, the instantaneous power can be written as
+\begin{align*}
+p(t) = P + S\cos(2\omega t + \varphi)
+\end{align*}
+where $P$ is the active power, $S$ is the apparent power, and $\varphi$ is the phase angle.
+Thus the oscillating part has the amplitude $S$, so
+\begin{align*}
+p_{\rm max} &= P + S \\
+p_{\rm min} &= P - S
+\end{align*}
+Insert the values:
+\begin{align*}
+p_{\rm max} &= 240.1~{\rm W} + 252.4~{\rm W}
+            = 492.5~{\rm W} \\
+p_{\rm min} &= 240.1~{\rm W} - 252.4~{\rm W}
+            = -12.3~{\rm W}
+\end{align*}
+The negative minimum value means that for a short time interval, reactive energy is fed back from the capacitor to the source.
+</WRAP>
+#@PathEnd_HTML@#
+</WRAP>
+<WRAP half column>
+#@ResultBegin_HTML~503~@#
+\begin{align*}
+p_{\rm max} &= 492.5~{\rm W} \\
+p_{\rm min} &= -12.3~{\rm W}
+\end{align*}
+#@ResultEnd_HTML@#
+</WRAP>
+</WRAP>
+#@TaskEnd_HTML@#
+#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~  AC Solenoid Branch of a Valve Driver: Parallel Resistor-Inductor Circuit                    #@TaskText_HTML@#
+<WRAP right>{{drawio>electrical_engineering_and_electronics_2:ACcontrolunitV01.svg}}</WRAP>
+An industrial valve driver contains an AC solenoid branch that can be modeled by a resistor in parallel with an inductor.
+The resistor represents electrical losses, while the inductive branch represents the magnetizing behavior of the actuator.
+For thermal design and grid-side power assessment, the RMS currents and powers of the branch shall be determined.
+Data:
+\begin{align*}
+u_1 &= \hat U_1 \cos(\omega t) \\
+\hat U_1 &= 170~{\rm V} \\
+f &= 60~{\rm Hz} \\
+R &= 220~{\rm \Omega} \\
+L &= 325~{\rm mH}
+\end{align*}
+. Determine the RMS values of the voltage $U_1$ and of the currents $I$, $I_R$, and $I_L$.
+<WRAP group>
+<WRAP half column rightalign>
+#@PathBegin_HTML~511~@#
+<WRAP leftalign>
+First convert the voltage amplitude into the RMS value:
+\begin{align*}
+U_1 &= \frac{\hat U_1}{\sqrt{2}}
+    = \frac{170~{\rm V}}{\sqrt{2}}
+    = 120.2~{\rm V}
+\end{align*}
+The angular frequency is
+\begin{align*}
+\omega &= 2\pi f
+       = 2\pi \cdot 60~{\rm s^{-1}}
+       = 377.0~{\rm s^{-1}}
+\end{align*}
+The inductive reactance is
+\begin{align*}
+\omega L &= 377.0~{\rm s^{-1}} \cdot 0.325~{\rm H}
+         = 122.5~{\rm \Omega}
+\end{align*}
+The resistor current is
+\begin{align*}
+I_R &= \frac{U_1}{R}
+    = \frac{120.2~{\rm V}}{220~{\rm \Omega}}
+    = 0.546~{\rm A}
+\end{align*}
+The inductor current is
+\begin{align*}
+I_L &= \frac{U_1}{\omega L}
+    = \frac{120.2~{\rm V}}{122.5~{\rm \Omega}}
+    = 0.981~{\rm A}
+\end{align*}
+Because $I_R$ and $I_L$ are perpendicular in the phasor diagram, the total current is
+\begin{align*}
+I &= \sqrt{I_R^2+I_L^2} \\
+  &= \sqrt{(0.546~{\rm A})^2 + (0.981~{\rm A})^2} \\
+  &= 1.123~{\rm A}
+\end{align*}
+</WRAP>
+#@PathEnd_HTML@#
+</WRAP>
+<WRAP half column>
+#@ResultBegin_HTML~511~@#
+\begin{align*}
+U_1 &= 120.2~{\rm V} \\
+I_R &= 0.546~{\rm A} \\
+I_L &= 0.981~{\rm A} \\
+I   &= 1.123~{\rm A}
+\end{align*}
+#@ResultEnd_HTML@#
+</WRAP>
+</WRAP>
+. What active, reactive, and apparent power does the circuit absorb?
