Circuit Theory/1Source Excitement/Example 7

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Series RL circuit for example 7

Given:

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

Prior Work calculating Steady State/Particular Solution[edit]

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[edit]

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[edit]

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[edit]

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[edit]

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

This agrees with the Laplace solution and simulation.