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

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

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

$A=e^C$

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

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

$A=e^c$