ssrs barcodelib FIRST-ORDER CIRCUITS in Software

Encoder QR Code JIS X 0510 in Software FIRST-ORDER CIRCUITS

FIRST-ORDER CIRCUITS
Recognizing Quick Response Code In None
Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications.
Encoding QR Code In None
Using Barcode generation for Software Control to generate, create QR Code ISO/IEC18004 image in Software applications.
v2 t
QR Code JIS X 0510 Scanner In None
Using Barcode scanner for Software Control to read, scan read, scan image in Software applications.
Creating Quick Response Code In Visual C#.NET
Using Barcode drawer for VS .NET Control to generate, create QR Code 2d barcode image in VS .NET applications.
R2 t= R1 C e u t R1
Printing QR In Visual Studio .NET
Using Barcode generation for ASP.NET Control to generate, create Denso QR Bar Code image in ASP.NET applications.
Make QR In VS .NET
Using Barcode generation for Visual Studio .NET Control to generate, create Quick Response Code image in .NET framework applications.
Alternate Approach
QR Code Printer In Visual Basic .NET
Using Barcode generator for .NET framework Control to generate, create QR Code 2d barcode image in .NET applications.
Creating Code 39 Full ASCII In None
Using Barcode creator for Software Control to generate, create Code 39 image in Software applications.
The unit step response may also be found by the Laplace transform method (see 16). EXAMPLE 7.17 Passive phase shifter. Find the relationship between v2 and v1 in the circuit of Fig. 7-45(a). Let node D be the reference node. Apply KCL at nodes A and B to nd KCL at node A: KCL at node B: C dvA vA v1 0 R dt d vB v1 vB C 0 dt R
DataMatrix Maker In None
Using Barcode drawer for Software Control to generate, create Data Matrix image in Software applications.
Making EAN13 In None
Using Barcode generator for Software Control to generate, create EAN-13 image in Software applications.
Subtracting the second equation from the rst and noting that v2 vA vB we get v2 RC dv2 dv v1 RC 1 dt dt
Create Barcode In None
Using Barcode maker for Software Control to generate, create barcode image in Software applications.
Universal Product Code Version A Encoder In None
Using Barcode printer for Software Control to generate, create GTIN - 12 image in Software applications.
EXAMPLE 7.18 Active phase shifter. Show that the relationship between v2 and v1 in the circuit of Fig. 7-45(b) is the same as in Fig. 7-45(a). Apply KCL at the inverting (node A) and non-inverting (node B) inputs of the op amp. KCL at node A: KCL at node B: vA v1 vA v2 0 R1 R1 vB v1 dv C B 0 R dt Substituting
USPS OneCode Solution Barcode Maker In None
Using Barcode generation for Software Control to generate, create USPS Intelligent Mail image in Software applications.
Making Bar Code In Objective-C
Using Barcode printer for iPad Control to generate, create barcode image in iPad applications.
From the op amp we have vA vB and from the KCL equation for node A, we have vA v1 v2 =2. the preceding values in the KCL at node B, we nd v2 RC dv2 dv v1 RC 1 dt dt
UCC - 12 Drawer In .NET Framework
Using Barcode creator for ASP.NET Control to generate, create EAN / UCC - 13 image in ASP.NET applications.
Code 3 Of 9 Creator In .NET Framework
Using Barcode printer for Visual Studio .NET Control to generate, create Code 3 of 9 image in .NET framework applications.
Solved Problems
Making Barcode In None
Using Barcode encoder for Online Control to generate, create bar code image in Online applications.
Bar Code Maker In Java
Using Barcode creator for Android Control to generate, create bar code image in Android applications.
7.1 At t 0 , just before the switch is closed in Fig. 7-20, vC 100 V. charge transients. Obtain the current and
UPC-A Supplement 5 Recognizer In None
Using Barcode reader for Software Control to read, scan read, scan image in Software applications.
Barcode Recognizer In VB.NET
Using Barcode Control SDK for Visual Studio .NET Control to generate, create, read, scan barcode image in VS .NET applications.
Fig. 7-20 With the polarities as indicated on the diagram, vR vC for t > 0, and 1=RC 62:5 s 1 . vC 0 vC 0 100 V. Thus, vR vC 100e 62:5t V i vR 0:25e 62:5t R A q CvC 4000e 62:5t mC Also,
FIRST-ORDER CIRCUITS
[CHAP. 7
In Problem 7.1, obtain the power and energy in the resistor, and compare the latter with the initial energy stored in the capacitor.
pR vR i 25e 125t W t t wR pR dt 25e 125t dt 0:20 1 e 125t
The initial stored energy is
2 W0 1 CV0 1 40 10 6 100 2 J 0:20 wR 1 2 2
In other words, all the stored energy in the capacitor is eventually delivered to the resistor, where it is converted into heat.
An RC transient identical to that in Problems 7.1 and 7.2 has a power transient pR 360e t=0:00001 Obtain the initial charge Q0 , if R 10
2 105 pR P0 e 2t=RC or or C 2 mF RC t wR pR dt 3:6 1 e t=0:00001 mJ
Then, wR 1 3:6 mJ
Q2 =2C, 0
from which Q0 120 mC.
The switch in the RL circuit shown in Fig. 7-21 is moved from position 1 to position 2 at t 0. Obtain vR and vL with polarities as indicated.
Fig. 7-21 The constant-current source drives a current through the inductance in the same direction as that of the transient current i. Then, for t > 0, i I0 e Rt=L 2e 25t vR Ri 200e
25t
25t
vL vR 200e
For the transient of Problem 7.4 obtain pR and pL .
pR vR i 400e 50t pL vL i 400e
50t
W W And, since
Negative power for the inductance is consistent with the fact that energy is leaving the element. this energy is being transferred to the resistance, pR is positive.
CHAP. 7]
FIRST-ORDER CIRCUITS
A series RC circuit with R 5 k
and C 20 mF has a constant-voltage source of 100 V applied at t 0; there is no initial charge on the capacitor. Obtain i, vR , vC , and q, for t > 0.
The capacitor charge, and hence vC , must be continuous at t 0: vC 0 vC 0 0 As t ! 1, vC ! 100 V, the applied voltage. from Section 6.10, The time constant of the circuit is  RC 10 1 s. Hence,
vC vC 0 vC 1 e t= vC 1 100e 10t 100 V The other functions follow from this. enters, vR vC 100 V, and so If the element voltages are both positive where the current
vR 100e 10t V v i R 20e 10t mA R q CvC 2000 1 e 10t
mC
The switch in the circuit shown in Fig. 7-22(a) is closed at t 0, at which moment the capacitor has charge Q0 500 mC, with the polarity indicated. Obtain i and q, for t > 0, and sketch the graph of q.
Fig. 7-22 The initial charge has a corresponding voltage V0 Q0 =C 25 V, whence vC 0 25 V. The sign is negative because the capacitor voltage, in agreement with the positive direction of the current, would be on the top plate. Also vC 1 50 V and  0:02 s. Thus, as in Problem 7.6, vC 75e 50t 50 V from which q CvC 1500e 50t 1000 mC i dq 75e 50t dt mA
The sketch in Fig. 7-22(b) shows that the charge changes from 500 mC of one polarity to 1000 mC of the opposite polarity.
Obtain the current i, for all values of t, in the circuit of Fig. 7-23.
For t < 0, the voltage source is a short circuit and the current source shares 2 A equally between the two 10-
resistors:
Copyright © OnBarcode.com . All rights reserved.