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FIRST-ORDER CIRCUITS
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[CHAP. 7
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Fig. 7-23 i t i 0 i 0 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 R 20
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. Consequently, as t ! 1, i ! 50=20 2:5 A. Then, by Sections 6.10 and 7.3, i t i 0 i 1 e Rt=L i 1 3:5e 100t 2:5 A By means of unit step functions, the two formulas may be combined into a single formula valid for all t: i t u t 3:5e 100t 2:5 u t A
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In Fig. 7-24(a), the switch is closed at t 0. vC , and vs for all times if is 2 mA.
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The capacitor has no charge for t < 0.
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For t < 0, iR 2 mA, iC vC 0, and vs 2 mA 5000
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10 V. For t > 0, the time constant is  RC 10 ms and iR 0 0; iR 1 2 mA, and iR 2 1 e 100t mA
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[See Fig. 7-24 b : [See Fig. 7-24 c : [See Fig. 7-24 e :
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vC 0 0; vC 1 2 mA 5 k
10 V, and vC 10 1 e 100t V iC 0 2 mA; iC 1 0, and iC 2e 100t mA [See Fig. 7-24 d : V vs 0 0; vs 1 2 mA 5 k
10 V, and vs 10 1 e 100t
In Fig. 7-25, the switch is opened at t 0.
Find iR , iC , vC , and vs .
For t < 0, the circuit is at steady state with iR 6 4 = 4 2 4 mA, iC 0, and vC vs 4 2 8 V. During the switching at t 0, the capacitor voltage remains the same. After the switch is opened, at t 0 , the capacitor has the same voltage vC 0 vC 0 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  2 3 103 2 10 6 0:01 s. The currents and voltages are vC 8e 100t V mA
iR iC vC =5000 8=5000 e 100t 1:6e 100t vs 6 mA 4 k
24 V since, for t > 0, all of the 6 mA goes through the 4-k
resistor.
The switch in the circuit of Fig. 7-26 is closed on position 1 at t 0 and then moved to 2 after one time constant, at t  250 ms. Obtain the current for t > 0.
It is simplest rst to nd the charge on the capacitor, since it is known to be continuous (at t 0 and at t ), and then to di erentiate it to obtain the current. For 0 t , q must have the form q Ae t= B
CHAP. 7]
FIRST-ORDER CIRCUITS
Fig. 7-24
Fig. 7-25
Fig. 7-26
FIRST-ORDER CIRCUITS
[CHAP. 7
From the assumption q 0 0 and the condition dq 20 V 40 mA i 0 0 dt 500
we nd that A B 10 mC, or q 10 1 e 4000t
mC
20
From (20), q  10 1 e mC; and we know that q 1 0:5 mF 40 V 20 mC. Hence, q, is determined for t !  as q q  q 1 e t  = q 1 71:55e 4000t 20 Di erentiating (20) and (21), dq i dt See Fig. 7-27.  40e 4000t 286:2e 4000t mA mA 0 < t <  t >  mC 21
Fig. 7-27
A series RL circuit has a constant voltage V applied at t 0.
At what time does vR vL
The current in an RL circuit is a continuous function, starting at zero in this case, and reaching the nal value V=R. Thus, for t > 0, i V 1 e t= R and vR Ri V 1 e t=
where  L=R is the time constant of the circuit.
Since vR vL V, the two voltages will be equal when vR 1 V 2
V 1 e t= 1 V 2 e t= 1 2 t ln 2  that is, when t 0:693. Note that this time is independent of V.
A constant voltage is applied to a series RL circuit at t 0. 20 V at 3.46 ms and 5 V at 25 ms. Obtain R if L 2 H.
Using the two-point method of Section 7-6.
The voltage across the inductance is
CHAP. 7]
FIRST-ORDER CIRCUITS
 and so
t2 t1 25 3:46 15:54 ms ln v1 ln v2 ln 20 ln 5 R L 2 128:7
 15:54 10 3
In Fig. 7-28, switch S1 is closed at t 0.
Switch S2 is opened at t 4 ms.
Obtain i for t > 0.
Fig. 7-28 As there is always inductance in the circuit, the current is a continuous function at all times. In the interval 0 t 4 ms, with the 100
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