+<WRAP group>
+<WRAP half column rightalign>
+#@PathBegin_HTML~512~@#
+<WRAP leftalign>
+The resistor absorbs only active power:
+\begin{align*}
+P &= \frac{U_1^2}{R}
+   = \frac{(120.2~{\rm V})^2}{220~{\rm \Omega}} \\
+  &= 65.7~{\rm W}
+\end{align*}
+The inductor absorbs only reactive power. For an inductor, reactive power is positive:
+\begin{align*}
+Q &= \frac{U_1^2}{\omega L}
+   = \frac{(120.2~{\rm V})^2}{122.5~{\rm \Omega}} \\
+  &= 117.9~{\rm var}
+\end{align*}
+The apparent power is
+\begin{align*}
+S &= U_1 I \\
+  &= 120.2~{\rm V}\cdot 1.123~{\rm A} \\
+  &= 135.0~{\rm VA}
+\end{align*}
+Check with the power triangle:
+\begin{align*}
+S &= \sqrt{P^2+Q^2}
+   = \sqrt{(65.7~{\rm W})^2+(117.9~{\rm var})^2} \\
+  &= 135.0~{\rm VA}
+\end{align*}
+</WRAP>
+#@PathEnd_HTML@#
+</WRAP>
+<WRAP half column>
+#@ResultBegin_HTML~512~@#
+\begin{align*}
+P &= 65.7~{\rm W} \\
+Q &= 117.9~{\rm var} \\
+S &= 135.0~{\rm VA}
+\end{align*}
+#@ResultEnd_HTML@#
+</WRAP>
+</WRAP>
+. Determine the maximum and minimum value of the instantaneous power.
+<WRAP group>
+<WRAP half column rightalign>
+#@PathBegin_HTML~513~@#
+<WRAP leftalign>
+For a sinusoidal AC circuit, the instantaneous power can be written as
+\begin{align*}
+p(t) = P + S\cos(2\omega t + \varphi_p)
+\end{align*}
+Thus the oscillating part has the amplitude $S$, so
+\begin{align*}
+p_{\rm max} &= P + S \\
+p_{\rm min} &= P - S
+\end{align*}
+Insert the values:
+\begin{align*}
+p_{\rm max} &= 65.7~{\rm W} + 135.0~{\rm W}
+            = 200.7~{\rm W} \\
+p_{\rm min} &= 65.7~{\rm W} - 135.0~{\rm W}
+            = -69.3~{\rm W}
+\end{align*}
+The negative minimum value means that magnetic energy stored in the inductor is temporarily fed back to the source.
+</WRAP>
+#@PathEnd_HTML@#
+</WRAP>
+<WRAP half column>
+#@ResultBegin_HTML~513~@#
+\begin{align*}
+p_{\rm max} &= 200.7~{\rm W} \\
+p_{\rm min} &= -69.3~{\rm W}
+\end{align*}
+#@ResultEnd_HTML@#
+</WRAP>
+</WRAP>
+#@TaskEnd_HTML@#
+#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~  AC Filter and Sensor Front-End: Power Flow in a Composite Reactive Network                    #@TaskText_HTML@#
+<WRAP right>{{drawio>electrical_engineering_and_electronics_2:ACsensorFrontendV01.svg}}</WRAP>
+An industrial AC sensor front-end contains a compensation capacitor, a series inductor, and an output branch made of a resistor and another inductor.
+For thermal design and reactive-power assessment, the active, reactive, and apparent powers at the input and at each component shall be determined.
+Data:
+\begin{align*}
+\omega C &= 0.01~{\rm S} \\
+\omega L_1 &= 50~{\rm \Omega} \\
+\omega L_2 &= 200~{\rm \Omega} \\
+R &= 100~{\rm \Omega} \\
+\hat U_1 &= 325~{\rm V}
+\end{align*}
+. Calculate the active, reactive, and apparent power at the input and at the individual components of the circuit.
