how to generate barcode in ssrs report Select mesh currents I1 and I2 as shown on the diagram and write the KVL equations in matrix form. in Software

Generate QR Code 2d barcode in Software Select mesh currents I1 and I2 as shown on the diagram and write the KVL equations in matrix form.

Select mesh currents I1 and I2 as shown on the diagram and write the KVL equations in matrix form.
Decode QR-Code In None
Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications.
QR Code ISO/IEC18004 Creator In None
Using Barcode printer for Software Control to generate, create Denso QR Bar Code image in Software applications.
MUTUAL INDUCTANCE AND TRANSFORMERS
Decoding QR Code ISO/IEC18004 In None
Using Barcode decoder for Software Control to read, scan read, scan image in Software applications.
QR Code Creator In C#.NET
Using Barcode drawer for Visual Studio .NET Control to generate, create Denso QR Bar Code image in .NET applications.
[CHAP. 14
QR Code 2d Barcode Generation In VS .NET
Using Barcode creation for ASP.NET Control to generate, create QR Code image in ASP.NET applications.
Quick Response Code Drawer In Visual Studio .NET
Using Barcode maker for Visual Studio .NET Control to generate, create QR image in VS .NET applications.
Fig. 14-30
Generating Quick Response Code In VB.NET
Using Barcode maker for VS .NET Control to generate, create Denso QR Bar Code image in .NET applications.
UPC-A Supplement 5 Maker In None
Using Barcode maker for Software Control to generate, create UCC - 12 image in Software applications.
Fig. 14-31  3 j1 3 j2 3 j2 8 j6  I1 I2   50 08 0 
Drawing GS1-128 In None
Using Barcode drawer for Software Control to generate, create GS1-128 image in Software applications.
Code128 Encoder In None
Using Barcode creation for Software Control to generate, create Code128 image in Software applications.
The impedances in Fig. 14-32 are selected to give the identical Z-matrix. Thus, since I1 and I2 pass through the common impedance, Zb , in opposite directions, Z12 in the matrix is Zb . Then, Zb 3 j2 . Since Z11 is to include all impedances through which I1 passes, 3 j1 Za 3 j2 from which Za j1 . Similarly, Z22 8 j6 Zb Zc and Zc 5 j4 .
Paint GTIN - 13 In None
Using Barcode encoder for Software Control to generate, create EAN-13 Supplement 5 image in Software applications.
DataMatrix Maker In None
Using Barcode creator for Software Control to generate, create Data Matrix 2d barcode image in Software applications.
Fig. 14-32
Generate EAN-8 Supplement 5 Add-On In None
Using Barcode creator for Software Control to generate, create EAN8 image in Software applications.
Data Matrix ECC200 Decoder In Visual Basic .NET
Using Barcode scanner for Visual Studio .NET Control to read, scan read, scan image in VS .NET applications.
14.15 For the transformer circuit of Fig. 14-11(b), k 0:96, R1 1:2 , R2 0:3 , X1 20 , X2 5 , ZL 5:0 36:878 , and V2 100 08 V. Obtain the coil emfs E1 and E2 , and the magnetizing current I .
Code 128C Creator In None
Using Barcode creator for Online Control to generate, create Code 128 Code Set C image in Online applications.
GS1 - 12 Creation In None
Using Barcode creation for Font Control to generate, create UPC-A Supplement 5 image in Font applications.
X11 1 k X1 1 0:96 20 0:8  X22 1 k X2 0:2  s p X1 2 XM k X1 X2 9:6  a X2
EAN-13 Maker In Java
Using Barcode encoder for Java Control to generate, create EAN-13 image in Java applications.
Bar Code Decoder In Visual Studio .NET
Using Barcode Control SDK for ASP.NET Control to generate, create, read, scan barcode image in ASP.NET applications.
CHAP. 14]
Painting Code 3 Of 9 In Java
Using Barcode encoder for Android Control to generate, create Code 39 Full ASCII image in Android applications.
Paint Bar Code In Java
Using Barcode generator for Java Control to generate, create barcode image in Java applications.
