generated from kenryuS/report-temp
316 lines
14 KiB
TeX
316 lines
14 KiB
TeX
\section{考察}
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\subsection{実験1}
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\cref{fig:v-i-r}より測定値はすべて$\pm 5\ \%$の抵抗値の誤差に収まっている．
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理想値の曲線は$I = \frac{V}{R}$なので測定値はオームの法則(\cref{equ:ohm})に従っているといえる．
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\subsubsection{抵抗器の制限について}
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オームの法則は実験で使用した抵抗器よりも低い抵抗値を持つものでも成り立つはずだが定格電力の制限で手順書では使用しなかった．
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試しに，実験で使用した抵抗値が$\frac{1}{10}$で定格電力が1/4 Wの抵抗を用いて同じ実験を行なった場合を考える．
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オームの法則により，3，6，9 Vでの電流と電力は\cref{tab:v-i-r-tenth}となる．
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\begin{table}[!ht]
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\centering
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\caption{Current and Power of Resistors with tenth of resistance}
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\label{tab:v-i-r-tenth}
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\begin{tabular}{c|r|r|r|r|r|r}
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\hline
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\multirow{2}{5em}{Voltage $[\text{V}]$} & \multicolumn{2}{c|}{$100 \ \Omega$} & \multicolumn{2}{c|}{$220 \ \Omega$} & \multicolumn{2}{c}{$330 \ \Omega$} \\
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\cline{2-7}
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& Current (mA) & Power (W) & Current (mA) & Power (W) & Current (mA) & Power (W) \\
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\hline
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3 & 30 & 0.09 & 13.64 & 0.041 & 9.09 & 0.027 \\
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6 & 60 & 0.36 & 27.27 & 0.16 & 18.18 & 0.11 \\
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9 & 90 & 0.81 & 40.91 & 0.37 & 27.27 & 0.25 \\
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\hline
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\end{tabular}
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\end{table}
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一部で1/4 = 0.25 W を超過してしまう条件がある．
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これらの値は理想的な抵抗を使用した場合なので現実ではかろうじて超過しなかったり，僅かながら超える条件があるだろう．
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抵抗器はその性質上，電力の一部を熱に変換して発熱しながら電流を制限する．
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定格電力を超えての使用は抵抗器が焼損・破裂する可能性があるので注意すること\supercite{resistor-overload-example}．
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\subsection{実験2}
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実験結果\cref{tab:exp2-res1}，\cref{tab:exp2-res2}より接点(b)での電流の総和はそれぞれ\cref{tab:current-in-b}となった．
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\begin{table}[!ht]
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\centering
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\caption{Applying Kirchhoff's Current Law at Point (b) in each Circuit}
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\label{tab:current-in-b}
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\begin{tabular}{c|c}
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\hline
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Circuit & Current $[\text{mA}]$ \\
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\hline
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(a) & 0.01 \\
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(b) & 0.80 \\
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\hline
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\end{tabular}
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\end{table}
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また，各回路の閉路abef，bcde，acdfでの電圧の和は\cref{tab:voltage-in-loops}となった．
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\begin{table}[!ht]
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\centering
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\caption{Applying Kirchhoff's Voltage Law on each Loop}
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\label{tab:voltage-in-loops}
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\begin{tabular}{c|c|c}
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\hline
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Loop & Circuit (a) $[\text{V}]$ & Circuit (b) $[\text{V}]$ \\
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\hline
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abef & 0.01 & 0.01 \\
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bcde & -0.005 & 0.007 \\
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acdf & 0.005 & -0.017 \\
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\hline
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\end{tabular}
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\end{table}
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これらから，実験回路はおおかたキルヒホッフの法則に従っているといえる．
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しかし，回路(b)の電流則と閉路acdfでは真の値である0からかなり離れてしまった．
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これには2つの実験回路での測定方法の差異や測定機器・抵抗値の誤差などが考えられる．
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\subsection{実験3}
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実験結果\cref{tab:exp3-res1}・\cref{tab:exp3-res2}と\cref{fig:cd-exp2-a}より，電流の向きに注意しながら重ね合わせると
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\begin{align}
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V_{R_1} &= 12.40 \ \text{V} - 1.71 \ \text{V} = 10.69 \ \text{V} \\
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V_{R_2} &= 2.61 \ \text{V} - 1.29 \ \text{V} = 1.32 \ \text{V} \\
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V_{R_3} &= 2.59 \ \text{V} + 1.70 \ \text{V} = 4.29 \ \text{V} \\
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I_{R_1} &= 3.81 \ \text{mA} - 0.52 \ \text{mA} = 3.29 \ \text{mA} \\
