811-2 The linear transformer R, R2 M II(R+jOlD-12 joM 11)43 L2 I2)ki0=-l1 jaM+I2(R2 jaL,+z primary sec ondary Zu=r+jal z2=r2+jal,+ZL jaM ( 1) oMI 2 0=-1njoM+I222(2) gam) M oM In 11 11 11 22 22 R22+22 R212 aM(R22-jx MR oMX 2 R22+X22R22+X 22 R1+
§11-2 The linear transformer = − + + + = + − • • • • • 0 ( ) ( ) 2 2 1 2 2 1 1 1 L s I j M I R j L Z V I R j L I j M L ZL Z11 = R1 + jL1 Z22 = R2 + j 2 + = − + = − • • • • • 0 (2) (1) 22 1 2 2 11 1 I j M I Z V s I Z I j M 22 1 2 Z j M I I • • = ' 11 1 ' 1 ' = Z11 + R1 + jX = Z + Z 22 2 11 1 ( ) Z j M Z I V Z s i n = = − • • 22 2 2 11 Z M Z = + 22 2 22 2 22 22 2 2 11 ( ) R X M R jX Z + − = + 22 2 22 2 22 2 2 22 2 22 2 22 2 2 11 R X j M X R X M R Z + − + = + 22 22 2 2 11 R jX M Z + = +
02M2R R reflected resis tance R22+X 2 22 MX R2+X2 reflected reac tance Z=R+jx reflected impedance R R L LX JoM/ JOMI z1+21 22
refletted reac tance R X M X X ' refletted resistance R X M R R ' 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 1 + = − + = Z R jX refletted impedance ' ' ' 1 = 1 + 1 ' 11 1 1 Z Z V I s + = • • 22 1 2 Z j M I I • • =
The t equivalent of the transformer In side(同侧): V1=jOL 11+joM 12 1) v2=jOMI1+jOL 12 (2) =11+12(3) M (3)→>2 1 L jOL I1+jaM(-11 jOL-MI1+joM I (3)→> i1=i-i2→(2) +L1-M3L2-M+ 2=joM(-12)+JaL 12 +M jaMI+jo( -/2 Out of side(异侧):(L1+M,L2+M,M
The T equivalent of the transformer 1 2 (1) 1 1 • • • V = jL I + jM I 2 (2) 2 2 1 • • • V = jM I + jL I 1 2 (3) • • • I = I + I (3) → 2 = − 1 → (1) • • • I I I • • = j L − M I 1+ jM I 1 ( ) (3) → 1 = − 2 → (2) • • • I I I 2 2 ( ) • • = jM I+ j L − M I 1 ( 1 ) 1 1 • • • • V = jL I + jM I− I 2 2 2 ( 2 ) • • • • V = jM I− I + jL I In side(同侧): Out of side (异侧): (L1+M, L2+M,-M)
Example 2: Find the three inductance appearing in the T equivalent of the transformer shown in Fig, if L2=4H and the input inductance at A-B is 6H with C-D open-circuited, and 2H with C-D short-circuited L+M Lr+M M 日 日 Solution: C-D open-circuited L,=4H LAD=L +MM6 L1=6H C-D short-circuited L=6+M+ 4+M)(-M) =2∴M=4H 4+M-M L+M=10H,L,+M=8H,-M=-4H
Example 2: Find the three inductance appearing in the T equivalent of the transformer shown in Fig., if L2=4H and the input inductance at A-B is 6H with C-D open-circuited, and 2H with C-D short-circuited. Solution: C-D open-circuited L2 = 4H LAB=L1+M-M=6 L1 = 6H C-D short-circuited M H M M M M LAB M 2 4 4 (4 )( ) 6 = = + − + − = + + L1+M=10H, L2+M= 8H, -M= - 4H
DP: Let li=0.1h.L=0,4h and M012H in the transformer of Fig. Find the input inductance at terminals A-B. If(a)c is connected to D; (b)C is connected to A, and d to B;(c)Cis connected to b. and d to a A M L+m l+ A M 0.22H0.52H M 0.12H Solution: (a):LAB=0.064H (b):LAB=0.0346H D AIM ): YD 002H0.28H L13(L2 +M LAB=0.0985H 0.12H 日
DP: Let L1=0.1H, L2=0.4H, and M=0.12H in the transformer of Fig.. Find the input inductance at terminals A-B. If (a) C is connected to D; (b) C is connected to A, and D to B; (c) C is connected to B, and D to A. Solution: (c): (a): LAB=0.064H (b): LAB=0.0346H LAB=0.0985H