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:

$I_{m}={\frac {V_{m}}{\sqrt {R^{2}+(L\omega )^{2}}}}$

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

Or numerically:

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

Calculating the transient/Homogeneous Solution

Need to find the transient/Homogeneous Solution to:

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

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:

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

divide through by A, cancel L's

$Re^{\frac {-t}{L/R}}-Re^{\frac {-t}{L/R}}{\overset {\underset {\mathrm {???} }{}}{=}}0$

It works, therefore it must be the solution:

$i(t)=i(t)_{s_{P}}+i(t)_{s_{H}}=599\cos(377t+3.30)+Ae^{\frac {-t}{0.0001}}$

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:

$v_{r}(t)+v_{L}(t)-V_{s}=0$

$R*i(t)_{s}+L*{di(t)_{s} \over dt}-V_{s}=0$

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.