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--- electrical_engineering_and_electronics_2:block04 [2026/04/21 01:48] – mexleadmin
+++ electrical_engineering_and_electronics_2:block04 [2026/04/21 02:12] (current) – mexleadmin
@@ Line 712: / Line 712: @@
 . Determine the complex input impedance of the circuit.
+<WRAP group>
+<WRAP half column rightalign>
+#@PathBegin_HTML~421~@#
+<WRAP leftalign>
+First determine the individual impedances:
+\begin{align*}
+\underline{Z}_C &= \frac{1}{j\,\omega C}
+= -j\,\frac{1}{\omega C}
+= -j\,100~{\rm \Omega} \\
+\underline{Z}_{L_1} &= j\,\omega L_1
+= j\,50~{\rm \Omega} \\
+\underline{Z}_{L_2} &= j\,\omega L_2
+= j\,200~{\rm \Omega}
+\end{align*}
+Now the parallel branch:
+\begin{align*}
+\underline{Z}_{R\parallel L_2}
+&= \frac{R \cdot \underline{Z}_{L_2}}{R+\underline{Z}_{L_2}} \\
+&= \frac{100 \cdot j\,200}{100+j\,200}
+= 80~{\rm \Omega} + j\,40~{\rm \Omega}
+\end{align*}
+Total input impedance:
+\begin{align*}
+\underline{Z}_{\rm in}
+&= \underline{Z}_C + \underline{Z}_{L_1} + \underline{Z}_{R\parallel L_2} \\
+&= (-j\,100~{\rm \Omega}) + (j\,50~{\rm \Omega}) + (80~{\rm \Omega}+j\,40~{\rm \Omega}) \\
+&= 80~{\rm \Omega} - j10~{\rm \Omega}
+\end{align*}
+In polar form:
+\begin{align*}
+\underline{Z}_{\rm in}
+= 80.6~{\rm \Omega}\angle -7.13^\circ
+\end{align*}
+</WRAP>
+#@PathEnd_HTML@#
+</WRAP>
+<WRAP half column>
+#@ResultBegin_HTML~421~@#
+\begin{align*}
+\underline{Z}_{\rm in} &= 80 - j10~{\rm \Omega} \\
+                       &= 80.6~{\rm \Omega}\angle -7.13^\circ
+\end{align*}
+#@ResultEnd_HTML@#
+</WRAP>
+</WRAP>
 . Calculate the input current by magnitude and phase.
+<WRAP group>
+<WRAP half column rightalign>
+#@PathBegin_HTML~422~@#
+<WRAP leftalign>
+\begin{align*}
+\underline{\hat I}_{\rm in}
+&= \frac{\underline{\hat U}_1}{\underline{Z}_{\rm in}}
+\end{align*}
+With $\underline{\hat U}_1 = 325~{\rm V}\angle 0^\circ$:
+\begin{align*}
+\left|\underline{\hat I}_{\rm in}\right|
+&= \frac{325}{80.6}
+= 4.03~{\rm A}
+\end{align*}
+Because the impedance angle is $-7.13^\circ$, the current angle is
+\begin{align*}
+\varphi_I = +7.13^\circ
+\end{align*}
+So the current leads the source voltage slightly.
+</WRAP>
+#@PathEnd_HTML@#
+</WRAP>
+<WRAP half column>
+#@ResultBegin_HTML~422~@#
+\begin{align*}
+\underline{\hat I}_{\rm in} = 4.03~{\rm A}\angle 7.13^\circ
+\end{align*}
+#@ResultEnd_HTML@#
+</WRAP>
+</WRAP>
 . Draw the phasor diagram for all currents and voltages.
+<WRAP group>
+<WRAP half column rightalign>
+#@PathBegin_HTML~423~@#
+<WRAP leftalign>
+Use the source voltage as reference:
+\begin{align*}
+\underline{\hat U}_1 = 325~{\rm V}\angle 0^\circ
+\end{align*}
+Input current:
+\begin{align*}
+\underline{\hat I}_{\rm in} = 4.03~{\rm A}\angle 7.13^\circ
+\end{align*}
+Voltage drops in the series elements:
+\begin{align*}
+\underline{\hat U}_C
+&= \underline{\hat I}_{\rm in}\,\underline{Z}_C
+= 403.1~{\rm V}\angle -82.9^\circ \\
+\underline{\hat U}_{L_1}
+&= \underline{\hat I}_{\rm in}\,\underline{Z}_{L_1}
+= 201.6~{\rm V}\angle 97.1^\circ
+\end{align*}
+Voltage across the parallel branch:
+\begin{align*}
+\underline{\hat U}_{R\parallel L_2}
+&= \underline{\hat I}_{\rm in}\,(80+j40) \\
+&= 360.6~{\rm V}\angle 33.7^\circ
+\end{align*}
+Branch currents:
+\begin{align*}
+\underline{\hat I}_R
+&= \frac{\underline{\hat U}_{R\parallel L_2}}{R}
+= 3.61~{\rm A}\angle 33.7^\circ \\
+\underline{\hat I}_{L_2}
+&= \frac{\underline{\hat U}_{R\parallel L_2}}{j200}
+= 1.80~{\rm A}\angle -56.3^\circ
+\end{align*}
+Check:
+\begin{align*}
+\underline{\hat I}_{\rm in}
+= \underline{\hat I}_R + \underline{\hat I}_{L_2}
+\end{align*}
+</WRAP>
+#@PathEnd_HTML@#
+</WRAP>
+<WRAP half column>
+#@ResultBegin_HTML~423~@#
+\begin{align*}
+\underline{\hat I}_{\rm in} &= 4.03~{\rm A}\angle 7.13^\circ \\
+\underline{\hat U}_C &= 403.1~{\rm V}\angle -82.9^\circ \\
+\underline{\hat U}_{L_1} &= 201.6~{\rm V}\angle 97.1^\circ \\
+\underline{\hat U}_{R\parallel L_2} &= 360.6~{\rm V}\angle 33.7^\circ \\
+\underline{\hat I}_R &= 3.61~{\rm A}\angle 33.7^\circ \\
+\underline{\hat I}_{L_2} &= 1.80~{\rm A}\angle -56.3^\circ
+\end{align*}
+{{drawio>electrical_engineering_and_electronics_2:diagramsolution423.svg}}
+#@ResultEnd_HTML@#
+</WRAP>
+</WRAP>
 #@TaskEnd_HTML@#