Electronics Handbook/Circuits/RL Series

Series RL Circuit

Circuit Analysis

Circuit Impedance

In Polar Coordinate

$Z=Z_R+Z_L$ = R/_0 + ω L/_90

Z = |Z|/_θ = $\sqrt{R^2+(\omega L{)}^2}$/_Tan^-1$\omega \frac{L}{R}$

In Rectangular Coordinate

$Z=Z_R+Z_L=R+j\omega L$

$Z=R+j\omega L=R(1+j\omega \frac{L}{R})$

Phase Angle of Difference Between Voltage and Current

In RL series circuit, only L is the component that depends on frequency . There is no difference between voltage and current on R . There is an angle difference between voltage and current by 90 degree . When connect R and L in series , there is a difference in angle between voltage and current from 0 to 90 degree which can be expressed as a mathematic formula below

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

$\omega =Tan\theta \frac{R}{L}$

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

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

Natural Response

$L\frac{dI}{dt}+IR=0$

$\frac{dI}{dt}=-I\frac{R}{L}$

$\int \frac{1}{I}dI=-\int \frac{L}{R}dt$

ln I = $(-\frac{L}{R}+c)$

I = $e^{(}-\frac{L}{R}+c)t$

I = $e^c{e}^{(}-\frac{L}{R}t)$

I = $A{e}^{(}-\frac{L}{R}t)$

Time Constant RL

τ = $\frac{L}{R}$

I = A $e^{-}(\frac{L}{R})t$

t	I(t)	% Io
0	A = eC = Io	100%
$\frac{1}{RC}$	.63 Io	63% Io
$\frac{2}{RC}$	Io
$\frac{3}{RC}$	Io
$\frac{4}{RC}$	Io
$\frac{5}{RC}$	.01 Io	10% Io

Current of the circuit is decreased exponentially with time . At time t = 0 I = I_o . At time t = R/L I = 63% I_o . At time t = 5 time time constant I = 10% I_o

I = A $e^{-}(\frac{L}{R})t$

Summary

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

$\frac{d}{dt}f(t)+\frac{1}{T}=0$

Which has one real root

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

$A=e^c$