# Circuit Theory/1Source Excitement/Example 7

Series RL circuit for example 7

Given:

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

## Prior Work calculating Steady State/Particular Solution

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

Where:

${\displaystyle I_{m}={\frac {V_{m}}{\sqrt {R^{2}+(L\omega )^{2}}}}}$
${\displaystyle \alpha =\phi -\angle arctan({\frac {L\omega }{R}})}$

Or numerically:

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

## Calculating the transient/Homogeneous Solution

Need to find the transient/Homogeneous Solution to:

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

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

Guess:

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

Finding the time constant:

${\displaystyle R*A*e^{\frac {-t}{\tau }}+{\frac {LA}{-\tau }}e^{\frac {-t}{\tau }}=0}$
${\displaystyle R-{\frac {L}{\tau }}=0}$
${\displaystyle \tau =L/R=.001}$

Now find see if it works:

${\displaystyle RAe^{\frac {-t}{L/R}}+LA(-{\frac {R}{L}})*e^{\frac {-t}{L/R}}{\overset {\underset {\mathrm {???} }{}}{=}}0}$
divide through by A, cancel L's
${\displaystyle Re^{\frac {-t}{L/R}}-Re^{\frac {-t}{L/R}}{\overset {\underset {\mathrm {???} }{}}{=}}0}$

It works, therefore it must be the solution:

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

Now must find the initial conditions.

## Determining the Constants

There are two constants. ${\displaystyle A}$ and ${\displaystyle C}$ 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.

${\displaystyle i(t)_{s}=i(t)_{s_{P}}+i(t)_{s_{H}}}$
${\displaystyle i(t)_{s}=15.9cos(377t+1.73)+A*e^{\frac {-t}{.001}}+C}$

There are two initial conditions that have to be true:

1. initial source voltage has a value at t=0: ${\displaystyle 120*{\sqrt {2}}\cos({\frac {2\pi }{3}})}$
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:

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

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

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

Evaluating this at t=0:

${\displaystyle A+C-2.58=0}$

The second equation comes from the loop:

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

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

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

So solving get:

${\displaystyle C=0}$
${\displaystyle A=2.5781}$

## Summary

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

This agrees with the Laplace solution and simulation.