\documentclass[titlepage]{article} \usepackage[czech]{babel} \usepackage[utf8x]{inputenc} \usepackage{ulem} \usepackage{graphicx} \author{Filip Munk, xmunkf00 (120076)} \title{Projekt do předmětu ITO} \date{\today} \begin{document} \maketitle %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Úloha první} Stanovte napětí \( U_{R6} \) a proud \( I_{R6} \). Použijte metodu postupného zjednodušování obvodu. \bigskip \begin{center} \begin{tabular}{|c|c|c|c|c|c|c|c|c|} \hline U [V] & \(R_{1} [\Omega] \) & \(R_{2} [\Omega]\) & \(R_{3} [\Omega]\) & \(R_{4} [\Omega]\) & \(R_{5} [\Omega]\) & \(R_{6} [\Omega]\) & \(R_{7} [\Omega] \) & \(R_{8} [\Omega] \) \\ \hline 130 & 380 & 420 & 330 & 440 & 450 & 650 & 410 & 275 \\ \hline \end{tabular} \end{center} \begin{figure}[h] \centering \includegraphics[scale=0.5]{img/u1d1.eps} \caption{Schema zapojení osporů} \end{figure} \subsection{Transformace trojúhelnku na hvězdu \(R_2,R_3,R_4\)} \[R_A=\frac{R_2 \times R_3}{R_2 + R_3 + R_4}\] \[R_B=\frac{R_2 \times R_4}{R_2 + R_3 + R_4}\] \[R_C=\frac{R_3 \times R_4}{R_2 + R_3 + R_4}\] \begin{figure}[h] \centering \includegraphics[scale=0.5]{img/u1d2.eps} \caption{Zapojení po transformaci trojúhelníku na hvězdu} \end{figure} \subsection{Tvorba a úprava vzorců} Nejprve si spočítáme celkový odpor R potá Proud na odporu \(R_6\) a nekonec napětí na tomto odporu. \subsubsection{Celkový odpor R} \[R_{1A}=R_1+R_A\] \[R_{57B}=R_5+R_7+R_B\] \[R_{6C}=R_6+R_C\] \[R_{567BC}=\frac{R_{57B}\times R_{6C}}{R_{57B}+R_{6C}}\] \[R_{15678ABC}=R_{1A}+R_{567BC}+R_8\] \[R= R_1+R_A+ \frac {\left(R_5+R_7+\left(R_B\right)\right)\times \left(R_6+\left(R_C\right)\right)} {R_5+R_7+\left(R_B\right)+R_6+\left(R_C\right)} +R_8=\] \bigskip \[=R_1+\left(\frac{R_2 \times R_3}{R_2 + R_3 + R_4}\right)+\] \[+\frac {\left(R_5+R_7+\left(\frac{R_2 \times R_4}{R_2 + R_3 + R_4}\right)\right)\times \left(R_6+\left(\frac{R_3 \times R_4}{R_2 + R_3 + R_4}\right)\right)} {R_5+R_7+\left(\frac{R_2 \times R_4}{R_2 + R_3 + R_4}\right)+R_6+\left(\frac{R_3 \times R_4}{R_2 + R_3 + R_4}\right)} +R_8 \] Poslední úpravu tohoto vzorce však nebudu používat, jelikož je nepohodlná pro dosazování a výpočty. \subsubsection{Hledání \(I_{R6}\) a \(U_{R6}\)} \[I=\frac{U}{R}\] Využijeme toho, že se napětí poměrem dělí na odporech zapojených do série a na paralelních větvích zůstává stejné. (Proud se chová přesně opačně) Z toho vypýlvá \[I_{R1A}=I_{R567BC}=I_{R8}=I\] \[U_{R567BC}=R_{R567BC}\times I\] \[U_{R6C}=U_{R57B}=U_{R567BC}\] \[I_{R6C}=\frac{U_{R6C}}{R_{6C}}\] \[I_{R6}=I_{RC}=I_{R6C}\] \[\uuline{I_{R6}}=I_{RC}=I_{R6C}\] \[\uuline{U_{R6}}=I_{R6}\times