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electrical_engineering_and_electronics_2:block04 [2026/04/21 01:31] – mexleadminelectrical_engineering_and_electronics_2:block04 [2026/04/21 02:12] (current) – mexleadmin
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 \begin{align*} \begin{align*}
 \left|{1}\over{\underline{Z}_{\rm in}} \right| &= 4.78~{\rm mS} \\ \left|{1}\over{\underline{Z}_{\rm in}} \right| &= 4.78~{\rm mS} \\
-\varphi_{1}\over{\underline{Z}_{\rm in}} &= \arctan\left(\frac{1.48}{4.55}\right) \approx 18.0^\circ+\varphi_{{1}\over{Z}} &= \arctan\left(\frac{1.48}{4.55}\right) \approx +18.0^\circ
 \end{align*} \end{align*}
  
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 \underline{Z}_{\rm in} &= \frac{1}{4.78~{\rm mS}} \\ \underline{Z}_{\rm in} &= \frac{1}{4.78~{\rm mS}} \\
                        &= 209.2~{\rm \Omega} \\                        &= 209.2~{\rm \Omega} \\
-\varphi_Z &= -\approx 18.0^\circ+\varphi_Z &= \approx -18.0^\circ
 \end{align*} \end{align*}
  
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 <WRAP leftalign> <WRAP leftalign>
 \begin{align*} \begin{align*}
-\underline{\hat I}_{\rm in} &= \underline{\hat U}_1 \,\underline{Y}_{\rm in}+\underline{\hat I}_{\rm in} &= \underline{\hat U}_1 \cdot {{1}\over{\underline{Z}_{\rm in}}}
 \end{align*} \end{align*}
  
Line 599: Line 599:
 \end{align*} \end{align*}
  
-The current phase is equal to the admittance phase:+The current phase is equal to the admittance phase (negative phase of impedance):
 \begin{align*} \begin{align*}
 \varphi_I &= 18.0^\circ \varphi_I &= 18.0^\circ
Line 617: Line 617:
  
 4. Draw the phasor diagram for all currents and voltages. 4. Draw the phasor diagram for all currents and voltages.
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~404~@#
 +<WRAP leftalign>
 +Choose the input voltage as reference:
 +\begin{align*}
 +\underline{\hat U}_1 &= 325~{\rm V}\angle 0^\circ
 +\end{align*}
 +
 +Branch currents:
 +\begin{align*}
 +\underline{\hat I}_R &= \frac{325}{220}~{\rm A}\angle 0^\circ
 += 1.48~{\rm A}\angle 0^\circ \\
 +\underline{\hat I}_C &= 325 \cdot \omega C~{\rm A}\angle 90^\circ
 += 0.48~{\rm A}\angle 90^\circ
 +\end{align*}
 +
 +Total current:
 +\begin{align*}
 +\underline{\hat I}_{\rm in}
 += \underline{\hat I}_R + \underline{\hat I}_C
 += 1.55~{\rm A}\angle 18.0^\circ
 +\end{align*}
 +
 +All branch voltages are equal in a parallel circuit:
 +\begin{align*}
 +\underline{\hat U}_R = \underline{\hat U}_C = \underline{\hat U}_1
 +\end{align*}
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~404~@#
 +\begin{align*}
 +\underline{\hat U}_1 &= \underline{\hat U}_R = \underline{\hat U}_C
 += 325~{\rm V}\angle 0^\circ \\
 +\underline{\hat I}_R &= 1.48~{\rm A}\angle 0^\circ \\
 +\underline{\hat I}_C &= 0.48~{\rm A}\angle 90^\circ \\
 +\underline{\hat I}_{\rm in} &= 1.55~{\rm A}\angle 18.0^\circ
 +\end{align*}
 +
 +{{drawio>electrical_engineering_and_electronics_2:diagramsolution404.svg}}
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
  
 #@TaskEnd_HTML@# #@TaskEnd_HTML@#
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 1. Determine the complex input impedance of the circuit. 1. 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>
  
 2. Calculate the input current by magnitude and phase. 2. 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>
  
 3. Draw the phasor diagram for all currents and voltages. 3. 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@# #@TaskEnd_HTML@#