60 lines
1.2 KiB
TeX
60 lines
1.2 KiB
TeX
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\subsection{Problem 15}
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Find the currents in the circuit, assuming all resistances are 10 Ohm.
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\begin{figure}[h]
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\centering
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\includegraphics[width=0.5\textwidth]{images/electro.png}
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\label{fig:electro}
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\end{figure}
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\subsubsection*{Solution}
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\begin{equation*}
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\left\{\begin{matrix}
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I_{L1}(R_1 + R_2) - I_{L3}(R_2) = -E_2\\
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I_{L2}(R_3 + R_4) - I_{L3}(R_3) = E_2\\
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-I_{L1}(R_2) - I_{L2}(R_3) + I_{L3}(R_2 + R_3) = -E_1
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\end{matrix}\right.
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\end{equation*}
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After substitution we have:
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\begin{equation*}
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\begin{amatrix}{1}{1}
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\matr{A} & \matr{b}
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\end{amatrix} =
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\begin{amatrix}{3}{1}
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20 & 0 & -10 & -10 \\
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0 & 20 & -10 & 10 \\
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-10 & -10 & 20 & -20
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\end{amatrix}
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\xrightarrow[magic]{}
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\begin{amatrix}{3}{1}
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1 & 0 & 0 & -1.5 \\
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0 & 1 & 0 & -0.5 \\
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0 & 0 & 1 & -2
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\end{amatrix}
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\end{equation*}
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\begin{equation*}
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\left\{\begin{matrix}
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I_1 = -I_{L3} \\
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I_2 = -I_{L1} + I_{L2} \\
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I_3 = I_{L1} - I_{L3} \\
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I_4 = -I_{L1} \\
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I_5 = I_{L2} - I_{L3} \\
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I_6 = -I_{L2}
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\end{matrix}\right.
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\end{equation*}
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\begin{equation*}
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\left\{\begin{matrix}
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I_1 = 2 \\
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I_2 = 1 \\
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I_3 = 0.5 \\
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I_4 = 1.5 \\
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I_5 = 1.5 \\
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I_6 = 0.5
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\end{matrix}\right.
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\end{equation*}
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