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barcode reader code in asp.net Figure 514 Abrupt change in capacitor voltage in Software
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Draw Leitcode In None Using Barcode drawer for Software Control to generate, create Leitcode image in Software applications. Bar Code Scanner In VB.NET Using Barcode reader for .NET framework Control to read, scan read, scan image in Visual Studio .NET applications. Known Quantities: Source current, IS ; inductor and resistor values Find: Inductor current at t = 0+ and as t Schematics, Diagrams, Circuits, and Given Data: IS = 10 mA Assumptions: The current source has been connected to the circuit for a very long time Figure 515 Printing Bar Code In .NET Framework Using Barcode generator for ASP.NET Control to generate, create barcode image in ASP.NET applications. Create Code128 In ObjectiveC Using Barcode printer for iPhone Control to generate, create Code 128 Code Set A image in iPhone applications. 5
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long time, the inductor acts as a short circuit, and iL (0 ) = IS Since all the current ows through the inductor, the voltage across the resistor must be zero At t = 0+ , the switch opens and we can state that iL (0+ ) = iL (0 ) = IS because of the continuity of inductor current The circuit for t 0 is shown in Figure 516, where the presence of the current iL (0+ ) denotes the initial condition for the circuit A qualitative sketch of the current as a function of time is also shown in Figure 516, indicating that the inductor current eventually becomes zero as t Comments: Note that the direction of the current in the circuit of Figure 516 is dictated
Analysis: At t = 0 , since the current source has been connected to the circuit for a very
iL(t) 10 mA
by the initial condition, since the inductor current cannot change instantaneously Thus, the current will ow counterclockwise, and the voltage across the resistor will therefore have the polarity shown in the gure Complete Solution of FirstOrder Circuits In this section, we illustrate the application of the principles put forth in the preceding sections by presenting a number of examples The rst example summarizes the complete solution of a simple RC circuit EXAMPLE 56 Complete Solution of FirstOrder Circuit
Problem
t=0 R + 12 V i(t) vC (t) _ C vC (0) = 5 V
Determine an expression for the capacitor voltage in the circuit of Figure 517
Solution
Known Quantities: Initial capacitor voltage; battery voltage, resistor and capacitor
values
Figure 517 Find: Capacitor voltage as a function of time, vC (t), for all t
C = 470 F; VB = 12 V
Assumptions: None
Schematics, Diagrams, Circuits, and Given Data: vC (t = 0 ) = 5 V; R = 1 k ; Analysis: We rst observe that the capacitor had previously been charged to an initial voltage of 5 V Thus, vC (t) = 5 V
t <0
At t = 0 the switch closes, and the circuit is described by the following differential equation, obtained by application of KVL: VB RC dvC (t) vC (t) = 0 dt t >0 t >0 1 dvC (t) 1 + vC (t) = VB dt RC RC
Part I
Circuits
In the above equation we recognize the following variables, with reference to equation 522: 1 = RC f (t) = t >0s VB x = vC RC The natural response of the circuit is therefore of the form: xN (t) = vCN (t) = Ke t/ = Ke t/RC while the forced response is of the form: xF (t) = vCF (t) = f (t) = VB t > 0 s t > 0 s, Thus, the complete response of the circuit is given by the expression x(t) = vC (t) = vCN (t) + vCF (t) = Ke t/RC + VB t >0s Now that we have the complete response, we can apply the initial condition to determine the value of the constant K At time t = 0, vC (0) = 5 = Ke 0/RC + VB K = 5 12 = 7 V We can nally write the complete response with numerical values: vC (t) = 7e t/047 + 12 V = vC T (t) + vCSS (t) = 12 1 e t/047 + 5e t/047 V = vCF (t) + vCN (t) The complete response described by the above equations is shown graphically in Figure 518 (a) and (b) 15 10 5 Volts Volts 0 5 10 0 02 04 06 08 1 12 14 16 18 2 Time, (s) (a) vC (t) vCT (t) vCSS (t) 15 10 5 0 5 10 0 02 04 06 08 1 12 14 16 18 2 Time, (s) (b) vC (t) vCN (t) vCF (t) t >0s
t >0s
Figure 518 (a) Complete, transient, and steadystate responses of the circuit of Figure 517 (b) Complete, natural, and forced responses of the circuit of Figure 517 Comments: Note how in Figure 518(a) the steadystate response vCSS (t) is simply equal to the battery voltage, while the transient response, vCT (t), rises from 7 V to 0 V exponentially In Figure 518(b), on the other hand, we can see that the energy initially stored in the capacitor decays to zero via its natural response, vCN (t), while the external forcing function causes the capacitor voltage to eventually rise exponentially to 12 V, as shown in the forced response, VCF (t) The example just completed, though based on a very simple circuit, illustrates all the steps required to complete the solution of a rstorder circuit The methodology applied in the example is summarized in a box, next

