# Electronics/RC transient

For a series RC consist of one resistor connected with one capacitor in a closed loop.

## Circuit Impedance

In Polar Form Z/_θ

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

In Complex Form Z(jω)

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

## Differential Equation of circuit at equilibrium

${\displaystyle C{\frac {dV}{dt}}+{\frac {V}{R}}=0}$
${\displaystyle {\frac {dV}{dt}}=-{\frac {1}{RC}}V}$
${\displaystyle {\frac {1}{V}}dV=-{\frac {1}{RC}}dt}$
${\displaystyle \int {\frac {1}{V}}dV=-\int {\frac {1}{RC}}dt}$
${\displaystyle \ln V=-{\frac {1}{RC}}t+C}$
${\displaystyle V=e^{(}{\frac {-t}{RC}}+C)=e^{C}\cdot e^{\frac {-t}{RC}}}$
${\displaystyle V=Ae^{\left({\frac {-t}{RC}}\right)}=Ae^{\left({\frac {-t}{T}}\right)}}$

## Time Constant

T = R C
t V(t) % Vo
0 A = eC = Vo 100%
1/RC .63 Vo 60% Vo
2/RC Vo
3/RC Vo
4/RC Vo
5/RC .01 Vo 10% Vo

## Angle Difference Between Voltage and Current

Current leads Voltage on an angle; Does this mean in measuring a angle of DC current causes a decrease in current? Let us take some time to understand the previous math in determing a factor of sinuous solution.

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

Change the value of R and C will change the value of Angle Difference, Angular Frequency, Frequency and Time

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

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## Application

When the switch is open, the initial voltage across the capacitor is zero. When the switch closes (which we will refer to as time zero) the capacitor charges via the resistor to ${\displaystyle V_{0}}$ .

When the switch is closed, the circuit must follow the relationship:

${\displaystyle V_{0}=v_{c}(t)+i_{c}(t)R}$
${\displaystyle V_{0}=v_{c}(t)+{\frac {dv_{c}(t)}{dt}}RC}$

which is derived by analysing the circuit using Kirchoff's Voltage Law.

By letting ${\displaystyle \tau =RC}$  and rearranging the equation:

${\displaystyle {\frac {dv_{c}(t)}{dt}}+{\frac {1}{\tau }}v_{c}(t)={\frac {1}{\tau }}V_{0}}$

This is a first order linear differential equation with integrating factor:

${\displaystyle e^{\int {\frac {1}{\tau }}dt}}$
${\displaystyle e^{\frac {t}{\tau }}}$

Multiplying both sides by the integrating factor:

${\displaystyle {\frac {dv_{c}(t)}{dt}}e^{\frac {t}{\tau }}+{\frac {1}{\tau }}v_{c}(t)e^{\frac {t}{\tau }}={\frac {1}{\tau }}V_{0}e^{\frac {t}{\tau }}}$

Note that:

${\displaystyle {\frac {d}{dt}}[e^{t/\tau }v_{c}(t)]={\frac {dv_{c}(t)}{dt}}e^{\frac {t}{\tau }}+{\frac {1}{\tau }}v_{c}(t)e^{\frac {t}{\tau }}}$

Substituting and integrating both sides:

${\displaystyle e^{\frac {t}{\tau }}v_{c}(t)=V_{0}e^{\frac {t}{\tau }}+K}$

where K is the integration constant.

When t=0

${\displaystyle v_{c}(0)=0}$

Therefore:

${\displaystyle K=-V_{0}}$

When t>0 this gives:

${\displaystyle e^{\frac {t}{\tau }}v_{c}(t)=V_{0}e^{\frac {t}{\tau }}-V_{0}}$
${\displaystyle v_{c}(t)=V_{0}-V_{0}e^{-{\frac {t}{\tau }}}}$
${\displaystyle v_{c}(t)=V_{0}(1-e^{-{\frac {t}{\tau }}})}$

when t<0:

${\displaystyle v_{c}(t)=0}$