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zen barcode ssrs FIRSTORDER CIRCUITS in Software
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QR Generation In Visual Basic .NET Using Barcode generation for .NET Control to generate, create QR Code ISO/IEC18004 image in .NET applications. Creating UCC  12 In None Using Barcode generator for Software Control to generate, create UPC A image in Software applications. A more complex circuit containing resistors, sources, and a single energy storage element may be converted to a Thevenin or Norton equivalent as seen from the two terminals of the inductor or capacitor. This reduces the complex circuit to a simple RC or RL circuit which may be solved according to the methods described in the previous sections. If a source in the circuit is suddently switched to a dc value, the resulting currents and voltages are exponentials, sharing the same time constant with possibly di erent initial and nal values. The time constant of the circuit is either RC or L=R, where R is the resistance in the Thevenin equivalent of the circuit as seen by the capacitor or inductor. Generate ANSI/AIM Code 39 In None Using Barcode generation for Software Control to generate, create Code 3 of 9 image in Software applications. Printing Data Matrix ECC200 In None Using Barcode creation for Software Control to generate, create Data Matrix image in Software applications. EXAMPLE 7.5 Find i, v, and i1 in Fig. 711(a). Barcode Creator In None Using Barcode generation for Software Control to generate, create barcode image in Software applications. Code 128B Generation In None Using Barcode generation for Software Control to generate, create Code 128 Code Set A image in Software applications. Fig. 711 The Thevenin equivalent of the circuit to the left of the inductor is shown in Fig. 711(b) with RTh 4 Painting ITF14 In None Using Barcode maker for Software Control to generate, create ITF14 image in Software applications. Generating GS1128 In Visual Basic .NET Using Barcode encoder for VS .NET Control to generate, create GS1 128 image in .NET applications. and vTh 3u t (V). The time constant of the circuit is L=RTh 5 10 3 =4 s 1:25 ms. The initial value of the inductor current is zero. Its nal value is i 1 Therefore, i 0:75 1 e 800t u t A v L di 3e 800t u t dt V i1 9 v 1 3 e 800t u t A 12 4 vTh 3V 0:75 A RTh 4 Scan Bar Code In Visual Basic .NET Using Barcode decoder for .NET framework Control to read, scan read, scan image in .NET framework applications. Create ANSI/AIM Code 39 In C# Using Barcode generation for Visual Studio .NET Control to generate, create Code 39 Full ASCII image in .NET applications. v can also be derived directly from its initial value v 0 9 6 = 12 6 3 V, its nal value v 1 0 and the circuit s time constant. EAN13 Creation In None Using Barcode maker for Office Excel Control to generate, create European Article Number 13 image in Microsoft Excel applications. EAN 13 Recognizer In VB.NET Using Barcode reader for VS .NET Control to read, scan read, scan image in Visual Studio .NET applications. CHAP. 7] Code 3 Of 9 Generator In .NET Framework Using Barcode generation for Visual Studio .NET Control to generate, create Code39 image in Visual Studio .NET applications. Code 39 Extended Creator In None Using Barcode encoder for Microsoft Word Control to generate, create Code 39 Extended image in Office Word applications. FIRSTORDER CIRCUITS
EXAMPLE 7.6 In Fig. 712 the 9mF capacitor is connected to the circuit at t 0. is v0 17 V. Find vA , vB , vC , iAB , iAC , and iBC for t > 0: At this time, capacitor voltage
