回路-09-4
RC直列交流回路を真面目に解く - 電圧を主体に考える
RとCが直列に並んだ交流回路を電圧を主体にまじめに考えます．式は，
\(\Large \displaystyle V = V_0 \cdot e^{j \omega t} = V_R (t)+ V_C (t) \)
\(\Large \displaystyle V_R (t) = R \cdot I(t) \)
\(\Large \displaystyle I(t) = C \cdot \frac{d}{dt} V_C (t) \)
となるので，
\(\Large \displaystyle V = R \cdot C \cdot \frac{d}{dt} V_C + V_C \)
となります．したがって，
\(\Large \displaystyle R \cdot C \cdot \frac{d}{dt} V_C (t) + V_C (t) = V_0 \cdot e^{j \omega t} \)
\(\Large \displaystyle \frac{d}{dt} V_C (t) =- \frac{1}{RC} V_C (t) + \frac{1}{RC}V_0 \cdot e^{j \omega t} \)
定数変化法を用いて，
\(\Large \displaystyle V_C (t) = A_0 \cdot e^{ - \frac{1}{RC} t} \)
\(\Large \displaystyle \frac{d}{dt} V_C (t) = A_0' \cdot e^{ - \frac{1}{RC} t} \color{blue}{- \frac{1}{RC}A_0 \cdot e^{ - \frac{1}{RC} t}} \)
\(\Large \displaystyle = \color{blue}{- \frac{1}{RC} V_C (t) } + \frac{1}{RC}V_0 \cdot e^{j \omega t} \)
\(\Large \displaystyle A_0' \cdot e^{ - \frac{1}{RC} t} = \frac{1}{RC}V_0 \cdot e^{j \omega t} \)
\(\Large \displaystyle A_0' = \frac{1}{RC}V_0 \cdot e^{(\frac{1}{RC} + j \omega) t} \)
\(\Large \displaystyle A_0 = \frac{1}{RC}V_0 \cdot \frac{1}{\frac{1}{RC} + j \omega}\cdot e^{(\frac{1}{RC} + j \omega) t} + D \)
\(\Large \displaystyle V_C (t) = \left( \frac{1}{RC}V_0 \cdot \frac{1}{\frac{1}{RC} + j \omega}\cdot e^{(\frac{1}{RC} + j \omega) t} + D\right) \cdot e^{ - \frac{1}{RC} t} \)
\(\Large \displaystyle = \frac{1}{RC}V_0 \cdot \frac{1}{\frac{1}{RC} + j \omega}\cdot e^{ j \omega t} + D \cdot e^{ - \frac{1}{RC} t} \)
\(\Large \displaystyle = V_0 \cdot \frac{1}{1 + j \omega RC}\cdot e^{ j \omega t} + D \cdot e^{ - \frac{1}{RC} t} \)
ここで，初期値は，
\(\Large \displaystyle V_C(0) = V_0; \)
となります．その理由は，
\(\Large \displaystyle R \cdot C \cdot \frac{d}{dt} V_C + V_C = V_0 \ e^{j \omega t} \)
電流の初期値は，I(0)=0，として，
\(\Large \displaystyle I(0) = 0 = C \left.\frac{d}{dt} V_C \right|_{t=0} ; \)
元の微分方程式に戻すと
\(\Large \displaystyle R \cdot C \cdot \left. \frac{d}{dt} V_C \right|_{t=0} + V_C(0) = V_0 \ e^{j \omega \cdot 0} = V_0 \)
となり，したがって，
\(\Large \displaystyle V_C(0) = V_0 \)
となります．
したがって，
\(\Large \displaystyle R \cdot C \cdot \frac{d}{dt} V_C (t) + V_C (t) = V_0 \cdot e^{j \omega t} \)
\(\Large \displaystyle \frac{d}{dt} V_C (t) =- \frac{1}{RC} V_C (t) + \frac{1}{RC}V_0 \cdot e^{j \omega t} \)
\(\Large \displaystyle V_C (t) = \frac{1}{RC}V_0 \cdot \frac{1}{\frac{1}{RC} + j \omega}\cdot e^{ j \omega t} + D \cdot e^{ - \frac{1}{RC} t} \)
