Circuit Theory/1Source Excitement/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:
 * $$i(t)_{s_P} = 15.9 cos(377t + 1.73)$$

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
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.