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so that vC (t) = 493 1 e t/1974 10
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0 < t < 1 s
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At the time when the switch opens again, t = 1 s, the capacitor voltage can be found to be vC (t = 1 s) = 493 V When the switch opens again, the capacitor will discharge through the load resistor, RL ; this discharge is described by the natural response of the circuit consisting of C and RL and is governed by the following values: vC (t = 1 s) = 493 V, off = RL C = 15 s We can directly write the natural solution as follows: vC (t) = vC (t = 1 10 6 ) e (t 1 10 = 493 e (t 1 10
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Figure 531 shows a plot of the solution for t > 0, along with the voltage pulse
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6 5 4 Volts 3 2 1 0 0 02 04 06 Time (10 5 s) 08 1
Figure 531 Coaxial cable pulse response
Comments Note that the voltage response shown in Figure 531 rapidly
reaches the desired value, near 5 volts, thanks to the very short charging time constant, on On the other hand, the discharging time constant, off , is signi cantly slower As the length of the cable is increased, however, on will increase, to the point that the voltage pulse may not rise suf ciently close to the desired 5-V value in the desired time While the numbers used in this example are somewhat unrealistic, you should remember that cable length limitations may exist in some applications because of the cable intrinsic capacitance and resistance Focus on Computer-Aided Tools The Matlab m- le containing the numerical analysis and plotting commands for this example may be found in the CD that accompanies this book An EWB solution is also enclosed
Check Your Understanding
51 Write the differential equation for the circuit shown in Figure 532 52 Write the differential equation for the circuit shown in Figure 533 53 Write the differential equation for the circuit shown in Figure 534
Part I
Circuits
C iC(t) +
+ v (t) _ S
iR(t) C R
+ v(t) _
iL(t) iS(t) L
iR(t) R
+ v(t) _
iS(t) R i(t) vR(t) _
Figure 533 Figure 532
54 It is instructive to repeat the analysis of Example 55 for a capacitive circuit For the circuit shown in Figure 535, compute the quantities vC (0 ) and iR (0+ ), and sketch the response of the circuit, that is, vC (t), if the switch opens at t = 0
2 t>0 iR 12 V 4 + vC _ C 15 V R2 C R1 R3 + vC (t) _
200
10 mA
1 k
001 F
1 k
55 The circuit of Figure 536 has a switch that can be used to connect and disconnect a battery The switch has been open for a very long time At t = 0, the switch closes, and then at t = 50 ms, the switch opens again Assume that R1 = R2 = 1,000 , R3 = 500 , and C = 25 F a Determine the capacitor voltage as a function of time b Plot the capacitor voltage from t = 0 to t = 100 ms 56 If the 10-mA current source is switched on at t = 0 in the circuit of Figure 537,
how long will it take for the capacitor to charge to 90 percent of its nal voltage
50
01 H
100
150
57 Find the time constant for the circuit shown in Figure 538 58 Repeat the calculations of Example 59 if the load resistance is 1,000 What is
the effect of this change
TRANSIENT RESPONSE OF SECOND-ORDER CIRCUITS
In many practical applications, understanding the behavior of rst- and secondorder systems is often all that is needed to describe the response of a physical system
5
Transient Analysis
to external excitation In this section, we discuss the solution of the second-order differential equations that characterize second-order circuits Deriving the Differential Equations for Second-Order Circuits A simple way of introducing second-order circuits consists of replacing the box labeled Circuit containing RL/RC combinations in Figure 53 with a combination of two energy-storage elements, as shown in Figure 539 Note that two different cases are considered, depending on whether the energy-storage elements are connected in series or in parallel Consider the parallel case rst, which has been redrawn in Figure 540 for clarity Practice and experience will eventually suggest the best method for writing the circuit equations At this point, the most sensible procedure consists of applying the basic circuit laws to the circuit of Figure 540 Start with KVL around the lefthand loop: vT (t) RT iS (t) vC (t) = 0 Then apply KCL to the top node, to obtain
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