148

FIRST-ORDER CIRCUITS

[CHAP. 7

Fig. 7-23 iðtÞ ¼ ið0 Þ ¼ ið0þ Þ ¼ 1 A For t > 0, the current source is replaced by an open circuit and the 50-V source acts in the RL series circuit ðR ¼ 20 Þ. Consequently, as t ! 1, i ! 50=20 ¼ 2:5 A. Then, by Sections 6.10 and 7.3, iðtÞ ¼ ½ðið0þ Þ  ið1ÞeRt=L þ ið1Þ ¼ 3:5e100t  2:5 ðAÞ By means of unit step functions, the two formulas may be combined into a single formula valid for all t: iðtÞ ¼ uðtÞ þ ð3:5e100t  2:5ÞuðtÞ

7.9

In Fig. 7-24(a), the switch is closed at t ¼ 0. vC , and vs for all times if is ¼ 2 mA.

ðAÞ

The capacitor has no charge for t < 0.

Find iR , iC ,

For t < 0, iR ¼ 2 mA, iC ¼ vC ¼ 0, and vs ¼ ð2 mAÞð5000 Þ ¼ 10 V. For t > 0, the time constant is  ¼ RC ¼ 10 ms and iR ð0þ Þ ¼ 0; iR ð1Þ ¼ 2 mA, and iR ¼ 2ð1  e100t Þ ðmAÞ

[See Fig. 7-24ðbÞ:

þ

vC ð0 Þ ¼ 0; vC ð1Þ ¼ ð2 mAÞð5 kÞ ¼ 10 V, and vC ¼ 10ð1  e100t Þ ðVÞ iC ð0þ Þ ¼ 2 mA; iC ð1Þ ¼ 0, and iC ¼ 2e100t

ðmAÞ

vs ð0þ Þ ¼ 0; vs ð1Þ ¼ ð2 mAÞð5 kÞ ¼ 10 V, and vs ¼ 10ð1  e100t Þ

7.10

In Fig. 7-25, the switch is opened at t ¼ 0.

[See Fig. 7-24ðcÞ:

[See Fig. 7-24ðdÞ: ðVÞ

[See Fig. 7-24ðeÞ:

Find iR , iC , vC , and vs .

For t < 0, the circuit is at steady state with iR ¼ 6ð4Þ=ð4 þ 2Þ ¼ 4 mA, iC ¼ 0, and vC ¼ vs ¼ 4ð2Þ ¼ 8 V. During the switching at t ¼ 0, the capacitor voltage remains the same. After the switch is opened, at t ¼ 0þ , the capacitor has the same voltage vC ð0þ Þ ¼ vC ð0 Þ ¼ 8 V. For t > 0, the capacitor discharges in the 5-k resistor, produced from the series combination of the 3-k and 2-k resistors. The time constant of the circuit is  ¼ ð2 þ 3Þð103 Þð2  106 Þ ¼ 0:01 s. The currents and voltages are vC ¼ 8e100t

ðVÞ

iR ¼ iC ¼ vC =5000 ¼ ð8=5000Þe100t ¼ 1:6e100t

ðmAÞ

vs ¼ ð6 mAÞð4 kÞ ¼ 24 V since, for t > 0, all of the 6 mA goes through the 4-k resistor.

7.11

The switch in the circuit of Fig. 7-26 is closed on position 1 at t ¼ 0 and then moved to 2 after one time constant, at t ¼  ¼ 250 ms. Obtain the current for t > 0. It is simplest ﬁrst to ﬁnd the charge on the capacitor, since it is known to be continuous (at t ¼ 0 and at t ¼ ), and then to diﬀerentiate it to obtain the current. For 0  t  , q must have the form q ¼ Aet= þ B

Mahmood_Nahvi_eBook_Schaum_s_Outlines_Theory_An
Mahmood_Nahvi_eBook_Schaum_s_Outlines_Theory_An