R2 i2 L2

Scan QR Code ISO/IEC18004 In NoneUsing Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications.

Printing Quick Response Code In NoneUsing Barcode encoder for Software Control to generate, create QR Code ISO/IEC18004 image in Software applications.

Fig. 14-6

Decode QR-Code In NoneUsing Barcode scanner for Software Control to read, scan read, scan image in Software applications.

Create Quick Response Code In Visual C#.NETUsing Barcode creation for VS .NET Control to generate, create QR Code JIS X 0510 image in VS .NET applications.

MUTUAL INDUCTANCE AND TRANSFORMERS

QR Generation In .NET FrameworkUsing Barcode printer for ASP.NET Control to generate, create QR Code image in ASP.NET applications.

Painting QR Code JIS X 0510 In VS .NETUsing Barcode generation for VS .NET Control to generate, create Quick Response Code image in .NET applications.

[CHAP. 14

Generating QR Code In VB.NETUsing Barcode creator for .NET framework Control to generate, create QR Code image in .NET framework applications.

ECC200 Generator In NoneUsing Barcode maker for Software Control to generate, create ECC200 image in Software applications.

while that of the active loop is R1 i1 L1 di1 di M 2 v1 dt dt

Drawing Code 128 Code Set B In NoneUsing Barcode encoder for Software Control to generate, create Code 128 Code Set A image in Software applications.

UPC A Maker In NoneUsing Barcode generation for Software Control to generate, create UPCA image in Software applications.

Writing the above equations in the s-domain with the initial conditions i1 0 i2 0 0 and eliminating I1 s , we nd H s response I s Ms 2 excitation V1 s L1 L2 M 2 s2 R1 L2 R2 L1 s R1 R2

GTIN - 128 Creator In NoneUsing Barcode maker for Software Control to generate, create UCC - 12 image in Software applications.

Drawing Bar Code In NoneUsing Barcode maker for Software Control to generate, create bar code image in Software applications.

and from the poles of H s we have the natural frequencies of i2 .

Printing UCC - 14 In NoneUsing Barcode printer for Software Control to generate, create ITF-14 image in Software applications.

Making Data Matrix ECC200 In VS .NETUsing Barcode generation for ASP.NET Control to generate, create DataMatrix image in ASP.NET applications.

DOT RULE

Drawing 2D Barcode In Visual C#Using Barcode generator for .NET framework Control to generate, create Matrix 2D Barcode image in .NET framework applications.

Bar Code Encoder In NoneUsing Barcode creation for Excel Control to generate, create bar code image in Excel applications.

The sign on a voltage of mutual inductance can be determined if the winding sense is shown on the circuit diagram, as in Figs. 14-4 and 14-5. To simplify the problem of obtaining the correct sign, the coils are marked with dots at the terminals which are instantaneously of the same polarity. To assign the dots to a pair of coupled coils, select a current direction in one coil and place a dot at the terminal where this current enters the winding. Determine the corresponding ux by application of the right-hand rule [see Fig. 14-7(a)]. The ux of the other winding, according to Lenz s law, opposes the rst ux. Use the right-hand rule to nd the natural current direction corresponding to this second ux [see Fig. 14-7(b)]. Now place a dot at the terminal of the second winding where the natural current leaves the winding. This terminal is positive simultaneously with the terminal of the rst coil where the initial current entered. With the instantaneous polarity of the coupled coils given by the dots, the pictorial representation of the core with its winding sense is no longer needed, and the coupled coils may be illustrated as in Fig. 14-7(c). The following dot rule may now be used: (1) (2) when the assumed currents both enter or both leave a pair of coupled coils by the dotted terminals, the signs on the M-terms will be the same as the signs on the L-terms; but if one current enters by a dotted terminal while the other leaves by a dotted terminal, the signs on the M-terms will be opposite to the signs on the L-terms.

UCC-128 Encoder In NoneUsing Barcode creation for Font Control to generate, create EAN / UCC - 13 image in Font applications.

DataMatrix Printer In JavaUsing Barcode printer for Eclipse BIRT Control to generate, create Data Matrix ECC200 image in BIRT applications.

Fig. 14-7 EXAMPLE 14.3 The current directions chosen in Fig. 14-8(a) are such that the signs on the M-terms are opposite to the signs on the L-terms and the dots indicate the terminals with the same instantaneous polarity. Compare this to the conductively coupled circuit of Fig. 14-8(b), in which the two mesh currents pass through the common element in opposite directions, and in which the polarity markings are the same as the dots in the magnetically coupled circuit. The similarity becomes more apparent when we allow the shading to suggest two black boxes.

Print Code 128A In JavaUsing Barcode drawer for Java Control to generate, create Code 128B image in Java applications.

Bar Code Creation In Objective-CUsing Barcode maker for iPhone Control to generate, create barcode image in iPhone applications.

ENERGY IN A PAIR OF COUPLED COILS The energy stored in a pair

The energy stored in a single inductor L carrying current i is 0.5Li2 J. of coupled coils is given by

CHAP. 14]

MUTUAL INDUCTANCE AND TRANSFORMERS

Fig. 14-8

1 1 2 2 W L1 i1 L2 i2 Mi1 i2 2 2

where L1 and L2 are the inductances of the two coils and M is their mutual inductance. The term Mi1 i2 in (9) represents the energy due to the e ect of the mutual inductance. The sign of this term is (a) positive if both currents i1 and i2 enter either at the dotted or undotted terminals, or (b) negative if one of the currents enters at the dotted terminal and the other enters the undotted end.

EXAMPLE 14.4 In a pair of coils, with L1 0:1 H and L2 0:2 H, at a certain moment, i1 4 A and p i2 10 A.p Find the total energy in the coils if the coupling coe cient M is (a) 0.1 H, (b) 2=10 H, (c) 0:1 H, and (d) 2=10 H. From (9), a b c d W 0:5 0:1 42 0:5 0:2 102 0:1 10 4 14:8 J W 16:46 J W 6:8 J W 5:14 J p 2=10 and perfect

The maximum and minimum energies occur in conjunction with perfect positive coupling M p negative coupling M 2=10 .