+<WRAP group>
+<WRAP half column rightalign>
+#@PathBegin_HTML~521~@#
+<WRAP leftalign>
+For power calculations, we use RMS values. Therefore, the source voltage is
+\begin{align*}
+U_1 &= \frac{\hat U_1}{\sqrt{2}}
+    = \frac{325~{\rm V}}{\sqrt{2}}
+    = 229.8~{\rm V}
+\end{align*}
+First determine the impedances of the individual elements:
+\begin{align*}
+\underline{Z}_C &= \frac{1}{j\omega C}
+= -j\frac{1}{\omega C}
+= -j100~{\rm \Omega} \\
+\underline{Z}_{L_1} &= j\omega L_1
+= j50~{\rm \Omega} \\
+\underline{Z}_{L_2} &= j\omega L_2
+= j200~{\rm \Omega}
+\end{align*}
+Now the parallel branch:
+\begin{align*}
+\underline{Z}_{R\parallel L_2}
+&= \frac{R\underline{Z}_{L_2}}{R+\underline{Z}_{L_2}} \\
+&= \frac{100\cdot j200}{100+j200}
+= 80+j40~{\rm \Omega}
+\end{align*}
+Thus the total input impedance is
+\begin{align*}
+\underline{Z}_{\rm in}
+&= \underline{Z}_C + \underline{Z}_{L_1} + \underline{Z}_{R\parallel L_2} \\
+&= (-j100) + (j50) + (80+j40) \\
+&= 80-j10~{\rm \Omega}
+\end{align*}
+The input current is therefore
+\begin{align*}
+\underline{I}
+&= \frac{\underline{U}_1}{\underline{Z}_{\rm in}}
+= \frac{229.8~{\rm V}}{80-j10~{\rm \Omega}} \\
+&= 2.828 + j0.354~{\rm A}
+\end{align*}
+Its magnitude is
+\begin{align*}
+I = 2.85~{\rm A}
+\end{align*}
+The complex input power is
+\begin{align*}
+\underline{S}_{\rm in}
+&= \underline{U}_1 \underline{I}^{\,*} \\
+&= 229.8 \cdot (2.828-j0.354)~{\rm VA} \\
+&= 650 - j81.25~{\rm VA}
+\end{align*}
+Hence
+\begin{align*}
+P_{\rm in} &= 650~{\rm W} \\
+Q_{\rm in} &= -81.25~{\rm var} \\
+S_{\rm in} &= \sqrt{P_{\rm in}^2+Q_{\rm in}^2}
+= 655.1~{\rm VA}
+\end{align*}
+Now calculate the powers of the individual components.
+For the capacitor:
+\begin{align*}
+\underline{S}_C
+&= \underline{U}_C \underline{I}^{\,*}
+= -j812.5~{\rm VA}
+\end{align*}
+So
+\begin{align*}
+P_C &= 0~{\rm W} \\
+Q_C &= -812.5~{\rm var} \\
+S_C &= 812.5~{\rm VA}
+\end{align*}
+For the inductor $L_1$:
+\begin{align*}
+\underline{S}_{L_1}
+&= \underline{U}_{L_1}\underline{I}^{\,*}
+= j406.25~{\rm VA}
+\end{align*}
+Thus
+\begin{align*}
+P_{L_1} &= 0~{\rm W} \\
+Q_{L_1} &= 406.25~{\rm var} \\
+S_{L_1} &= 406.25~{\rm VA}
+\end{align*}
+The voltage across the parallel branch is
+\begin{align*}
+\underline{U}_{R\parallel L_2}
+&= \underline{I}\,\underline{Z}_{R\parallel L_2} \\
+&= (2.828+j0.354)(80+j40) \\
+&= 212.13 + j141.42~{\rm V}
+\end{align*}
+The resistor current is
+\begin{align*}
+\underline{I}_R
+= \frac{\underline{U}_{R\parallel L_2}}{R}
+= 2.121 + j1.414~{\rm A}
+\end{align*}
+The inductor current is
+\begin{align*}
+\underline{I}_{L_2}
+= \frac{\underline{U}_{R\parallel L_2}}{j200}
+= 0.707 - j1.061~{\rm A}
+\end{align*}
+For the resistor:
+\begin{align*}
+\underline{S}_R
