Circuit's Impedance

In Rectangular coordinate

• ${\displaystyle Z=Z_{R}+Z_{C}}$

${\displaystyle Z=R+{\frac {1}{j\omega C}}={\frac {1}{j\omega C}}(1+j\omega T)}$

${\displaystyle T=RC}$

In Polar coordinate

• ${\displaystyle Z=Z_{R}+Z_{C}}$

${\displaystyle Z=R\angle 0+{\frac {1}{\omega C}}\angle -90=|Z|\angle \theta ={\sqrt {R^{2}+({\frac {1}{\omega C}})^{2}}}\angle Tan^{-}1{\frac {1}{\omega RC}}}$

${\displaystyle Tan\theta ={\frac {1}{\omega RC}}={\frac {1}{2\pi fRC}}={\frac {t}{2\pi RC}}}$

The value of ${\displaystyle \theta ,\omega ,f}$ depend on the value of R and C . Therefore, when the value of R or C changed the value of Phase angle difference between Current and Voltage, Frequency, and Angular of Frequency also change

${\displaystyle \omega ={\frac {1}{Tan\theta RC}}}$

${\displaystyle f={\frac {1}{2\pi Tan\theta RC}}}$

${\displaystyle t=2\pi Tan\theta RC}$

Circuit's Response

Natural Response of the cicuit can be obtained by setting the differential equation of the circuit to zero

${\displaystyle L{\frac {di}{dt}}+iR=0}$

${\displaystyle {\frac {di}{dt}}=-{\frac {R}{L}}i}$

${\displaystyle \int {\frac {di}{i}}=-{\frac {R}{L}}\int dt}$

${\displaystyle Lni=-({\frac {R}{L}})t+e^{c}}$

${\displaystyle i=Ae^{-}({\frac {R}{L}})t}$

${\displaystyle i=Ae^{-}({\frac {t}{T}})}$

${\displaystyle A=e^{c}={\frac {V}{R}}}$

${\displaystyle T=RC}$

The natural response of the circuit is an exponential decrease