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I_{R_2} &= -2.61 \ \text{mA} + 1.32 \ \text{mA} = -1.29 \ \text{mA} \\
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I_{R_3} &= 1.19 \ \text{mA} + 0.78 \ \text{mA} = 1.97 \ \text{mA}
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\end{align}
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それぞれの誤差率は\cref{tab:exp3-exp2-diff}の通りである．
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\begin{table}[!ht]
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\centering
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\caption{Percentage Difference of Experiment \# 3 from Experiment \# 2 on Circuit (a)}
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\label{tab:exp3-exp2-diff}
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\begin{tabular}{c|c}
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\hline
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Measurement & Difference $(\%)$ \\
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\hline
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$V_{R_1}$ & 0 \\
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$V_{R_2}$ & +1.54 \\
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$V_{R_3}$ & -0.23 \\
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$I_{R_1}$ & +0.30 \\
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$I_{R_2}$ & -1.53 \\
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$I_{R_3}$ & -0.51 \\
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\hline
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\end{tabular}
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\end{table}
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この誤差は前節の測定方法の差異が影響していると思われる．
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\subsubsection{電源の除去法について}
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この重ね合わせの理を適用する際の電源の除去法は電圧源と電流源で違ってくる．
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電圧源は短絡除去，電流源は開放除去である．
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これは\cref{fig:v-c-s-removal}のように電圧源は直列接続，電流源は並列接続であるため，電圧・電流をなくし，抵抗値を変化させないような除去を行なっている．
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\begin{figure}[tbh]
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\centering
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\begin{minipage}[h]{0.9\linewidth}
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\centering
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\begin{circuitikz}
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\draw (0,0) to [battery1={$E$},invert] ++(0,2) to [R={$R_i$}] ++(0,2);
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\draw (0,0) to [short, -o] ++(2,0);
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\draw (0,4) to [short, -o] ++(2,0);
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\draw (3.25,2) node {\Huge $\Rightarrow$};
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\draw (5,0) -- (5,2) to [R={$R_i$}] ++(0,2);
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\draw (5,0) to [short, -o] ++(2,0);
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\draw (5,4) to [short, -o] ++(2,0);
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\end{circuitikz}
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\subcaption{Voltage Source}
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\label{fig:vs-removal}
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\end{minipage}
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\begin{minipage}[h]{0.9\linewidth}
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\centering
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\begin{circuitikz}
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\draw (0,0) to [isourceAM={$I$}] ++(0,2);
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\draw (2,0) to [R={$R_i$}] ++(0,2);
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\draw (0,0) to [short, -*] ++(2,0) to [short, -o] ++(2,0);
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\draw (0,2) to [short, -*] ++(2,0) to [short, -o] ++(2,0);
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\draw (5,1) node {\Huge $\Rightarrow$};
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\draw (8,0) to [R={$R_i$}] ++(0,2);
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\draw (6,0) to [short, o-*] ++(2,0) to [short, -o] ++(2,0);
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\draw (6,2) to [short, o-*] ++(2,0) to [short, -o] ++(2,0);
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\end{circuitikz}
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\subcaption{Current Source}
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\label{fig:cs-removal}
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\end{minipage}
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\caption{Removal of Voltage and Current Source}
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\label{fig:v-c-s-removal}
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\end{figure}
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\subsection{実験4}
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実験結果\cref{tab:exp4-res}から誤差率\cref{tab:exp4-diff}を求める．
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\begin{table}[!ht]
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\centering
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\caption{Percentage Difference of Original and Equivalent Circuit of Experiment \# 4}
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\label{tab:exp4-diff}
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\begin{tabular}{c|c}
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\hline
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Measurement & Difference $[\%]$ \\
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\hline
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Voltage & -0.73 \\
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Current & -0.38 \\
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\hline