+&= \underline{U}_{R\parallel L_2}\underline{I}_R^{\,*}
+= 650 + j0~{\rm VA}
+\end{align*}
+So
+\begin{align*}
+P_R &= 650~{\rm W} \\
+Q_R &= 0~{\rm var} \\
+S_R &= 650~{\rm VA}
+\end{align*}
+For the inductor $L_2$:
+\begin{align*}
+\underline{S}_{L_2}
+&= \underline{U}_{R\parallel L_2}\underline{I}_{L_2}^{\,*}
+= j325~{\rm VA}
+\end{align*}
+Thus
+\begin{align*}
+P_{L_2} &= 0~{\rm W} \\
+Q_{L_2} &= 325~{\rm var} \\
+S_{L_2} &= 325~{\rm VA}
+\end{align*}
+</WRAP>
+#@PathEnd_HTML@#
+</WRAP>
+<WRAP half column>
+#@ResultBegin_HTML~521~@#
+\begin{align*}
+P_{\rm in} &= 650~{\rm W} \\
+Q_{\rm in} &= -81.25~{\rm var} \\
+S_{\rm in} &= 655.1~{\rm VA}
+\end{align*}
+\begin{align*}
+P_C &= 0~{\rm W}, & Q_C &= -812.5~{\rm var}, & S_C &= 812.5~{\rm VA} \\
+P_{L_1} &= 0~{\rm W}, & Q_{L_1} &= 406.25~{\rm var}, & S_{L_1} &= 406.25~{\rm VA} \\
+P_R &= 650~{\rm W}, & Q_R &= 0~{\rm var}, & S_R &= 650~{\rm VA} \\
+P_{L_2} &= 0~{\rm W}, & Q_{L_2} &= 325~{\rm var}, & S_{L_2} &= 325~{\rm VA}
+\end{align*}
+#@ResultEnd_HTML@#
+</WRAP>
+</WRAP>
+. Verify the validity of the following reactive-power balance:
+\[
+Q = Q_C + Q_{L_1} + Q_{L_2}
+\]
+<WRAP group>
+<WRAP half column rightalign>
+#@PathBegin_HTML~522~@#
+<WRAP leftalign>
+Insert the reactive powers found above:
+\begin{align*}
+Q_C &= -812.5~{\rm var} \\
+Q_{L_1} &= 406.25~{\rm var} \\
+Q_{L_2} &= 325~{\rm var}
+\end{align*}
+Then
+\begin{align*}
+Q_C + Q_{L_1} + Q_{L_2}
+&= -812.5 + 406.25 + 325 \\
+&= -81.25~{\rm var}
+\end{align*}
+But the input reactive power is
+\begin{align*}
+Q_{\rm in} = -81.25~{\rm var}
+\end{align*}
+Therefore,
+\begin{align*}
+Q_{\rm in} = Q_C + Q_{L_1} + Q_{L_2}
+\end{align*}
+So the reactive-power balance is fulfilled.
+</WRAP>
+#@PathEnd_HTML@#
+</WRAP>
+<WRAP half column>
+#@ResultBegin_HTML~522~@#
+\begin{align*}
+Q_C + Q_{L_1} + Q_{L_2}
+= -812.5 + 406.25 + 325
+= -81.25~{\rm var}
+\end{align*}
+\begin{align*}
+Q_{\rm in} = -81.25~{\rm var}
+\end{align*}
+\begin{align*}
+Q_{\rm in} = Q_C + Q_{L_1} + Q_{L_2}
+\end{align*}
+#@ResultEnd_HTML@#
+</WRAP>
+</WRAP>
+#@TaskEnd_HTML@#
+#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~  Real Current-Sense Choke: Series Model of an Industrial Coil                    #@TaskText_HTML@#
+A small current-sense choke in an industrial electronics module is used at low frequency for filtering and current shaping.
+In practice, the coil is not ideal: besides its inductance, it also has a winding resistance.
+Therefore, the real coil is modeled as a series connection of an inductance and an ohmic resistance.
+Data:
+\begin{align*}
+L_{\rm sp} &= 2.5~{\rm mH} \\
+R_{\rm sp} &= 100~{\rm m\Omega} \\
+I_{\rm sp} &= 0.5~{\rm A} \\
+f &= 50~{\rm Hz}
+\end{align*}
+. Draw the circuit and place all current and voltage phasors in the phasor diagram.
