Circuit Theory/1Source Excitement/Example 7

Series RL circuit for example 7

Given:

$V_s(t)=120\sqrt{2}cos(377t+120^{\circ })$

Prior Work calculating Steady State/Particular Solution

Have already found the steady state/Particular solution:

$i(t)=I_{m}cos(\omega t+\alpha )$

Where:

$I_m=\frac{V_m}{\sqrt{R^2+(L\omega {)}^2}}$

$\alpha =\varphi -\mathrm{\angle }arctan(\frac{L\omega }{R})$

Or numerically:

$i(t{)}_{s_P}=15.9cos(377t+1.73)$

Calculating the transient/Homogeneous Solution

Need to find the transient/Homogeneous Solution to:

$R*i(t)+L*\frac{d}{dt}i(t)=0$

There is no V_S ... this makes the homogeneous solution easy!

Guess:

$i(t{)}_{s_H}=A*e^{\frac{-t}{\tau }}$

Finding the time constant:

$R*A*e^{\frac{-t}{\tau }}+\frac{LA}{-\tau }{e}^{\frac{-t}{\tau }}=0$

$R-\frac{L}{\tau }=0$

$\tau =L/R=.001$

Now find see if it works:

$RA{e}^{\frac{-t}{L/R}}+LA(-\frac{R}{L})*e^{\frac{-t}{L/R}}\stackrel{\mathrm{?}\mathrm{?}\mathrm{?}}{=}0$

divide through by A, cancel L's

$R{e}^{\frac{-t}{L/R}}-R{e}^{\frac{-t}{L/R}}\stackrel{\mathrm{?}\mathrm{?}\mathrm{?}}{=}0$

It works, therefore it must be the solution:

$i(t)=i(t{)}_{s_P}+i(t{)}_{s_H}=599\cos (377t+3.30)+A{e}^{\frac{-t}{0.0001}}$

Now must find the initial conditions.

Determining the Constants

There are two constants. $A$ and $C$ come from any homogeneous solution to a non-homogeneous differential equation equation. These were not ignored in the steady state phasor solutions earlier, the fact that they were not being computed was pointed out.

$i(t{)}_s=i(t{)}_{s_P}+i(t{)}_{s_H}$

$i(t{)}_s=15.9cos(377t+1.73)+A*e^{\frac{-t}{.001}}+C$

There are two initial conditions that have to be true:

1. initial source voltage has a value at t=0: $120*\sqrt{2}\cos (\frac{2\pi }{3})$
2. initial current through the inductor has at t=0 has to be 0, thus the current throughout the entire series circuit is 0 at t=0

Finding two initial conditions

mupad and matlab code to find constants A and C .. /code/

Two equations are necessary to find A and C.

Initially the current through the inductor and the entire circuit is going to be zero:

$i(0_{-})=0$, thus $i(0_{+})=0$.

This means that setting t=0, have one equation:

$i(t{)}_s=15.9cos(377t+1.73)+A*e^{\frac{-t}{.001}}+C=0$

Evaluating this at t=0:

$A+C-2.58=0$

The second equation comes from the loop:

$v_r(t)+v_L(t)-V_s=0$

$R*i(t{)}_s+L*\frac{di(t{)}_s}{dt}-V_s=0$

Substituting for i(t)_S and V(t)_S, taking the differential and then evaluating at t=0, get:

$1.86*10^{-9}-10.0*C=0$

So solving get:

$C=0$

$A=2.5781$

Summary

$i(t)=15.9cos(377t+1.73)+2.58*e^{\frac{-t}{.001}}$

This agrees with the Laplace solution and simulation. Template:BookCat