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\end{tabular}
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\end{table}
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比較的小さな誤差に収まったが，可変抵抗の抵抗値が少し触れるだけで変化してしまうため設定が難しく，誤差が出てしまった．
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\subsubsection{テブナンの定理の証明}
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\cref{fig:thevenin-proof-open-circuit}のような回路$N$を考える．
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この回路には複数の電圧源・電流源があり内部インピーダンスは$Z_0$である．
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そして，この回路の開放電圧を$V_0$とする．
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次に\cref{fig:thevenin-proof-load}のように負荷インピーダンス$Z_L$を接続する．
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この時，回路には電流$I$が流れる．
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そして\cref{fig:thevenin-proof-ec}を考える．
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この回路は負荷インピーダンスだけでなく，$V_0$と同じ電圧を持つ2つの電源を互いに打ち消し合うように接続する．
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重ね合わせの理を適用して\cref{fig:thevenin-proof-d1}と\cref{fig:thevenin-proof-d2}のように電圧源$V_1$，$V_2$をそれぞれ独立させる．
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電流$I$は$I_1$と$I_2$の和として表わせれる．
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\cref{fig:thevenin-proof-d1}では点Aでの電位が等しいため電流が流れない，よって$I_1 = 0$である．
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\cref{fig:thevenin-proof-d2}では回路Nの電圧源を短絡，電流源を開放して内部インピーダンス$Z_0$を得る．
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この時，$I_2$はオームの法則より\cref{equ:thevenin-proof-d2-I2}で表わせれる．
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\begin{equation}\label{equ:thevenin-proof-d2-I2}
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I_2 = \frac{V_2}{Z_0 + Z_L} = \frac{V_0}{Z_0 + Z_L}
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\end{equation}
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結果的に$I_1$と$I_2$の和である電流$I$は$0 + \frac{V_0}{Z_0 + Z_L}$で\cref{equ:thevenin}が得られる．
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\cref{fig:thevenin-proof-d2}の回路を変形し電圧源となる部分を抜き出したのが\cref{fig:thevenin-proof-evs}である\supercite{ac-theory:thevenin}．
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\begin{figure}[tbh]
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\centering
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\begin{minipage}[h]{0.45\linewidth}
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\centering
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\begin{circuitikz}
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\draw (0,0) node[fourport] (N) {$N$};
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\draw ($(N.center) + (0,-0.5)$) node[vsourceAMshape,scale=0.5,rotate=180](Vi){};
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\draw ($(N.center) + (0,0)$) node[isourceAMshape,scale=0.5](Ci){};
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\draw (Vi.right) -- ++(-0.25,0);
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\draw (Vi.left) -- ++(0.25,0);
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\draw (Ci.left) -- ++(-0.25,0);
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\draw (Ci.right) -- ++(0.25,0);
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\draw ($(N.center) + (0,0.25)$) node[above] {$Z_0$};
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\draw (N.port3) to [short, -o] ++(1,0) node[above]{A} coordinate (A);
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\draw (N.port2) to [short, -o] ++(1,0) node[right]{B} coordinate (B);
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\draw[->] ($(B) + (0,0.1)$) -- ($(A) + (0,-0.1)$);
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\draw ($(A)!0.5!(B)$) node[right]{$V_0$};
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\end{circuitikz}
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\subcaption{Open Circuit}
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\label{fig:thevenin-proof-open-circuit}
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\end{minipage}
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\begin{minipage}[h]{0.45\linewidth}
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\centering
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\begin{circuitikz}
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\draw (0,0) node[fourport] (N) {$N$};
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\draw ($(N.center) + (0,-0.5)$) node[vsourceAMshape,scale=0.5,rotate=180](Vi){};
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\draw ($(N.center) + (0,0)$) node[isourceAMshape,scale=0.5](Ci){};
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\draw (Vi.right) -- ++(-0.25,0);
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\draw (Vi.left) -- ++(0.25,0);
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\draw (Ci.left) -- ++(-0.25,0);
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\draw (Ci.right) -- ++(0.25,0);
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\draw ($(N.center) + (0,0.25)$) node[above] {$\dot{Z_0}$};
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\draw (N.port3) to [short, -o, i={$I$}] ++(1,0) node[above]{A} coordinate (A);
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\draw (N.port2) to [short, -o] ++(1,0) node[right]{B} coordinate (B);
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\draw (A) -- ++(1,0) to [R={$Z_L$}] ++(0,-2) -- ++(-1,0) -- (B);
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\end{circuitikz}
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\subcaption{Connected to Load}
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\label{fig:thevenin-proof-load}
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\end{minipage}
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\begin{minipage}[h]{0.45\linewidth}
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\centering
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\begin{circuitikz}
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\draw (0,0) node[fourport] (N) {$N$};
