811-3 The ideal transformer An ideal transformer is a useful approximation of a very tightly coupled transformer in which the coefficient of coupling is almost unity and both the primary and secondary inductive reactance are extremely large in comparison with the terminating impedance (kxl,joL>>Eiland jOL2>>ZL 2 turns ratio 1=I1jOL-12 jaM (1)0=-I1jaM +I2(,+jOL,)(2) lIam 12M2 (2)→I Z1+/2→>(①) V1=jal+ Z+jOL
§11-3 The ideal transformer An ideal transformer is a useful approximation of a very tightly coupled transformer in which the coefficient of coupling is almost unity and both the primary and secondary inductive reactance are extremely large in comparison with the terminating impedance. ( ) L L ZL k 1, jL1 Z and j 2 turns ratio N N a a N N L L = − − = = 1 2 2 2 1 2 2 1 2 ( ) ( ) 2 (1) 1 V1 I 1 jL I jM • • • = − 0 ( ) (2) 2 = − I 1 jM + I 2 ZL + jL • • (2) (1) 2 1 2 → + → = • • Z j L I j M I L 2 2 2 1 1 1 1 Z j L I M V I j L L + = + • • •
2 V1=l1jOL,+ 10-M 1:a L +jaL? =JOL, 0M Z+JOL L 0+L,L JoL Zte L jOLIZL-OLL2+O'l,Lr jOLZL ZL+jaL L tya J If the amplifier internal impedance is If a100 4000Q2 and the speaker impedance is 8 Q2, L Z=20k then we desire that 8 4000 →a =22.4 224
2 2 2 1 1 1 1 Z j L I M V I j L L + = + • • • 2 2 2 1 1 1 Z j L M j L I V Z L i n + = = + • • 2 1 2 2 1 2 2 1 Z j L j L Z L L L L L L + − + = 2 1 2 1 2 1 a Z L L j L Z Z Z j L j L Z L L L L L = + = + = 1 • I − + • V 1 2 a ZL If a=100 ZL=20k Zin=2 If the amplifier internal impedance is 4000Ω and the speaker impedance is 8 Ω, then we desire that 22.4 8 1 4000 2 2 = = → a = a a ZL 22.4 2 1 = N N 2 1 2 2 1 Z j L L L j L L + = +
The relationship between I1 and I2 1:a 十 I .0=-l1jOM +12(ZL+jaL,) 2 = JOM M√L1L ZL+joLy l2 2 2 l ,N, or N,I=N2I2
The relationship between 1 2 . • • I and I 0 ( ) 2 = − I 1 jM + I 2 ZL + jL • • 2 1 2 Z j L j M I I L + = • • 2 2 1 1 2 1 1 2 • • • • = or N I = N I N N I I L a L L L L L M 1 2 1 2 1 2 2 = = = = N a N or 1 2 1 =
The relationship between V1 and v2 IIZ=l 2 2 N 2 =a and v22=vilI The primary and secondary complex volt-amperes are equal
The relationship between 1 2 . • • V and V 2 2 2 1 1 1 a Z V I Z V I Z I L L i n • • • • • = = = 2 1 2 1 2 a Z I I Z V V L L • • • • = 2 2 1 1 1 2 1 2 • • • • • • = = a and V I =V I N N V V The primary and secondary complex volt-amperes are equal. 1 2 2 • • = I I a 2 1 1 2 1 2 • • = = = = I I N N a a a
Example 2: Find the turns ratio a for the transformer in the circuit of Fig. so that:(a) the power delivered to the 10 resistor is 25 percent of that delivered to the 2 resistdQ2 for any independent sinusoidal source connected to x-y:(b) 2=s when an independent source Vs is connected to x-y 2Q Solution: 1 10=0.25l,×2 10gl2 n20 V20 a=447 2I1+V 2 =a or vI=-v2 1=a12=a 10 2×a 2 2 (V2=Vs)→2 10 5 a2-5a+5=0→a=3.62ora=1.38