+<WRAP group>
+<WRAP half column rightalign>
+#@PathBegin_HTML~531~@#
+<WRAP leftalign>
+The real coil is modeled as a series connection of
+\begin{align*}
+R_{\rm sp} \text{ and } L_{\rm sp}
+\end{align*}
+Because this is a series circuit, the same current flows through both elements:
+\begin{align*}
+I_R = I_L = I_{\rm sp} = 0.5~{\rm A}
+\end{align*}
+Choose the current as the reference phasor:
+\begin{align*}
+\underline{I}_{\rm sp} = 0.5~{\rm A}\angle 0^\circ
+\end{align*}
+Then:
+\begin{align*}
+\underline{U}_R &\text{ is in phase with } \underline{I}_{\rm sp} \\
+\underline{U}_L &\text{ leads } \underline{I}_{\rm sp} \text{ by } 90^\circ \\
+\underline{U}_{\rm sp} &= \underline{U}_R + \underline{U}_L
+\end{align*}
+So in the phasor diagram:
+\begin{align*}
+\underline{I}_{\rm sp} &: \text{horizontal to the right} \\
+\underline{U}_R &: \text{same direction as } \underline{I}_{\rm sp} \\
+\underline{U}_L &: \text{vertical upward} \\
+\underline{U}_{\rm sp} &: \text{diagonal sum of } \underline{U}_R \text{ and } \underline{U}_L
+\end{align*}
+</WRAP>
+#@PathEnd_HTML@#
+</WRAP>
+<WRAP half column>
+#@ResultBegin_HTML~531~@#
+\begin{align*}
+\underline{I}_{\rm sp} &= 0.5~{\rm A}\angle 0^\circ \\
+\underline{U}_R &\parallel \underline{I}_{\rm sp} \\
+\underline{U}_L &\text{ leads by }90^\circ \\
+\underline{U}_{\rm sp} &= \underline{U}_R+\underline{U}_L
+\end{align*}
+#@ResultEnd_HTML@#
+</WRAP>
+</WRAP>
+. Calculate the complex input impedance.
+<WRAP group>
+<WRAP half column rightalign>
+#@PathBegin_HTML~532~@#
+<WRAP leftalign>
+First calculate the angular frequency:
+\begin{align*}
+\omega &= 2\pi f
+= 2\pi \cdot 50~{\rm s^{-1}}
+= 314.16~{\rm s^{-1}}
+\end{align*}
+Then the inductive reactance is
+\begin{align*}
+X_L = \omega L_{\rm sp}
+= 314.16~{\rm s^{-1}} \cdot 2.5\cdot 10^{-3}~{\rm H}
+= 0.785~{\rm \Omega}
+\end{align*}
+The complex input impedance of the real coil is
+\begin{align*}
+\underline{Z}_{\rm sp}
+&= R_{\rm sp} + jX_L \\
+&= 0.100 + j0.785~{\rm \Omega}
+\end{align*}
+Its magnitude is
+\begin{align*}
+\left|\underline{Z}_{\rm sp}\right|
+&= \sqrt{0.100^2+0.785^2} \\
+&= 0.792~{\rm \Omega}
+\end{align*}
+</WRAP>
+#@PathEnd_HTML@#
+</WRAP>
+<WRAP half column>
+#@ResultBegin_HTML~532~@#
+\begin{align*}
+\underline{Z}_{\rm sp} &= 0.100 + j0.785~{\rm \Omega} \\
+\left|\underline{Z}_{\rm sp}\right| &= 0.792~{\rm \Omega}
+\end{align*}
+#@ResultEnd_HTML@#
+</WRAP>
+</WRAP>
+. How large is the voltage across the coil, and what is the phase-shift angle?
+<WRAP group>
+<WRAP half column rightalign>
+#@PathBegin_HTML~533~@#
+<WRAP leftalign>
+The voltage across the winding resistance is
+\begin{align*}
+U_R &= I_{\rm sp} R_{\rm sp}
+= 0.5~{\rm A}\cdot 0.100~{\rm \Omega}
+= 0.050~{\rm V}
+\end{align*}
+The voltage across the inductance is
+\begin{align*}
+U_L &= I_{\rm sp} X_L
+= 0.5~{\rm A}\cdot 0.785~{\rm \Omega}
+= 0.3927~{\rm V}
+\end{align*}
+The total coil voltage is the vector sum:
+\begin{align*}
+U_{\rm sp}
+&= I_{\rm sp}\left|\underline{Z}_{\rm sp}\right| \\
+&= 0.5~{\rm A}\cdot 0.792~{\rm \Omega}
+= 0.396~{\rm V}
+\end{align*}
+The phase-shift angle between coil voltage and current is
+\begin{align*}
+\varphi &= \arctan\left(\frac{X_L}{R_{\rm sp}}\right)
+= \arctan\left(\frac{0.785}{0.100}\right)
+= 82.74^\circ
+\end{align*}
+So the coil voltage leads the current by about $82.7^\circ$.
+</WRAP>
+#@PathEnd_HTML@#
+</WRAP>
+<WRAP half column>
+#@ResultBegin_HTML~533~@#
+\begin{align*}
+U_R &= 0.050~{\rm V} \\
+U_L &= 0.3927~{\rm V} \\
+U_{\rm sp} &= 0.396~{\rm V} \\
+\varphi &= 82.74^\circ
+\end{align*}
+#@ResultEnd_HTML@#
+</WRAP>
+</WRAP>
+. Calculate the active, reactive, and apparent power absorbed by the coil, and determine the power factor.