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\draw ($(N.center) + (0,-0.5)$) node[vsourceAMshape,scale=0.5,rotate=180](Vi){};
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\draw ($(N.center) + (0,0)$) node[isourceAMshape,scale=0.5](Ci){};
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\draw (Vi.right) -- ++(-0.25,0);
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\draw (Vi.left) -- ++(0.25,0);
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\draw (Ci.left) -- ++(-0.25,0);
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\draw (Ci.right) -- ++(0.25,0);
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\draw ($(N.center) + (0,0.25)$) node[above] {$\dot{Z_0}$};
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\draw (N.port3) to [short, -o, i={$I$}] ++(1,0) node[above]{A} coordinate (A);
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\draw (N.port2) to [short, -o] ++(1,0) node[right]{B} coordinate (B);
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\draw (A) to [battery1,l={$V_1=V_0$}] ++(1,0) to [R={$Z_L$}] ++(0,-2) to [battery1,l={$V_2=V_0$},invert] ++(-1,0) -- (B);
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\end{circuitikz}
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\subcaption{Equivalent Circuit}
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\label{fig:thevenin-proof-ec}
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\end{minipage}
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\begin{minipage}[h]{0.45\linewidth}
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\centering
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\begin{circuitikz}
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\draw (0,0) node[fourport] (N) {$N$};
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\draw ($(N.center) + (0,-0.5)$) node[vsourceAMshape,scale=0.5,rotate=180](Vi){};
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\draw ($(N.center) + (0,0)$) node[isourceAMshape,scale=0.5](Ci){};
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\draw (Vi.right) -- ++(-0.25,0);
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\draw (Vi.left) -- ++(0.25,0);
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\draw (Ci.left) -- ++(-0.25,0);
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\draw (Ci.right) -- ++(0.25,0);
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\draw ($(N.center) + (0,0.25)$) node[above] {$\dot{Z_0}$};
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\draw (N.port3) to [short, -o, i={$I_1$}] ++(1,0) node[above]{A} coordinate (A);
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\draw (N.port2) to [short, -o] ++(1,0) node[right]{B} coordinate (B);
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\draw (A) to [battery1,l={$V_1=V_0$}] ++(1,0) to [R={$Z_L$}] ++(0,-2) -- ++(-1,0) -- (B);
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\end{circuitikz}
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\subcaption{Decomposition 1}
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\label{fig:thevenin-proof-d1}
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\end{minipage}
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\begin{minipage}[h]{0.45\linewidth}
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\centering
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\begin{circuitikz}
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\draw (0,0) node[fourport] (N) {$N$};
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\draw ($(N.center) + (0,-0.5)$) node[shortshape,scale=0.5,rotate=180](Vi){};
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\draw ($(N.center) + (0,0)$) node[openshape,scale=0.5](Ci){};
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\draw (Vi.right) -- ++(-0.25,0);
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\draw (Vi.left) -- ++(0.25,0);
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\draw (Ci.left) to [short, o-] ++(-0.25,0);
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\draw (Ci.right) to [short, o-] ++(0.25,0);
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\draw ($(N.center) + (0,0.25)$) node[above] {$\dot{Z_0}$};
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\draw (N.port3) to [short, -o, i={$I_2$}] ++(1,0) node[above]{A} coordinate (A);
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\draw (N.port2) to [short, -o] ++(1,0) node[right]{B} coordinate (B);
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\draw (A) -- ++(1,0) to [R={$Z_L$}] ++(0,-2) to [battery1,l={$V_2=V_0$},invert] ++(-1,0) -- (B);
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\end{circuitikz}
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\subcaption{Decomposition 2}
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\label{fig:thevenin-proof-d2}
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\end{minipage}
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\begin{minipage}[h]{0.45\linewidth}
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\centering
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\begin{circuitikz}
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\draw (0,0) to [battery1={$V_0$},invert] ++(0,1.5) to [R={$Z_0$}] ++(0,1.5);
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\draw (0,3) to [short, -o] ++(2,0) node[below]{A};
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\draw (0,0) to [short, -o] ++(2,0) node[above]{B};
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\end{circuitikz}
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\vspace{2em}
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\subcaption{Thevenin's Equivalent Voltage Source}
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\label{fig:thevenin-proof-evs}
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\end{minipage}
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\caption{}
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\end{figure}
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