8 11-1 Mutual inductance i() U=r ah () 91 p21s911 12 22 q12q22
dt di = L §11-1 Mutual inductance 1 i 11 21 2 i 12 22 12 22 + (t) − i(t) L 21 11 L1 L2 L1 L2
M A current i, at Li produces an open circuit voltage u,at L ○L32U y21=M21U2= dv21=M21 t M A current i, at L2 produces an 十 open circuit voltage U,atLIs L Y12=M12L2 0,dv12 M d t 12 dt The mutual inductance: M12-M2IM(H) The coefficient of coupling: k=M ≤1 (i2→DC.U,U2=0) 1三
A current i1 at L1 produces an open circuit voltage at L2 . 2 21 21 1 = M i A current i2 at L2 produces an open circuit voltage at L1 . 1 12 12 2 = M i The mutual inductance: M12=M21=M(H) The coefficient of coupling: 1 1 2 = L L M k ( . , 0) If i 1, i 2 DC 1 2 = − + 2 1 i 2 i − + 1 L1 L2 M L1 L2 M dt di M dt d 1 21 21 2 = = dt di M dt d 2 12 12 1 = =
TX1 V1 311Vag HVdc 100 PARAMETERS: PARAMETERS coupling =1 150mV 100mV 50mV 0.4 0.6 1.0 口V(R1:2,R1:1)
coupling 0 0.2 0.4 0.6 0.8 1.0 V(R1:2,R1:1) 0V 50mV 100mV 150mV V+ PARAMETERS: L2 = 1m V- TX1 {L2} R1 100 V1 311Vac 0Vdc PARAMETERS: coupling = 1 R2 1k
“ dot convention” If,i,(entering two coils) produce flux linkages which are increase each other, then the two terminals(l, 2 or 1 2) are defined as“ dot terminals”(“ same name terminals”) i↑ p, 12 2 2
“dot convention” L2 2 i 1 i L1 If i1 ,i2 (entering two coils) produce flux linkages which are increase each other, then the two terminals(1, 2 or 1’ , 2 ’ ) are defined as “dot terminals”(“same name terminals”) . 1 i 11 21 L1 L2 2 i 12 22 1 1' 2 2
Determine“ dot terminals” Li-source L2--voltage-meter K--close, --upscale. The terminal connected (+ polarity of the source and the terminal connected (+ polarity fv are defined as“ dot terminals
V + − 1 i L1 L2 L1 --source L2 --voltage-meter K--close, V --upscale. The terminal connected (+) polarity of the source and the terminal connected (+) polarity of are defined as “ dot terminals” V Determine “dot terminals”
Determine the sign of the mutual voltage A current entering the dotted terminal of one coil produces an open-circuit voltage between the terminals of the second coil which is sensed in the direction indicated by a positive voltage reference at the dotted terminal of this second coil V2 O=M M 十 十 M dt
Determine the sign of the mutual voltage dt di M 1 2 = dt di M 1 2 = − dt di M 1 2 = − dt di M 1 2 = − + 2 1 i M − + 2 1 i M − + 2 1 i M − + 2 1 i M A current entering the dotted terminal of one coil produces an open-circuit voltage between the terminals of the second coil which is sensed in the direction indicated by a positive voltage reference at the dotted terminal of this second coil
If a nonzero current will be flowing in each of the coils. 十 十 2 2 i d i U,=L1-+M M dt dt dt di L,-2+M i 0= 2 L,-+M dt At a real frequency ((sinusoidal) V1=jOL I1+joM 12 VI=jOL, l1-jaM /2 V2=jaL2 12+joM I1 2=-jOL2 /2+joM I1
If a nonzero current will be flowing in each of the coils. dt di M dt di L 2 1 2 = 2 + dt di M dt di L 1 2 1 = 1 + dt di M dt di L 1 2 1 = 1 − dt di M dt di L 2 1 2 = − 2 + At a real frequency (sinusoidal) 1 2 1 1 • • • V = jL I + jM I 2 1 2 2 • • • V = jL I + jM I 2 1 2 2 • • • V = − jL I + jM I 1 2 1 1 • • • V = jL I − jM I 2 i 1 L2 L − + 1 − + 2 1 i M 2 i 1 L2 L − + 1 − + 2 1 i M
Example 2: Find each mesh current 1F (5+门7)1-j912+j2/3=V1 j9/+j1612-j8/3=0 1 TH 26H m-11s j211-j812+(3+j6)3=0 mlsh1:5/1+j7(1-I2)+j2(13-12)=V1 mesh2:j7(12-In)-jl2+j6(2-3)+j2(12-13)+j2(12-n)=0 mesh3:j6(3-12)+3l3+j2(11-12)=0
Example 2: Find each mesh current. − + + = − + − = + − + = • • • • • • • • • • 2 8 3 6 0 9 16 8 0 5 7 9 2 1 2 3 1 2 3 1 2 3 1 j I j I ( j )I j I j I j I ( j )I j I j I V 1: 5 1 7( 1 2 ) 2( 3 2 ) 1 • • • • • • mesh I + j I − I + j I − I =V 2 : 7( 2− 1 ) − 1 2+ 6( 2− 3 )+ 2( 2− 3 )+ 2( 2− 1 ) = 0 • • • • • • • • • mesh j I I j I j I I j I I j I I 3 : 6( 3− 2 ) + 3 3+ 2( 1− 2 ) = 0 • • • • • mesh j I I I j I I
Example 4: the parallel combination of the mutual inductance coils. (a) in side(同侧): +“1,解12 JOL 11+jaM /2 L13L2 V=jaMI+jaL 12 (2) 1+12 (3) M jal jaL2-jaM JaLI-JaM jaL1 j@M oLi+aM )212+0MD jaM jaL i=i1+i2=1o(1+2-=2M0i 0-L L2+02M2 oiL2+oM 2 2 L M J jO(L1+ l2 -2M) L1+L2-2M oL eg Leg= L1+L2-2M (b) out of side(异侧):" LL-M L1+L2+2M
Example 4: the parallel combination of the mutual inductance coils. 1 2 (1) 1 • • • V = jL I + jM I 2 (2) 2 1 • • • V = jM I + jL I 1 2 (3) • • • I = I + I 2 1 2 1 j M j L j L j M V j L V j M I • • • = • • − + − = V L L M j L j M I 2 2 1 2 2 1 2 • • • • − + + − = + = V L L M j L L M I I I 2 2 1 2 2 1 2 1 2 ( 2 ) ( 2 ) 1 2 2 2 1 2 2 j L L M L L M I V Z + − − + = = • • L L M L L M L eq 2 ' 1 2 2 1 2 + − − = L L M L L M L eq 2 '' 1 2 2 1 2 + + − = (a) in side (同侧): (b) out of side (异侧): 1 • I 2 • I • I − + • V 1 L2 L M • − + − = V L L M j L j M 2 2 1 2 2 2 L eq j L L M L L M j ' 1 2 2 2 1 2 = + − − =
