ssrs barcodelib In the circuit of Fig. 8-42 assume R 2 k in Software

Generating QR in Software In the circuit of Fig. 8-42 assume R 2 k

EXAMPLE 8.15 In the circuit of Fig. 8-42 assume R 2 k
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, C 10 nF, and R2 R1 . By substituting the element values in H s found in Example 8.14 we obtain V2 2 V1 4 10 10 s2 2 10 5 s 1 d 2 v2 dv 5 104 2 25 108 v2 5 109 v1 dt dt2 The response of the preceding equation for t > 0 to v1 u t is v2 2 e t 2 cos !t 2:31 sin !t 2 3:055e t cos !t 130:98 where 25 000 and ! 21 651 rad/s.
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EXAMPLE 8.16 Find conditions in the circuit of Fig. 8-42 for sustained oscillations in v2 t (with zero input) and nd the frequency of oscillations. In Example 8.14 we obtained V2 k V1 R2 C 2 s2 3 k RCs 1
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HIGHER-ORDER CIRCUITS AND COMPLEX FREQUENCY
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[CHAP. 8
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For sustained oscillations the roots of the characteristic equation in Example 8.14 should be imaginary numbers. This happens when k 3 or R2 2R1 , in which case ! 1=RC.
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8.1 A series RLC circuit, with R 3 k
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, L 10 H, and C 200 mF, has a constant-voltage source, V 50 V, applied at t 0. (a) Obtain the current transient, if the capacitor has no initial charge. (b) Sketch the current and nd the time at which it is a maximum.
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a R 150 s 1 2L !2 0 1 500 s 2 LC q 2 !2 148:3 s 1 0
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The circuit is overdamped > !0 . s1 1:70 s 1 and s2 298:3 s 1
i A1 e 1:70t A2 e 298:3t Thus, at t 0 ,
Since the circuit contains an inductance, i 0 i 0 0; also, Q 0 Q 0 0. KVL gives di di V 5 A=s or 0 0 L 0 V dt dt 0 L Applying these initial conditions to the expression for i, 0 A1 1 A2 1 5 1:70A1 1 298:3A2 1 from which A1 A2 16:9 mA. i 16:9 e 1:70t e 298:3t (b) For the time of maximum current, di 0 28:73e 1:70t 5041:3e 298:3t dt Solving by logarithms, t 17:4 ms. See Fig. 8-18. mA
Fig. 8-18
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HIGHER-ORDER CIRCUITS AND COMPLEX FREQUENCY
A series RLC circuit, with R 50
; L 0:1 H; and C 50 mF, has a constant voltage V 100 V applied at t 0. Obtain the current transient, assuming zero initial charge on the capacitor.
R 250 s 1 2L !2 0 1 2:0 105 s 2 LC q 2 !2 j370:8 rad=s 0
This is an oscillatory case < !0 , and the general current expression is i e 250t A1 cos 370:8 t A2 sin 370:8t The initial conditions, obtained as in Problem 8.1, are i 0 0
di 1000 A=s dt 0 Then A
and these determine the values: A1 0, A2 2:70 A.
i e 250t 2:70 sin 370:8t
Rework Problem 8.2, if the capacitor has an initial charge Q0 2500 mC.
Everything remains the same as in Problem 8.2 except the second initial condition, which is now di Q0 di 100 2500=50 500 A=s V or 0 L dt C dt 0:1
The initial values are half those in Problem 8.2, and so, by linearity, i e 250t 1:35 sin 370:8t A
A parallel RLC network, with R 50:0
, C 200 mF, and L 55:6 mH, has an initial charge Q0 5:0 mC on the capacitor. Obtain the expression for the voltage across the network.
1 8:99 104 s 2 LC q Since !2 > 2 , the voltage function is oscillatory and so !d !2 2 296 rad/s. 0 0 expression is !2 0 v e 50t A1 cos 296t A2 sin 296t With Q0 5:0 10 3 C, V0 25:0 V. At t 0, v 25:0 V. Then, A1 25:0. 1 50 s 1 2RC
The general voltage
dv 50e 50t A1 cos 296t A2 sin 296t 296e 50t A1 sin 296t A2 cos 296t dt At t 0, dv=dt V0 =RC !d A2 A1 , from which A2 4:22. v e
50t
Thus,
25:0 cos 296t 4:22 sin 296t V
In Fig. 8-19, the switch is closed at t 0. t > 0.
Obtain the current i and capacitor voltage vC , for
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
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