# Practical Electronics/Series RC

## Series RC

A circuit of two components R and C connected in series

## Circuit analysis

### Circuit's Impedance

In Rectangular coordinate

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

In Polar coordinate

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

The value of $\theta ,\omega ,f$  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

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

### Circuit's Response

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

$L{\frac {di}{dt}}+iR=0$
${\frac {di}{dt}}=-{\frac {R}{L}}i$
$\int {\frac {di}{i}}=-{\frac {R}{L}}\int dt$
$Lni=-({\frac {R}{L}})t+e^{c}$
$i=Ae^{-}({\frac {R}{L}})t$
$i=Ae^{-}({\frac {t}{T}})$
$A=e^{c}={\frac {V}{R}}$
$T=RC$

The natural response of the circuit is an exponential decrease