\(\Large \displaystyle V_C(0)=V_0; \)
\(\Large \displaystyle V_C(0)=V_0 = \frac{1}{RC}V_0 \cdot \frac{1}{\frac{1}{RC} + j \omega}\cdot e^{ j \omega \cdot 0} + D \cdot e^{ - \frac{1}{RC} \cdot 0} \)
\(\Large \displaystyle = V_0 \cdot \frac{1}{1 + j \omega RC} + D = V_0 \)
\(\Large \displaystyle D = V_0-V_0 \cdot \frac{1}{1 + j \omega RC} \)
\(\Large \displaystyle V_C (t) = \left( V_0 \cdot \frac{1}{1 + j \omega RC}\cdot e^{(\frac{1}{RC} + j \omega) t} +V_0- V_0 \cdot \frac{1}{1 + j \omega RC} \right) \cdot e^{ - \frac{1}{RC} t} \)
\(\Large \displaystyle =V_0 \cdot \frac{1}{1 + j \omega RC} \left( e^{j \omega t} - e^{ - \frac{1}{RC} t}\right) + V_0 \cdot e^{ - \frac{1}{RC} t} \)
\(\Large \displaystyle I(t) = C \cdot \frac{d}{dt} V_C (t) \)
\(\Large \displaystyle =V_0 \cdot \frac{C}{1 + j \omega RC} \left( j \omega \ e^{j \omega t} + \frac{1}{RC} e^{ - \frac{1}{RC} t}\right) - V_0 \frac{1}{RC} e^{ - \frac{1}{R} t} \)
\(\Large \displaystyle =V_0 \cdot \frac{j \omega C}{1 + j \omega RC} e^{j \omega t} + \frac{1}{1 + j \omega RC}\frac{1}{R} e^{ - \frac{1}{RC} t} - V_0 \frac{1}{R} e^{ - \frac{1}{RC} t} \)
\(\Large \displaystyle =V_0 \cdot \frac{j \omega C}{1 + j \omega RC} e^{j \omega t} + V_0 \frac{1 - 1 - j \omega RC}{1 + j \omega RC}\frac{1}{R} e^{ - \frac{1}{RC} t} \)
\(\Large \displaystyle =V_0 \cdot \frac{j \omega C}{1 + j \omega RC} e^{j \omega t} + V_0\frac{ - j \omega C}{1 + j \omega RC} e^{ - \frac{1}{RC} t} \)
\(\Large \displaystyle =V_0 \cdot \frac{j \omega C}{1 + j \omega RC} \left( e^{j \omega t} - e^{ - \frac{1}{RC} t} \right) \)
微分して電流にしてから初期値を入れると，
\(\Large \displaystyle I(t) = C \cdot \frac{d}{dt} V_C (t) \)
\(\Large \displaystyle =V_0 \cdot \frac{j \omega C}{1 + j \omega RC}\cdot e^{ j \omega t} - \frac{D}{R} \cdot e^{ - \frac{1}{RC} t} \)
\(\Large \displaystyle I(0)=0; \)
\(\Large \displaystyle I(0) = V_0 \cdot \frac{j \omega C}{1 + j \omega RC}\cdot e^{ j \omega \cdot 0} - \frac{D}{R} \cdot e^{ - \frac{1}{RC} \cdot 0} \)
\(\Large \displaystyle V_0 \cdot \frac{j \omega C}{1 + j \omega RC} =\frac{D }{R} \)
\(\Large \displaystyle D = V_0 \cdot \frac{j \omega RC}{1 + j \omega RC} \)
\(\Large \displaystyle I(t) = V_0 \cdot \frac{j \omega C}{1 + j \omega RC}\cdot (e^{ j \omega t} - e^{ - \frac{1}{RC} t}) \)
\(\Large \displaystyle = V_0 \cdot \frac{j \omega C}{1+j \omega RC } \left( e^{ j \omega t} - e^{- \frac{1}{RC}t} \right) \)