Example 2: Find the turns ratio a for the transformer in the circuit of Fig. so that: (a) the power delivered to the 10 resistor is 25 percent of that delivered to the 2 resistor for any independent sinusoidal source connected to x-y: (b) when an independent source is connected to x-y. V V s • • 2 = V s • Solution: ( ) 10 0.25 2 2 1 2 a I2 = I 20 1 1 20 1 1 2 2 1 2 2 = → = = I a I I I a = 4.47 ( ) 2 1 1 • • • b V s = I +V 1 2 1 2 1 • • • • = = V a a or V V V 10 2 1 2 • • • = = V I a I a a V V V s a 2 2 10 2 • • • = + 5 5 0 3.62 1.38 2 a − a + = → a = or a = a aV V V V s V 2 2 2 2 5 ( ) • • • • • = → = + − + s V •
V)=100 (A)=2A 电感线圈 (W)=120W f= 50Hz Find R and L P R =120/4=309 2 Z|==100/2=50 oL=√502-304=40 L=40/314=0.127H
−+ L •I R •V 电感线圈 V A W . 50 ( ) 120 ( ) 2 ( ) 100 Find R and L f Hz W W A A V V = === = = 120 / 4 = 30 2 IP R 50 30 40 2 2 L = − = | |= = 100/ 2 = 50 IV Z L = 40/ 314 = 0.127H (1)
(2) R=50 C=5uF L=0.4HRr=100 =220√2cos500V Find Ps and es 400 2=RL+joL=100+j200 oc 320+1240Z=R+2122=370+1240=4433 Z1+Z Z1+22 Z12 =Wz=0.5∠-33° Ⅰ=200∠3.9° Z1+Z 2 11=H1/Z1=0.5293.9°12=11/Z2=0.894∠-595° S=VI=923+j60VAS=Rr2=125 j100S2=1I2=80+160
− + L • I R C s V • 1 • V L R 1 • I 2 • I (2) . 220 2 cos 500 0.4 100 50 5 s s s L Find P and Q tV L H R R C F = = = = = 400 100 200 1 1 2 j Z R j L j C Z = − j = − = L + = + 320 240 1 2 1 2 j Z Z Z Z = + + = + = + = + 370 240 441 33 1 2 1 2 j Z Z Z Z Z R = = − • • I V/ Z 0.5 33 = + = • • 200 3.9 1 2 1 2 1 I Z Z Z Z V = = = = − • • • • I 1 V1 / Z1 0.5 93.9 I 2 V1 / Z2 0.894 59.5 100 80 160 92.3 60 12.5 1 2 _ 1 1 2 _ 1 2 _ _ S V I j S V I j S V I j VA SR RI s s = = − = = + = = + = = • • • • • •
(3) Z1=10+j50Z2=400+1000 Z2 B B=?,I2 leads v90° 12+B1 Z111+Z212=Z1(12+BI2)+ s=Z1+BZ1+22 2 leads vs 90 Re[Z1+B21+22=0 10+B10+400=0B=-41 B I2 leads v。60 S=(1+B)Z1+ 50(+B)+1000 10(1+B)+400tan(-60°) B=-2615
= = + = + • • ?, 90 10 50 400 1000 2 1 2 Vs I leads Z j Z j • • • • • • • • • 1 = 2 + 2 = 1 1 + 2 2 = 1 2 + 2 + 2 2 I I I V Z I Z I Z (I I ) Z I s 1 1 2 2 Z Z Z I Vs = + + • • 2 90 Re[ 1 + 1 + 2 ] = 0 • • I leadsVs Z Z Z 10 + 10 + 400 = 0 = −41 (3) • 1 I = • • ?, I 2 leadsVs 60 1 2 2 (1 ) Z Z I Vs = + + • • tan( 60 ) = −26.15 10(1 ) 400 50(1 ) 1000 = − + + + +
(4) 109+254302V 1i2+o=501/s,1=2.5∠40A 25220m Find us(t). j500×20×10×2.5∠40° =1∠130°4 25 1=12+IL=1∠1300+25∠409=-0.643+j0.767+1915+607 1.272+j2374=2.69∠618° =10/1+25∠-30°+500×20×10-°Ir=35.34∠58.5° U、(t)==35342cos(500t+58.5°)A
LI • ( ). 500 1/ , 2.5 40 . Find t If s I A s L = = • • 1 I • 2 I 25 500 20 10 2.5 40 3 2 = • − j I = + = + • • • I 1 I 2 I L 1 130 2.5 40 • − • • Vs = I + − + j I L 3 10 1 25 30 500 20 10 = 35.3458.5 s (t) = = 35.34 2 cos(500t + 58.5) A = 1130 A = 1.272 + j2.374 = 2.6961.8 = −0.643 + j0.767 + 1.915 + j1.607 (4)
6 (5)Find or and o Y=igI 1 F 2 H 36 366+j2 j1+6-12=/1+6 j2a 36+(2a) 3636+(2a)236+(2a) 2 O=rReaC=1 3636+(2a 1Ω Jat 361+j2 F 2H 1,1-j2a 1-j2a Ja+ J 361+(2a 36 20 Ja 十 L 6(2a)2(2a) CR, L C @=@rReaC=12 (>1)
Find and Q. r 6 2 1 36 1 j Y j + = + 2 36 (2 ) 6 2 36 1 + − = + j j 2 2 36 (2 ) 2 36 (2 ) 6 36 1 + − + + = + j j r = 3 Q = r ReqC = 1 (5) 1 2 1 36 1 j Y j + = + 2 1 (2 ) 1 2 36 1 + − = + j j 2 (2 ) 1 2 36 1 j j − = + r = 3 2 0 36 (2 ) 2 36 1 2 = + − 2 2 (2 ) 2 (2 ) 1 36 1 j = j + − Q = r ReqC = 12
