Geometry & Recognition

다음 회로에서 교류 전원이 $\mathcal{E}(t) = V_0 \sin(\omega t)$주어진 경우, a-b 사이의 전위차는?

풀이: $C_1 R_1$으로 흐르는 전류와 $R_2C_2$로 흐르는 전류는 각각 

\begin{align} i_1(t)& = I_1 \sin(\omega t + \varphi_1), \quad \tan \varphi_1 = \frac{1}{\omega R_1C_1} \\ I_1 &= \frac{V_0}{\sqrt{R_1^2 + \frac{1}{(\omega C_1)^2} }  } = \frac{\omega C_1 V_0}{\sqrt{1+ \alpha^2}},\qquad \alpha= \omega R_1C_1= \omega R_2 C_2 \\ i_2(t) &= I_2 \sin(\omega t+ \varphi_2) , \quad \tan \varphi_2 = \frac{1}{\omega R_2 C_2}\\ I_2 &= \frac{V_0 }{\sqrt{R_2^2 + \frac{1}{(\omega C_2)^2 } }}=\frac{ \omega C_2 V_0}{\sqrt{1 + \alpha^2 }} \end{align}

따라서, 

\begin{align} \varphi_1 &= \varphi_2 = \varphi = \arctan \frac1\alpha = \frac{\pi}{2} - \arctan \alpha  \quad     \\ v_{R_1}(t) &=  R_1 I_1 \sin(\omega t + \varphi) = \frac{\alpha V_0}{\sqrt{1+ \alpha^2} }\sin(\omega t +\varphi) =\cos\varphi  \sin(\omega t +\varphi) \\   v_{C_2} (t) &= X_{C_2} I_2 \sin\left( \omega t  -(\tfrac{\pi}{2}-\varphi)\right) =- \frac{ V_0}{\sqrt{1+ \alpha^2} }\cos(\omega t + \varphi) = -\sin\varphi\cos(\omega t+ \varphi) \end{align}이고,

\begin{align}v_{ab}(t) &=v_{R_1}(r)-v_{C_2}(t) \\ &= V_0 \left[ \cos(\varphi) \sin(\omega t +\varphi) + \sin \varphi \cos(\omega t +\varphi) \right] \\ &= V_0 \sin  (\omega t + 2\varphi)= V_0 \sin\left(\omega t + \pi - 2 \arctan (\omega RC)\right)\end{align}.