R_{6}\] \subsection{Výpočet a dosazení} \subsubsection{celkový odpor} \[R_A=\frac{420 \times 330}{420 + 330 + 440}=\frac{1980}{17} \Omega\] \[R_B=\frac{420 \times 440}{420 + 330 + 440}=\frac{2640}{17} \Omega\] \[R_C=\frac{330 \times 440}{420 + 330 + 440}=\frac{10890}{119} \Omega\] \[R_{1A}=380+\frac{1980}{17} =\frac{8440}{17} \Omega\] \[R_{57B}=450+410+\frac{2640}{17}=\frac{17260}{17} \Omega\] \[R_{6C}=650+\frac{10890}{119}=\frac{88240}{119}\Omega\] \[R_{567BC}=\frac{\frac{17260}{17}\times \frac{88240}{119}}{\frac{17260}{17}+\frac{88240}{119}}=\frac{1121900}{2831} \Omega \] \[R=R_{15678ABC}=\frac{8440}{17}+\frac{1121900}{2831}+275=\frac{56200865}{48127}\Omega\] \subsubsection{Výpočet Celkového Proudu I)} \[I=\frac{130}{\frac{56200865}{48127}}=\frac{1251302}{11240173} A\] \subsubsection{Výpočet \(I_{R6}\)} \[U_{R567BC}=\frac{1121900}{2831}\times \frac{1251302}{11240173}=\frac{495879800}{11240173} V\] \[I_{R6}=\frac{\frac{495879800}{11240173}}{\frac{88240}{119}}=\frac{210748915}{3542260234} A\] \[\uuline{I_{R6}=59,495mA}\] \subsubsection{Výpočet \(U_{R6}\)} \[U_{R6}=650\times \frac{210748915}{3542260234}=\frac{495879800}{11240173} V\] \[\uuline{U_{R6}=44,1167V}\] %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \newpage%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Úloha druhá} Stanovte napětí U\(_{R6}\) a proud I\(_{R6}\). Použijte metodu Theveninovy věty. \bigskip \begin{center} \begin{tabular}{|c|c|c|c|c|c|c|c|} \hline U\(_{1}\) [V] & U\(_{2}\) [V] & \(R_{1} [\Omega] \) & \(R_{2} [\Omega]\) & \(R_{3} [\Omega]\) & \(R_{4} [\Omega]\) & \(R_{5} [\Omega]\) & \(R_{6} [\Omega]\) \\ \hline 250 & 50 & 525 & 620 & 210 & 415 & 230 & 130 \\ \hline \end{tabular} \end{center} \begin{figure}[h] \centering \includegraphics[scale=0.5]{img/u2d1.eps} \caption{Schéma zapojení s vyznačenými smyčkovými proudy} \end{figure} \subsection{Tvorba rovnic} Nejprve musíme odpojit odpor \(R_6\) a nalézt \(R_i\) a \(U_i\). Vytvořit Theveninovi náhradní schema. využijeme zde smyčkových proudů \(I_a\) a \(I_b\). \subsubsection{Hledání \(R_i\) a \(U_i\)} \begin{figure}[h] \centering \includegraphics[scale=0.5]{img/u2d2.eps} \caption{Takto bude vypadat obvod, se kterým budeme pracovat} \end{figure} Spočítáme \(R_i\) celkový odpor vůči odpojenému odporu \(R_6\) \[R_i=\frac{\left(\frac{R_1 \times R_2}{R_1+R_2}+R_3+R_4\right) \times R_5}{\frac{R_1\times R_2}{R_1+R_2}+R_3+R_4 + R_5}\] \bigskip Vyvtoříme rovnice pro smyčkové proudy. \[R_1 \times I_a+(I_a - I_b) \times R_2 = U_1 - U_2\] \[R_2 \times (I_b-I_a)+R_3 \times I_b+R_4 \times I_b+R_5 \times I_b=U_2\] \[U_i=U_5=R_5 \times I_b\] \subsubsection{Náhradní schema a hledání napětí U\(_{R6}\) a proud I\(_{R6}\)} \begin{figure}[h] \centering \includegraphics[scale=0.5]{img/u2d3.eps} \caption{Náhradní schema Theveninova zapojení} \end{figure} Podle Ohmova zákona spočítáme \(I_{R6} \) a \(U_{R6}\) \[I_{R6}=\frac{U_I}{(R_i+R_6)}\] \[U_{R6}=R_6 \times I_{R6}\] \subsection{Dosazení a výpočet rovnic} \[R_i=\frac{\left(\frac{525 \times 620}{525+620}+210+415\right) \times 230}{\frac{525\times 620}{525+620}+210+415 + 230}\] \[R_i=\frac{9578350}{52179} \Omega\] \[525 \times I_a+(I_a - I_b) \times 620 = 250 - 50\] \[620 \times (I_b-I_a)+210 \times I_b+415 \times I_b+230 \times I_b=50\] \bigskip \[I_a=\frac{200+620 \times I_b}{1145}\] \[620 \times (I_b-\frac{200+620 \times I_b}{1145})+210 \times I_b+415 \times I_b+230 \times I_b=50\] \[I_b=\frac{50+\frac{24800}{229}}{\frac{260865}{229}}=\uuline{\frac{7250}{52179}} A\] \[U_i=U_5=230\times \frac{7250}{52179}=\frac{1667500}{52179} V \] \bigskip \[I_{R6}=\frac{\frac{1667500}{52179}}{\left(\frac{9578350}{52179} +130\right)}\] \[U_{R6}=R_6 \times I_{R6}\] \bigskip \[\uuline{I_{R6}=101,9153mA}\] \[\uuline{U_{R6}=13,2489V}\] %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \newpage%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Úloha třetí} Stanovte napětí \(U_{R4}\) a proud \(I_{R4}\). Použijte metodu uzlových napětí. \bigskip \begin{center} 0\begin{tabular}{|c|c|c|c|c|c|c|c|} \hline \(U_1\) [V] & \(U_2\) [V] & \(I\) [A] & \(R_1\) [ \(\Omega\) ] & \(R_2\) [ \(\Omega\) ] & \(R_3\) [ \(\Omega\) ] &\(R_4\) [ \(\Omega\) ] & \(R_5\) [ \(\Omega\) ] \\ \hline 150 & 120 & 0.80 & 490 & 450 & 610 & 340 & 340 \\ \hline \end{tabular} \end{center} \subsection{Tvorba rovnic pro uzlová napětí} Musíme stanovit referenční uzel, po té určit směry proudů a podle nich vytvořit rovnice uzlových napětí. \subsubsection{Obecné rovnice} \[I_{R1}+I_{R3}=I+I_{R2} \] \[I=I_{R3}+I_{R4}+I_{R5} \] \subsubsection{Rovnice pro jednotlivé proudy} Jednotlive proudy podle Ohmova zákona. Proud se rovná součtu mapětí ve větvy dělený potřebným odporem. \[I_{R1}=- \frac{U_A-U_1}{R_1} \] \[I_{R2}=\frac{U_A}{R/2} \] \[I_{R3}=- \frac{U_A-U_B}{R_3} \] \[I_{R4}=\frac{U_B}{R_4} \] \[I_{R5}=\frac{U_B-U-2}{R_2} \] \subsubsection{Dosazení jednotlivých proudů} \[\frac{U_A-U_1}{R_1}+\left(- \frac{U_A-U_B}{R_3} \right)=I+\frac{U_A}{R_2}\] \[I=- \frac{U_A-U_B}{R_3} + \frac{U_B}{R_4}+\frac{U_B-U_2}{R_2} \] \subsubsection{Výpočet \(U_{R4}\) a \(I_{R4}\)} \[-\frac{U_A-U_1}{R_1}+\left(- \frac{U_A-U_B}{R_3} \right)=I+\frac{U_A}{R_2}\] \[I=- \frac{U_A-U_B}{R_3} + \frac{U_B}{R_4}+\frac{U_B-U_2}{R_5} \] nyni si vyjadrime si \[U_A\] a dosadime ho do druhe rovnice. \bigskip \[U_{R4}=U_B\] \[I_{R4}=\frac{U_{R4}}{R_4}\] \subsection{Dosazení a výpočet} \[-\left(\frac{11}{2989} \right) \times U_A+\frac{15}{49}+\left(\frac{1}{610}\right) \times U_B = 0.8+\left(\frac{1}{450} \right) \times U_A\] \[0.8 = - \left( \frac{1}{610}\right) \times U_A+ \left( \frac{39}{5185}\right) \times U_b-\frac{6}{17}\] \[U_A = -43,74960679 V\] \[I_{R4}=\frac{143,7468806}{340}\] \[\uuline{U_{R4}=U_B = 143,7468 V}\] \[\uuline{I_{R4}=422,7849mA}\] %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \newpage%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Úloha čtvrtá} Pro napájecí napětí platí: \(u= U \times sin(2 \pi f t)\). Ve vztahu pro napětí na kondenzítoru \(u_C=U_C \times sin(2 \pi f t + \varphi_C)\) určete \(\left|U_C\right|\) a \(\varphi_C.\) Použijte metodu zjednodušování obvodu \bigskip {\begin{center} \begin{tabular}{|c|c|c|c|c|c|c|} \hline \(U\) [V] & \(R_1\) [\(\Omega\)] & \(R_2\) [\(\Omega\)] & \(L_1\) [mH] &\(L_2\) [mH] & \(C\) [\(\mu\)F] & \(f\) [Hz] \\ \hline 45 & 145 & 165 & 430 & 500 & 315 & 50 \\ \hline \end{tabular} \end{center} \subsection{Vytváření rovnic} Nejprve spořítáms celkovu impedanci obvodu a z ní celkový proud. poté pomocí Ohmova zákona Zjistíme napětí na cívce \(U_C\) \[\omega = 2 \pi f\] \subsubsection{Jednotlivé impedance} \[X_C = -j \frac{1}{\omega C}\] \[X_{L1}=j \omega L_1\] \[X_{L2}=j \omega L_2\] \subsubsection{Zjednodušování obvodu a celková impedance} \[Z_{R1L2}=R_1+X_{L2}\] \[Z_{R1L2C}=\frac{Z_{R1L2}\times X_C}{Z_{R1L2}+X_C}\] \[Z=R2+X_{L1}+X_{L2}\] \subsubsection{Výpočet celkového proudu, napětí \(U_C\) a úhel \(\varphi_C\)} \[i=\frac{U}{Z}\] \[U_{ZR1L2C}=Z_{R1L2C}\times i\] \[U_C=U_{ZR1L2C}\] \[\varphi_C=arctan\left(\frac{U_{C(imaginarni)}}{U_{C(realna)}}\right)*\frac{180}{\pi}\] \subsection{Dosazení do vzorců a výpočet konkrétních hodnot} \[X_C=-\frac{1}{100\pi}j\] \[X_{L1}=43\pi j \Omega\] \[X_{L2}=50\pi j \Omega\] \[Z_{R1L2}=145+50\pi j \Omega\] \[Z_{R1L2C}=\frac{145+50\pi j \left( -\frac{1}{100\pi} \right)}{145+50\pi j + \left( -\frac{1}{100\pi} \right)}\] \[Z_{R1L2C}=0,3473499708-10,45715581j \Omega\] \[Z=145+0,3473499708-10,45715581j+50\pi j \Omega\] \[Z=165,3473500+123,6313283j \Omega\] \[i=\frac{54}{165,3473500+123,6313283j}A\] \[i=0,1735516610-,1308153656j A\] \[U_{ZR1L2C}=0,3473499708-10,45715581j \times 0,1735516610-,1308153656j V\] \[U_C=U_{ZR1L2C}=-1,307673496-1,860295473j