PRACTICAL MATLAB® FOR ENGINEERS PRACTICAL MATLAB

442 Practical MATLAB® Applications for Engineers

>> loop_diffeqs

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********* Matlab Symbolic Results **********

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The loop currents I1(s) and I2(s) are:

s + 20

I1(s) = 100 -------------------

2

(s + 40 s + 300) s

1000

I2(s) = -------------------

2

(s + 40 s + 300) s

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The loop currents i1(t) and i2(t) (in amp) are:

i1(t) =

-5/3*exp(-30*t)-5*exp(-10*t)+20/3

i2(t) =

10/3-5*exp(-10*t)+5/3*exp(-30*t)

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The initial currents: i1(t=0) and i2(t=0) (in amps)

using the initial value theorem are evaluated below:

I1 _ 0 = 0

I2 _ 0 = 0

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The final currents: i1(t=inf) and i2(t=inf) (in amps)

using the final value theorem are:

I1 _ inf = 20/3

I2 _ inf = 10/3

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Note that at t = ∞, the two inductors become shorts, then i 1 (∞) = 100 V/15 Ω = 20/3 A,

and i 2 (∞) = i 1 (∞)/ 2 = 10/3 A.

Observe that the analytical solutions agree completely with the MATLAB results (see

Figure 4.86).

0

0

1

2

3

4

5

6

7

0.1 0.2 0.3 0.4

time (sec)

0 0.1 0.2 0.3 0.4

time (sec)

i(t) in amps^1 i(t) in amps^2

3.5

3

2

1

0

0.5

1.5

2.5

i 1 (t) versus t i 2 (t) versus t

FIGURE 4.86
Current plots of i 1 (t) and i 2 (t) of Example 4.21.