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154

FIRST-ORDER CIRCUITS

[CHAP. 7

dip ¼ 500A sin 500t þ 500B cos 500t dt Substituting these expressions for ip and dip =dt into (27) and expanding the right-hand side, 500A sin 500t þ 500B cos 500t þ 250A cos 500t þ 250B sin 500t ¼ 530:3 cos 500t þ 530:3 sin 500t Now equating the coefficients of like terms, 500A þ 250B ¼ 530:3

and

500B þ 250A ¼ 530:3

Solving these simultaneous equations, A ¼ 0:4243 A, B ¼ 1:273 A. ip ¼ 0:4243 cos 500t þ 1:273 sin 500t ¼ 1:342 sin ð500t  0:322Þ ðAÞ i ¼ ic þ ip ¼ ke250t þ 1:342 sin ð500t  0:322Þ

and At t ¼ 0, i ¼ 0.

Applying this condition, k ¼ 0:425 A, and, finally, i ¼ 0:425e250t þ 1:342 sin ð500t  0:322Þ

7.20

ðAÞ

ðAÞ

For the circuit of Fig. 7-33, obtain the current iL , for all values of t.

Fig. 7-33 For t < 0, the 50-V source results in inductor current 50=20 ¼ 2:5 A. The 5-A current source is applied for t > 0. As t ! 1, this current divides equally between the two 10- resistors, whence iL ð1Þ ¼ 2:5 A. The time constant of the circuit is ¼

0:2  103 H 1 ¼ ms 20  100

and so, with t in ms and using iL ð0þ Þ ¼ iL ð0 Þ ¼ 2:5 A, iL ¼ ½iL ð0þ Þ  iL ð1Þet= þ iL ð1Þ ¼ 5:0e100t  2:5

ðAÞ

Finally, using unit step functions to combine the expressions for t < 0 and t > 0, iL ¼ 2:5uðtÞ þ ð5:0e100t  2:5ÞuðtÞ

7.21

ðAÞ

The switch in Fig. 7-34 has been in position 1 for a long time; it is moved to 2 at t ¼ 0. the expression for i, for t > 0.

Obtain

With the switch on 1, ið0 Þ ¼ 50=40 ¼ 1:25 A. With an inductance in the circuit, ið0 Þ ¼ ið0þ Þ. after the switch has been moved to 2, ið1Þ ¼ 10=40 ¼ 0:25 A. In the above notation, B ¼ ið1Þ ¼ 0:25 A and the time constant is  ¼ L=R ¼ ð1=2000Þ s.

A ¼ ið0þ Þ  B ¼ 1:00 A Then, for t > 0,

i ¼ 1:00e2000t þ 0:25

ðAÞ

Long

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Mahmood_Nahvi_eBook_Schaum_s_Outlines_Theory_An  
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