V\] \[\left|U_{ZR1L2C}\right|=\sqrt{-1,307673496^2-1,860295473j^2}\] \[\varphi_C = arctan\left(\frac{-1,860295473}{-1,307673496} \right)*\frac{180}{\pi}\] \[\uuline{\left|U_{C}\right|=2,2739V}\] \[\uuline{\varphi_C =58,1093^\circ}\] %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \newpage%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Úloha pátá} Pro napájecí napětí platí: \(u_1= U_1 \times sin(2 \pi f t)\) a \(u_2= U_2 \times sin(2 \pi f t)\). Ve vztahu pro napětí na kondenzítoru \(u_C=U_C \times sin(2 \pi f t + \varphi_{L_2})\) určete \(\left|U_{L2}\right|\) a \(\varphi_{L2}.\) Použijte metodu smyčkových proudů. \bigskip {\begin{center} \begin{tabular}{|c|c|c|c|c|c|c|c|c|} \hline \(U_2\) [V] & \(U_1\) [V] & \(R_1\) [\(\Omega\)] & \(R_2\) [\(\Omega\)] & \(L_1\) [mH] &\(L_2\) [mH] & \(C_1\) [\(\mu\)F] & \(C_2\) [\(\mu\)F] & \(f\) [Hz] \\ \hline 35 & 55 & 125 & 140 & 120 & 100 & 200 & 105 & 70 \\ \hline \end{tabular} \end{center} \subsection{Vytváření rovnic} Nejprve vytvorime jednodlive impedance, pote vytvorime smyckove proudy \[\omega = 2 \times \pi \times f)\] \subsubsection{Impedance} \[X_{L1} = j\omega \times L_1\] \[X_{L2} = j\omega \times L_2\] \[X_{C1} = \frac{-j}{\omega \times C_1}\] \[X_{C2} = \frac{-j}{\omega \times C_2}\] \subsubsection{Smyčkové proudy} \[X_{C1} \times I_a+X_{L2} \times \left( I_a-I_c \right) +R1 \times \left( I_a-I_b \right) +X_{L1} \times \left( I_a-I_b \right) = -U_2\] \[X_{L1} \times \left( I_b-I_a \right) +R_1 \times \left( I_b-I_a \right) +X_{C2} \times \left( I_b-I_c \right) = U_1\] \[X_{C2} \times \left( I_c-I_b \right) +X_{L2} \times \left( I_c-I_a \right) +R_2 \times I_c = 0\] \bigskip Nyní máme smyčkové proudy, do kterých pozdeji dosadíme a jednotlivě spočítáme. \subsubsection{Úhel \(\varphi\) a napětí\(U_{L2}\) na \(L_2\)} \[U_{L2} = X_{L2} \times (I_a-I_c)\] \subsection{Dosazení a výpočet} \[\omega = 439,8229716\] \[X_{L1} = 0,05277875659 j\] \[X_{L2} = 43,98229716 j\] \[X_{C1} = -11,36821022 j\] \[X_{C2} = -21,65373375 j\] \[-11,36821022 j I_a + 43,98229716 j \left(I_a - Ic\right) + 125 I_a - 125 I_b + 0,05277875659 j \left(I_a - I_b\right) = -55\] \[0,05277875659 j \left(I_b - I_a\right) + 125 I_b - 125 I-a - 21,65373375 j \left(I_b - I_c\right) = 35\] \[-21,65373375 j \left(I_c - I_b\right) + 43,98229716 j \left(I_c - I_a\right) + 140 I_c = 0\] \[I_a = -0,5646719983 + 1,631688462 j\] \[I_b = -0,5665550178 + 1,581244317 j \] \[I_c = -0,2753526961 - 0,04585221415 j\] \[U_{L2} = 43,98229716 j \times (-0,5646719983 + 1,631688462 j\-\left(-0,2753526961 - 0,04585221415 