Fig. 712 Apply KCL at nodes A, B, and C for t > 0 to nd voltages in term of i: 1 1 1 1 1 or 6vA 3vB vC 0 vA vB vC 0 Node A: 2 3 6 2 6 1 1 1 1 v 103 i vC 0 or 2vA 3vB vC 4 103 i Node B: vA 2 2 4 B 4 1 1 1 1 1 or 2vA 3vB 6vC 0 Node C: vA vB v 0 6 4 4 6 12 C Solving (11), (12), and (13) simultaneously, vA 7 103 i 3 vB 34 103 i 9 vC 8 103 i 3 (11) (12) (13) The circuit as seen by the capacitor is equivalent to a resistor R vB =i 34=9 k
. The capacitor discharges its initial voltage V0 in an exponential form with a time constant RC 34 103 9 10 6 0:034 s. For t > 0, 9 the voltages and currents are vB V0 e t= 17e 1000t=34 i C V A V A A A dvB 9 17 10 3 =34 e 1000t=34 4:5 10 3 e 1000t=34 dt V vC 8 103 i 12e 1000t=34 3
vA 7 103 i 10:5e 1000t=34 3 vAB vA vB 6:5e 1000t=34 vAC vA vC 1:5e vBC vB vC 5e 1000t=34 1000t=34
V V
iAB vAB =2000 3:25 10 3 e 1000t=34 iAC vAC =6000 0:25 10 3 e 1000t=34 iBC vBC =4000 1:25 10 e 3 1000t=34 All voltages and currents are exponential functions and have the same time constant. For simplicity, it is customary to use units of V, mA, k , and ms for voltage, current, resistance, and time, respectively, so that the multipliers 1000 and 10 3 can be omitted from the equations as summarized below. vA 10:5e t=34 vB 17e vC 12e t=34 t=34 t=34
V V V mA
vAB 6:5e t=34 vAC 1:5e vBC 5e
t=34 t=34
V V
iAB 3:25e t=34 iAC 0:25e iBC 1:25e
t=34 t=34
mA mA mA
i 4:5e
FIRSTORDER CIRCUITS
[CHAP. 7
DC STEADY STATE IN INDUCTORS AND CAPACITORS
As noted in Section 7.1, the natural exponential component of the response of RL and RC circuits to step inputs diminishes as time passes. At t 1, the circuit reaches steady state and the response is made of the forced dc component only. Theoretically, it should take an in nite amount of time for RL or RC circuits to reach dc steady state. However, at t 5, the transient component is reduced to 0.67 percent of its initial value. After passage of 10 time constants the transient component equals to 0.0045 percent of its initial value, which is less than 5 in 100,000, at which time for all practical purposes we may assume the steady state has been reached. At the dc steady state of RLC circuits, assuming no sustained oscillations exist in the circuit, all currents and voltages in the circuit are constants. When the voltage across a capacitor is constant, the current through it is zero. All capacitors, therefore, appear as open circuits in the dc steady state. Similarly, when the current through an inductor is constant, the voltage across it is zero. All inductors therefore appear as short circuits in the dc steady state. The circuit will be reduced to a dcresistive case from which voltages across capacitors and currents through inductors can be easily found, as all the currents and voltages are constants and the analysis involves no di erential equations. The dc steadystate behavior presented in the preceding paragraph is valid for circuits containing any number of inductors, capacitors, and dc sources. EXAMPLE 7.7 Find the steadystate values of iL , vC1 , and vC2 in the circuit of Fig. 713(a). When the steady state is reached, the circuit will be as shown in Fig. 713(b). The inductor current and capacitor voltages are obtained by applying KCL at nodes A and B in Fig. 713(b). Thus, Node A: Node B: vA vA vB vA 18 vB 3 3 6 6 vB vB vA vB 18 vA 0 12 6 6 or or 2vA vB 0 4vA 5vB 36 Solving for vA and vB we nd vA 6 V and vB 12 V. By inspection of Fig. 713(b), we have iL 2 mA, vC1 8 V, and vC2 6 V. EXAMPLE 7.8 Find i and v in the circuit of Fig. 714. At t 0, the voltage across the capacitor is zero. Its nal value is obtained from dc analysis to be 2 V. The time constant of the circuit of Fig. 714, as derived in Example 7.6, is 0.034 s. Therefore, v 2 1 e 1000t=34 u t i C dv 9 10 2 10 1000t=34 e u t dt 34 6 3

