how to generate barcode in ssrs report Rework Problem 16.10 by writing the voltage function as v 100e j500t V 25 in Software

Generator QR Code in Software Rework Problem 16.10 by writing the voltage function as v 100e j500t V 25

16.11 Rework Problem 16.10 by writing the voltage function as v 100e j500t V 25
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Now V s 100= s j500 , and the s-domain equation is 5I s 0:01sI s 100 s j500 or I s 104 s j500 s 500
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CHAP. 16]
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THE LAPLACE TRANSFORM METHOD
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Using partial fractions, I s and inverting, i 10 j10 e j500t 10 j10 e 500t 14:14e j 500t =4 10 j10 e 500t A 26 10 j10 10 j10 s j500 s 500
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The actual voltage is the imaginary part of (25); hence the actual current is the imaginary part of (26). i 14:14 sin 500t =4 10e 500t A
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16.12 In the series RLC circuit shown in Fig. 16-10, there is no initial charge on the capacitor. If the switch is closed at t 0, determine the resulting current.
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The time-domain equation of the given circuit is di 1 Ri L dt C Because i 0 0, the Laplace transform of (27) is RI s sLI s 1 V I s sC s 28 t i  d V
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or Hence,
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2I s 1sI s I s
1 50 I s 0:5s s
(29) (30)
50 50 s2 2s 2 s 1 j s 1 j j25 j25 s 1 j s 1 j
Expanding (30) by partial fractions, I s 31
and the inverse Laplace transform of (31) gives i j25fe 1 j t e 1 j t g 50e t sin t A
Fig. 16-10
Fig. 16-11
16.13 In the two-mesh network of Fig. 16-11, the two loop currents are selected as shown. Write the sdomain equations in matrix form and construct the corresponding circuit.
Writing the set of equations in the time domain,   t 1 di 5i1 Q0 i1  d 5i2  and 10i2 2 2 5i1  2 dt 0 Taking the Laplace transform of (32) to obtain the corresponding s-domain equations,
32
THE LAPLACE TRANSFORM METHOD
[CHAP. 16
5I1 s
Q0 1 I s 5I2 s V s 2s 2s 1
10I2 s 2sI2 s 2i2 0 5I1 s V s
33
When this set of s-domain equations is written in matrix form,      5 1=2s 5 V s Q0 =2s I1 s V s 2i2 0 5 10 2s I2 s the required s-domain circuit can be determined by examination of the Z s , I s , and V s matrices (see Fig. 16-12).
Fig. 16-12
Fig. 16-13
16.14 In the two-mesh network of Fig. 16-13, nd the currents which result when the switch is closed.
The time-domain equations for the network are 10i1 0:02 di1 di 0:02 2 100 dt dt di2 di1 0:02 5i2 0:02 0 dt dt
34
Taking the Laplace transform of set (34), 10 0:02s I1 s 0:02sI2 s 100=s From the second equation in set (35) we nd  I2 s I1 s  s s 250 36 5 0:02s I2 s 0:02sI1 s 0 35
which when substituted into the rst equation gives   s 250 10 3:33 I1 s 6:67 s s 166:7 s s 166:7 Inverting (37), i1 10 3:33e 166:7t Finally, substitute (37) into (36) and obtain   1 I2 s 6:67 s 166:7 A
37
whence
i2 6:67e 166:7t
16.15 Apply the initial- and nal-value theorems in Problem 16.14.
The initial value of i1 is given by    s 250 i1 0 lim sI1 s lim 6:667 6:67 A s!1 s!1 s 166:7
CHAP. 16]
THE LAPLACE TRANSFORM METHOD
and the nal value is    s 250 i1 1 lim sI1 s lim 6:67 10 A s!0 s!0 s 166:7 The initial value of i2 is given by   i2 0 lim sI2 s lim 6:667
s!1 s!1
 s 6:67 A s 166:7
and the nal value is    s i2 1 lim sI2 s lim 6:67 0 s!0 s!0 s 166:7 Examination of Fig. 16-13 veri es each of the preceding initial and nal values. At the instant of closing, the inductance presents an in nite impedance and the currents are i1 i2 100= 10 5 6:67 A. Then, in the steady state, the inductance appears as a short circuit; hence, i1 10 A, i2 0.
16.16 Solve for i1 in Problem 16.14 by determining an equivalent circuit in the s-domain.
In the s-domain the 0.02-H inductor has impedance Z s 0:02s. Therefore, the equivalent impedance of the network as seen from the source is   0:02s 5 s 166:7 Z s 10 15 0:02s 5 s 250 and the s-domain equivalent circuit is as shown in Fig. 16-14. The current is then     V s 100 s 250 s 250 I1 s 6:67 Z s s 15 s 166:7 s s 166:7 This expression is identical with (37) of Problem 16.14, and so the same time function i1 is obtained.
Fig. 16-14
Fig. 16-15
16.17 In the two-mesh network shown in Fig. 16-15 there is no initial charge on the capacitor. Find the loop currents i1 and i2 which result when the switch is closed at t 0.
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