j\right)\] \[U_{L2} =-73,78209251-12,72492752 j\] \[\left|U_{L2}\right| = \sqrt{-73,78209251^2+ 12,72492752^2 j}\] \[\varphi=arctan \left(\frac{12,72492752}{73,78209251} \times \frac{180}{\pi} \right)\] \bigskip \subsubsection{Výsledné hodnoty} \[\uuline{\left|U_{L2}\right| =74,8713 V }\] \[\uuline{\varphi=9,7853^\circ }\] %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \newpage%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Úloha šestá} Sestavte diferenciální rovnici popisující chování obcodu na obrázku, dále ji upravte dosazením hodnot parametrů. Vypočítejte analytické řešení \(i_L = f(t)\). Proveďte kontrolu výpočtu dosazením do sestavené diferenciální rovnice. \bigskip {\begin{center} \begin{tabular}{|c|c|c|} \hline \(L\) [H] & \(R\) [\( \Omega\)] & \(i_L(0)\) [A] \\ \hline 10 & 20 & 3 \\ \hline \end{tabular} \end{center} \bigskip \subsection{reseni diferecnialni rovnice} \[i_L 0=3\] \[i_L (\infty)=0\] \[i_{partikular}=i(\infty)\] \subsubsection{Rovnice pro smyčkový proud v obvodu z které budeme vycházet} \[s_1 : R_1 I_1 + \frac{di}{dt}=0\] \[\lambda L + R=0\] \subsubsection{Obecné a partikulátní řešení} \[\lambda = -\frac{R}{L}\] \[i_{Lo} =Ke^{\lambda t}\] \[i_{p} = 0\] \subsubsection{Diferenciální rovnice} \[i_L (t) = i_{Lo}+i_{Lp}=Ke^{\lambda t}+0\] \[i_L (t) = i_{Lo}+i_{Lp}=Ke^{-\frac{R}{L} t}+0\] \subsubsection{(t=0) pro výpočet K} \[i_L(t)=Ke^{\lambda 0} =K \times 1=K\] \[i_L(0)=4\] \[K=4\] \subsubsection{dosazení hodnot a zkouška} \[i_L (0)=Ke^{-\frac{R}{L} t}+0=3 \times e^{-\frac{20}{10}t}\] \[\uuline{i_L (0)=3 \times e^{-2t}}\] \bigskip \[R \times i_L + L\frac{di}{dt}=0\] \[20 \times 3 \times e^{-\frac{7}{3}t}-10 \times\frac{2 \times 3 \times e^{-2t}}{1}=0\] \[60 \times e^{2t}=60 \times e^{2t} \Rightarrow 0=0\] %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \newpage%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Tabulka s výsledky} \begin{center} \begin{tabular}{|c|c||c|c|} \hline Úloha & Var. & \multicolumn{2}{|c|}{Výsledky}\\ \hline \hline 1 & G & \(U_{R6}\)& \(I_{R6}\) \\ & & \(44,1167V\) & \(59,495mA\) \\ \hline 2 & A & \(U_{R6}\) & \(I_{R6}\) \\ & & \(13,2489V\) & \(101,9153mA\) \\ \hline 3 & B & \(U_{R4}\) & \(I_{R4}\) \\ & & \(143,7468 V\)& \(422,7849mA\) \\ \hline 4 & G & \(\left| U_C \right|\)& \(\varphi_C\) \\ & & \(2,2739V\) & \(58,1093^\circ \)\\ \hline 5 & A & \(\left| U_{L2} \right|\)& \(\varphi_{L2}\) \\ & & \(74,8713 V\) & \(9,7853^\circ\)\\ \hline 6 & B & \multicolumn{2}{|c|}{Rovnice} \\ & & \multicolumn{2}{|c|}{ \(i_L (0)=3 \times e^{-2t}\)} \\ \hline \end{tabular} \end{center} \end{document}