\(\Large \displaystyle = V_0 \cdot \frac{j \omega C + \omega^2 R C^2}{1+ (\omega RC)^2 } \left( e^{ j \omega t} - e^{- \frac{1}{RC}t} \right) \)
\(\Large \displaystyle = V_0 \cdot \frac{ \frac{j}{\omega C} + R }{\frac{1}{ (\omega C)^2}+ R^2 } \left( e^{ j \omega t} - e^{- \frac{1}{RC}t} \right) \)
となり，一致します．
後の展開は同じですが，きちんと書くと，
\(\Large \displaystyle = V_0 \cdot \frac{R + \frac{j}{ \omega C}}{ \sqrt{R^2 + \frac{1}{ (\omega C)^2 }}} \cdot \frac{1 }{{ \sqrt{R^2 + \frac{1}{ (\omega C)^2 }}}} \cdot \left( e^{ j \omega t} - e^{- \frac{1}{ RC} t} \right) \)
ここで，複素数の項を指数に置き換える．
\(\Large \displaystyle cos \ \theta = \frac{R }{ \sqrt{R^2 + \frac{1}{ (\omega C)^2 }}}\)
\(\Large \displaystyle j \cdot sin \ \theta = \frac{ \frac{j}{ \omega C}}{ \sqrt{R^2 + \frac{1}{ (\omega C)^2 }}}\)
\(\Large \displaystyle e^{ j \ \theta} =cos \ \theta + j \cdot sin \ \theta \)
より，
\(\Large \displaystyle \frac{R + \frac{j}{ \omega C}}{ \sqrt{R^2 + \frac{1}{ (\omega C)^2 }}} = e^{ j \ \theta} \)
ここで，
\(\Large \displaystyle tan \ \theta = \frac{1} {\omega RC} \)
となります．したがって，
\(\Large \displaystyle I(t) = V_0 \cdot \frac{R + \frac{j}{ \omega C}}{ \sqrt{R^2 + \frac{1}{ (\omega C)^2 }}} \cdot \frac{1 }{{ \sqrt{R^2 + \frac{1}{ (\omega C)^2 }}}} \cdot \left( e^{ j \omega t} - e^{- \frac{1}{ RC} t} \right) \)
\(\Large \displaystyle = V_0 \cdot e^{ j \ \theta} \cdot \frac{1 }{{ \sqrt{R^2 + \frac{1}{ (\omega C)^2 }}}} \cdot \left( e^{ j \omega t} - e^{- \frac{1}{ RC} t} \right)\)
\(\Large \displaystyle = V_0 \cdot \frac{1 }{{ \sqrt{R^2 + \frac{1}{ (\omega C)^2 }}}} \cdot \left\{ e^{ j (\omega t + \theta)} - e^{ j \ \theta} \cdot e^{- \frac{1}{ RC} t} \right\}\)
となります．ここで，三角関数に戻すために，虚数部分のみ取り出すと，
\(\Large \displaystyle Im [ I(t) ] = Im \left[V_0 \cdot \frac{1 }{{ \sqrt{R^2 + \frac{1}{ (\omega C)^2 }}}} \cdot e^{ j (\omega t + \theta)} \right]
- Im \left[V_0 \cdot \frac{1 }{{ \sqrt{R^2 + \frac{1}{ (\omega C)^2 }}}} \cdot e^{ j \ \theta} \cdot e^{- \frac{1}{ RC} t} \right] \)
\(\Large \displaystyle = V_0 \cdot \frac{1 }{{ \sqrt{R^2 + \frac{1}{ (\omega C)^2 }}}} \cdot sin (\omega t + \theta)
- V_0 \cdot \frac{1 }{{ \sqrt{R^2 + \frac{1}{ (\omega C)^2 }}}} \cdot sin \ \theta \cdot e^{- \frac{1}{ RC} t} \)
十分な時間が経てば，第二項は0となるので，
\(\Large \displaystyle I(t) \sim \frac{V_0 }{{ \sqrt{R^2 + \frac{1}{ (\omega C)^2 }}}} \cdot sin (\omega t + \theta) \)
\(\Large \displaystyle \theta = tan^{-1} \ \frac{1} {\omega RC} \)
となり，インピーダンスを用いた結果と一致します．
次は，０９－５．RC直列交流回路をラプラス変換で真面目に解く - 電圧を主体に考える，です．
