barcode reader code in asp.net Figure 514 Abrupt change in capacitor voltage in Software

Make Denso QR Bar Code in Software Figure 514 Abrupt change in capacitor voltage

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

5

iL(0+)

iL(t)

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

iL(t)

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 First-Order 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

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,

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

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)

Figure 518 (a) Complete, transient, and steady-state 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 steady-state 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 rst-order circuit The methodology applied in the example is summarized in a box, next