Electronics Handbook/Circuits/RC Series

Circuit's Configuration

Circuit Analysis

Circuit's Impedance

In Polar coordinate

$Z=Z_{R}+Z_{C}$

Z = R /_0 + ( 1 / ωC ) /_ - 90

Z = = |Z|/_θ = ${\sqrt {R^{2}+({\frac {1}{\omega C}})^{2}}}$ /_ Tan^-1 ${\frac {1}{\omega RC}}$

In Rectangular coordinate

$Z=Z_{R}+Z_{C}$

Z = $R+{\frac {1}{j\omega C}}={\frac {1+j\omega RC}{j\omega C}}$

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

Phase Angle Difference Between Voltage and Current

There is a difference in angle Between Voltage and Current . Current leads Voltage by an angle θ

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

The difference in angle between Voltage and Current relates to the value of R , C and the Angular of Frequency ω which also relates to f and t . Therefore when change the value of R or C , the angle difference will be changed and so are ω , f , t

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

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

$t=2\pi RCTan\theta$

Natural Response

$C{\frac {dV}{dt}}+{\frac {V}{R}}=0$

${\frac {dV}{dt}}=-{\frac {1}{RC}}V$

${\frac {1}{V}}dV=-{\frac {1}{RC}}dt$

$\int {\frac {1}{V}}dV=-\int {\frac {1}{RC}}dt$

ln V = $-{\frac {1}{RC}}+C$

$V=e^{-}({\frac {1}{RC}})t+C$

$V=Ae^{-}({\frac {1}{T}})t$

$A=e^{C}$

T = RC

Summary

In summary, RL series circuit has a first order differential equation of voltage

${\frac {d}{dt}}f(t)+{\frac {t}{T}}=0$

Which has one real root

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

$A=e^{c}$