MUTUAL INDUCTANCE AND TRANSFORMERS
Now a circuit of the form Fig. 14-11(a) can be constructed, starting from the phasor voltage-current relationship at the load, and working back through E2 to E1 . V2 100 08 20 36:878 A ZL 5:0 36:878 E2 I2 R2 jX22 V2 20 36:878 0:3 j0:2 100 08 107:2 j0:4 V I2 E1 aE2 214:4 j0:8 V E I 2 0:042 j11:17 A jXM
14.16 For the linear transformer of Problem 14.15, calculate the input impedance at the terminals where V1 is applied.
Method 1 Completing the construction begun in Problem 14.15, 1 I 0:042 j11:17 10 36:878 18:93 65:138 A a 2 V1 I1 R1 jX11 E1 18:93 65:138 1:2 j0:8 214:4 j0:8 I1 I 238:2 3:628 V Therefore, Zin Method 2 By (14a) of Example 14.5, Zin 1:2 j0:8 22 Method 3 By (14b) of Example 14.5, Zin 1:2 j20 9:6 2 0:3 j5 5:0 36:878 1:2 j20 4:80 j8:94 12:58 61:538  j4:8 0:3 j0:2 5:0 36:878 0:3 j5:0 5:0 36:878 V1 238:2 3:628 12:58 61:518  I1 18:93 65:138
114:3 123:258 12:58 61:508  9:082 61:758
14.17 In Fig. 14-33, three identical transformers are primary wye-connected and secondary delta-connected. A single load impedance carries current IL 30 08 A. Given
Ib2 20 08 A Ia2 Ic2 10 08 A
and N1 10N2 100, nd the primary currents Ia1 , Ib1 , Ic1 . The ampere-turn dot rule is applied to each transformer. N1 Ia1 N2 Ia2 0 N1 Ib1 N2 Ib2 0 N1 Ic1 N2 Ic2 0 or or or 10 10 08 1 08 A 100 10 Ib1 20 08 2 08 A 100 10 Ic1 10 08 1 08 A 100 Ia1
MUTUAL INDUCTANCE AND TRANSFORMERS
[CHAP. 14
Fig. 14-33 The sum of the primary currents provides a check: Ia1 Ib1 Ic1 0
14.18 For the ideal autotransformer shown in Fig. 14-34, nd V2 , Icb , and the input current I1 .
a V2 V1 100 08 V a 1 N1 1 N2 2 IL Iab V2 10 608 A ZL I L 6:67 608 A a 1
Icb IL Iab 3:33 608 A
Fig. 14-34
14.19 In Problem 14.18, nd the apparent power delivered to the load by transformer action and that supplied by conduction.
Scond 1 V2 I 1 100 08 6:67 608 333 608 VA ab 2 2 Strans aScond 167 608 VA
CHAP. 14]
MUTUAL INDUCTANCE AND TRANSFORMERS
14.20 In the coupled circuit p Fig. 14-35, nd the input admittance Y1 I1 =V1 and determine the of current i1 t for v1 2 2 cos t.
Fig. 14-35 Apply KVL around loops 1 and 2 in the s-domain. V1 sI1 sI2 I1 I2 s I2 I1 s
0 sI1 2s 1 I2 Eliminating I2 in these equations results in
I1 2s2 s 1 3 V1 s s2 5s 1 p For s j, the input admittance is Y1 1 j =4 2=4 458. Therefore, i1 t cos t 458 . Y1
14.21 Find the input impedance Z1 V1 =I1 in the coupled circuit of Fig. 14-36.
Fig. 14-36 Apply KVL around loops 1 and 2 in the s-domain. ( V1 sI1 1 sI2 2 I1 I2 3 1 0 1 sI1 1 sI2 2 I1 I2 12 sI2 3 4 ( or The result is I2 I1 and Z1 V1 2 s I1 3 The input impedance is V1 2 s I1 2 1 s I2 3 0 2 1 s I1 2 1 s I2 3 3
The current through the resistor is I1 I2 0 and the resistor has no e ect on Z1 . purely inductive.
Copyright © OnBarcode.com . All rights reserved.