+<WRAP group>
+<WRAP half column rightalign>
+#@PathBegin_HTML~534~@#
+<WRAP leftalign>
+The active power is dissipated only in the resistance:
+\begin{align*}
+P &= I_{\rm sp}^2 R_{\rm sp} \\
+  &= (0.5~{\rm A})^2 \cdot 0.100~{\rm \Omega} \\
+  &= 0.0250~{\rm W}
+\end{align*}
+The reactive power is taken by the inductance:
+\begin{align*}
+Q &= I_{\rm sp}^2 X_L \\
+  &= (0.5~{\rm A})^2 \cdot 0.785~{\rm \Omega} \\
+  &= 0.196~{\rm var}
+\end{align*}
+The apparent power is
+\begin{align*}
+S &= U_{\rm sp} I_{\rm sp} \\
+  &= 0.396~{\rm V}\cdot 0.5~{\rm A} \\
+  &= 0.198~{\rm VA}
+\end{align*}
+The power factor is
+\begin{align*}
+\lambda = \cos\varphi = \frac{P}{S}
+= \frac{0.0250}{0.198}
+= 0.126
+\end{align*}
+</WRAP>
+#@PathEnd_HTML@#
+</WRAP>
+<WRAP half column>
+#@ResultBegin_HTML~534~@#
+\begin{align*}
+P &= 0.0250~{\rm W} \\
+Q &= 0.196~{\rm var} \\
+S &= 0.198~{\rm VA} \\
+\lambda &= \cos\varphi = 0.126
+\end{align*}
+#@ResultEnd_HTML@#
+</WRAP>
+</WRAP>
+. Determine the loss factor and the loss angle.
+<WRAP group>
+<WRAP half column rightalign>
+#@PathBegin_HTML~535~@#
+<WRAP leftalign>
+For a real inductor, the loss factor is
+\begin{align*}
+d = \frac{P}{Q} = \frac{R_{\rm sp}}{X_L}
+\end{align*}
+Thus,
+\begin{align*}
+d &= \frac{0.100}{0.785}
+= 0.127
+\end{align*}
+The loss angle $\delta$ is related to the loss factor by
+\begin{align*}
+\tan\delta = d
+\end{align*}
+Hence,
+\begin{align*}
+\delta &= \arctan(d)
+= \arctan(0.127)
+= 7.26^\circ
+\end{align*}
+As a check:
+\begin{align*}
+\delta = 90^\circ - \varphi
+= 90^\circ - 82.74^\circ
+= 7.26^\circ
+\end{align*}
+</WRAP>
+#@PathEnd_HTML@#
+</WRAP>
+<WRAP half column>
+#@ResultBegin_HTML~535~@#
+\begin{align*}
+d &= 0.127 \\
+\delta &= 7.26^\circ
+\end{align*}
+#@ResultEnd_HTML@#
+</WRAP>
+</WRAP>
+. Draw the power triangle and indicate the loss angle.
+<WRAP group>
+<WRAP half column rightalign>
+#@PathBegin_HTML~536~@#
+<WRAP leftalign>
+In the power triangle:
+\begin{align*}
+P &= 0.0250~{\rm W} \qquad \text{horizontal axis} \\
+Q &= 0.196~{\rm var} \qquad \text{vertical axis upward} \\
+S &= 0.198~{\rm VA} \qquad \text{hypotenuse}
+\end{align*}
+The angle between $S$ and the horizontal axis is
+\begin{align*}
+\varphi = 82.74^\circ
+\end{align*}
+The loss angle is the complementary angle:
+\begin{align*}
+\delta = 7.26^\circ
+\end{align*}
+So in the sketch, mark $\delta$ between the vertical reactive-power axis and the apparent-power vector.
+</WRAP>
+#@PathEnd_HTML@#
+</WRAP>
+<WRAP half column>
+#@ResultBegin_HTML~536~@#
+\begin{align*}
+P &= 0.0250~{\rm W} \\
+Q &= 0.196~{\rm var} \\
+S &= 0.198~{\rm VA} \\
+\varphi &= 82.74^\circ \\
+\delta &= 7.26^\circ
+\end{align*}
+#@ResultEnd_HTML@#
+</WRAP>
+</WRAP>
+#@TaskEnd_HTML@#
 ===== Embedded resources =====
 <WRAP column half>