# 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

$i(t) = I_m cos(\omega t + \alpha)$

Where:

$I_m = \frac{V_m}{\sqrt{R^2 + (L\omega)^2}}$
$\alpha = \phi -\angle arctan(\frac{L\omega}{R})$

Or numerically:

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

## 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 VS ... this makes the homogeneous solution easy!

Guess:

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

Finding the time constant:

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

Now find see if it works:

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

It works, therefore it must be the solution:

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

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.

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

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:

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

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

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

This agrees with the Laplace solution and simulation.