回路-08-2
・L回路を真面目に解く
\(\Large I(t) = \frac{1}{ L} \int V(t) \ dt = \frac{1}{ L} \int V_0 \ e^{ i \omega t} \ dt = \frac{1}{ j \omega L} V_0 \ e^{ i \omega t} + D\)
\(\Large I(0) = 0 \)
\(\Large I(0) = \frac{1}{ j \omega L} V_0 \ e^{ j \omega \cdot 0} + D \rightarrow D= -\frac{1}{ j \omega L} V_0 \)
\(\Large I(t) = \frac{1}{ j \omega L} V_0 \ ( e^{ j \omega t} - 1)
=
\frac{j}{ \omega L} V_0 \ ( 1 - e^{ i \omega t} )
=
\frac{j}{ \omega L} V_0 - \frac{j}{ \omega L} V_0 \ e^{ j \omega t}
=
\frac{j}{ \omega L} V_0 - \frac{1}{ \omega L} \cdot e^{j \frac{\pi}{2} } V_0 \ e^{ j \omega t} \)
\(\Large I(t) = \frac{j}{ \omega L} V_0 - \frac{1}{ \omega L} \ V_0 \ e^{ j (\omega t+\frac{\pi}{2})} \)
\(\Large Im[I(t)] =
\displaystyle \frac{1}{ \omega L} V_0 - \frac{1}{ \omega L} \ V_0 \ sin \left( \omega t+\frac{\pi}{2} \right)
=
\frac{j}{ \omega L} V_0 \left\{ 1 - sin \left( \omega t+\frac{\pi}{2} \right) \right\} \)
ここで，
\(\Large - sin \theta = sin ( \theta - \pi) \)
より，
\(\Large I(t) = \displaystyle \frac{V_0}{ \omega L} \left\{ 1 + sin \left( \omega t-\frac{\pi}{2} \right) \right\} \)
となり，三角関数で解いた結果と一致します．