# Practical Electronics/Series RL

## Series RCEdit

Electric circuit of two components R and L connected in series

## Circuit analysisEdit

### Circuit's ImpedanceEdit

In Rectangular coordinate

• $Z = Z_R + Z_L$
$Z = R + \omega L = \frac{1}{R} (1 + j\omega T)$
$T = \frac{L}{R}$

In Polar coordinate

• $Z = Z_R + Z_L$
$Z = R \angle 0 + \omega L \angle 90 = |Z| \angle \theta = \sqrt{R^2 + (\omega L)^2} \angle Tan^-1 \omega \frac{L}{R}$
$Tan \theta = \omega \frac{L}{R} = 2 \pi f \frac{L}{R} = \frac{t}{2 \pi \frac{L}{R}}$

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

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

### Circuit's ResponseEdit

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

$C \frac{dV}{dt} + \frac{V}{R} = 0$
$\frac{dV}{dt} = -\frac{1}{RC} V$
$\int \frac{dV}{V} = -\frac{1}{RC} \int dt$
$Ln V = -(\frac{1}{RC})t + e^c$
$V = A e^ -(\frac{1}{RC})t$
$V = A e^ -(\frac{t}{T})$
$V = e^c = IR$
$T = \frac{L}{R}$

The natural reponse of the circuit is an exponential decrease

## SummaryEdit

Circuit Series RL
Configuration
Impedance $Z = Z_R + Z_L= R + j\omega L = \frac {1}{R} (1 + j\omega T)$
$T = \frac{L}{R}$
Diferenial Equation $C\frac{dV}{dt} + \frac{V}{R} = 0$
Root of the equation $V = A e^(-\frac{t}{T})$
$Z\angle\theta$ $\sqrt{R^2 + (\omega L)^2} \angle Tan ^-1 \omega \frac{L}{C}$
Phase Angle Difference between Voltage and Current $Tan \theta = \omega \frac{L}{R}$
$\omega$ $\omega = \frac{1}{Tan\theta} \frac{L}{R}$
$f$ $\omega = \frac{Tan\theta}{2 \pi} \frac{L}{R}$
$t$ $t = \frac{2 \pi}{Tan\theta} \frac{R}{L}$