ssrs barcodelib shorted out and a time constant  0:1 H = 50 in Software

Generation QR Code in Software shorted out and a time constant  0:1 H = 50

shorted out and a time constant  0:1 H = 50
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2 ms, i starts at zero and builds toward 100 V 2A 50
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even though it never gets close to that value. Hence, as in Problem 7.12 A 0 t 4 22
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i 2 1 e t=2 wherein t is measured in ms. In particular,
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i 4 2 1 e 2 1:729 A In the interval t ! 4 ms, i starts at 1.729 A and decays toward 100=150 0:667 A, with a time constant 0:1=150 2 ms. Therefore, with t again in ms, 3 i 1:729 0:667 e t 4 = 2=3 0:667 428:4e 3t=2 0:667 A t ! 4 23
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In the circuit of Fig. 7-29, the switch is closed at t 0, when the 6-mF capacitor has charge Q0 300 mC. Obtain the expression for the transient voltage vR .
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The two parallel capacitors have an equivalent capacitance of 3 mF. Then this capacitance is in series with the 6 mF, so that the overall equivalent capacitance is 2 mF. Thus,  RCeq 40 ms. At t 0 , KVL gives vR 300=6 50 V; and, as t ! 1, vR ! 0 (since i ! 0). Therefore, vR 50 e t= 50e t=40 in which t is measured in ms. V
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Fig. 7-29
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Fig. 7-30
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FIRST-ORDER CIRCUITS
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[CHAP. 7
In the circuit shown in Fig. 7-30, the switch is moved to position 2 at t 0. at t 34:7 ms.
After the switching, the three inductances have the equivalent Leq Then  5=200 25 ms, and so, with t in ms, i 6e t=25 and A i2   5 i 2e t=25 15 A 10 5 10 5H 6 15
Obtain the current i2
i2 34:7 2e 34:7=25 A 0:50 A
In Fig. 7-31, the switch is closed at t 0. t > 0.
Obtain the current i and capacitor voltage vC , for
Fig. 7-31 As far as the natural response of the circuit is concerned, the two resistors are in parallel; hence,  Req C 5
2 mF 10 ms By continuity, vC 0 vC 0 0. Furthermore, as t ! 1, the capacitor becomes an open circuit, leaving 20
in series with the 50 V. That is, i 1 50 2:5 A 20 vC 1 2:5 A 10
25 V
Knowing the end conditions on vC , we can write vC vC 0 vC 1 e t= vC 1 25 1 e t=10 wherein t is measured in ms. The current in the capacitor is given by iC C and the current in the parallel 10-
resistor is i10
Hence, vC 2:5 1 e t=10 10
A dvC 5e t=10 dt A V
i iC i10
2:5 1 e t=10 A
The problem might also have been solved by assigning mesh currents and solving simultaneous di erential equations.
The switch in the two-mesh circuit shown in Fig. 7-32 is closed at t 0. and i2 , for t > 0.
Obtain the currents i1
CHAP. 7]
FIRST-ORDER CIRCUITS
Fig. 7-32 10 i1 i2 5i1 0:01 di1 100 dt 10 i1 i2 5i2 100 24 25
From (25), i2 100 10i1 =15.
Substituting in (24), di1 833i1 3333 dt 26
The steady-state solution (particular solution) of (26) is i1 1 3333=833 4:0 A; hence i1 Ae 833t 4:0 A The initial condition i1 0 i1 0 0 now gives A 4:0 A, so that i1 4:0 1 e 833t A and i2 4:0 2:67e 833t A
Alternate Method
When the rest of the circuit is viewed from the terminals of the inductance, there is equivalent resistance Req 5 5 10 8:33
Then 1= Req =L 833 s 1 . At t 1, the circuit resistance is RT 10 5 5 12:5
so that the total current is iT 100=12:5 8 A. And, at t 1, this divides equally between the two 5-
resistors, yielding a nal inductor current of 4 A. Consequently, iL i1 4 1 e 833t A
A series RL circuit, with R 50
and L 0:2 H, has a sinusoidal voltage v 150 sin 500t 0:785 V applied at t 0. Obtain the current for t > 0.
The circuit equation for t > 0 is di 250i 750 sin 500t 0:785 dt 27
The solution is in two parts, the complementary function (ic ) and the particular solution ip , so that i ic ip . The complementary function is the general solution of (27) when the right-hand side is replaced by zero: ic ke 250t . The method of undetermined coe cients for obtaining ip consists in assuming that ip A cos 500t B sin 500t since the right-hand side of (27) can also be expressed as a linear combination of these two functions. Then
FIRST-ORDER CIRCUITS
[CHAP. 7
dip 500A sin 500t 500B cos 500t dt Substituting these expressions for ip and dip =dt into (27) and expanding the right-hand side, 500A sin 500t 500B cos 500t 250A cos 500t 250B sin 500t 530:3 cos 500t 530:3 sin 500t Now equating the coe cients of like terms, 500A 250B 530:3 and 500B 250A 530:3
Solving these simultaneous equations, A 0:4243 A, B 1:273 A. ip 0:4243 cos 500t 1:273 sin 500t 1:342 sin 500t 0:322 A and At t 0, i 0. i ic ip ke 250t 1:342 sin 500t 0:322 Applying this condition, k 0:425 A, and, nally, i 0:425e 250t 1:342 sin 500t 0:322 A A
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