《电路》（英文版）16- 55 Tellegen''s(特勒根) theorem

§16-5 Tellegen's(特勒根) theorem Tellegen’ s theore1: Consider a lumped network n consisting of N+l nodes and B branches, let Dn=ob1b2ObBI, b=1b1 b2"IbBl Where Dbk and ibk denote the voltage across and current through branch b for k=1.2...B. Then i=0 or bkbk =0 roo U=4 ∴Dnin=0or>U bkbk =0 k=1 The sum of the powers delivered to or absorbed by all the ranches of a given lumped network is equal to zero

§16-5 Tellegen’s(特勒根) theorem Tellegen’s theorem 1: Consider a lumped network n consisting of N+1 nodes and B branches, let . Where T b b b bB T b b b bB [ ] ,i [i i i ] = 1 2 = 1 2 • •     denote the voltage across and current through branch bk for k = 1,2…B. Then bk bk  and i 0 0 1 =  = = • • • b k B k b b k T b  i or  i Proof: • • • = n T  b A  • • •  = A T n T b  • • • • •  = b T b n T b  i  Ai i or b T b • • •   = 0 0 1  = = bk B k bk  i The sum of the powers delivered to or absorbed by all the branches of a given lumped network is equal to zero. • 0

Tellegen's theorem 2 Consider two networks n and n made up of lumped two terminal elements whose graphs are identical. Denote their corresponding branch voltage and current vectors Db, lb ∧ and Db,'b; Then D, in=0 and U,ib=0 for all t. Proof: for the n 6s r T:=Un 0 For the n Ai=0 A=A AL=0 We obtain U, i,=0 for allt. and unib=0 for allt

Proof: for the n • • • = n T  b A  • • • or = A T n T b  • • • • •  = b T b n T b  i  Ai ^ ^ For the n ^ • • • Ai b = 0 ^^ • • A = A ^ • • • = 0 b Ai ^ We obtain ^ i 0 for all t. b T b • • •  = ^ and i 0 for all t. b b • • •  = T • 0 ;Then i 0 and i 0 for all t. b b b T b • • • • • •  =  = Consider two networks n and made up of lumped two￾terminal elements whose graphs are identical. Denote their corresponding branch voltage and current vectors ^ n • • b b and  , i ^ ^ ^ ^ T • • b b  , i Tellegen’s theorem 2:

+U 八 +U2i2+U3i+U4i1+D5i;+U61 55 Note that the conclusion cannot be interpreted as a conservation of power; it is merely a mathematical relationship that exists between the branch voltages of one circuit and the branch currents of another circuit with the same topology(拓扑)

Note that the conclusion cannot be interpreted as a conservation of power; it is merely a mathematical relationship that exists between the branch voltages of one circuit and the branch currents of another circuit with the same topology(拓扑). ^ 6 i ^ 2 i ^ 1 i −1 + 5 + 6 ^ 3 i + 2 + 3 +4 ^ 4 i = 0 ^ 5 i + −  6 + 2 − + 4 − − +  1 − +  3 − +  5 6 i ^ 1 i ^ 5 i 4 ^ i ^ 2 i ^ 3 i ^

Example: ForR,=R2=2, ifUs=8V, G=2A, 02=2V; forR1=1.4,R2=0.8,U,=9,i1=3A. Find u2 11 R +∑R 十 R Solution:1=8-2×2=4J,i1=24;U2=2J 14 U1=9-3×1.4=48,i1=34;2=2,i2=U2/0.8 U1+U2l2+ ∑ U11+U2l2+ ∑ 八 B U,i1+0242+2 0i+22 k=3 k=3 八 4×3+2×U2/0.8=-4.8×2+U2×1 八 U,=1.6V

Example: 1.4, 0.8, 9 , 3 . . 2, 8 , 2 , 2 ; 1 2 1 2 1 2 1 2     for R R V i A Find For R R i f V i A V s s = = = = = = = = = ^ ^ ^ Solution: 1 = 8− 22 = 4V, i 1 = 2A;2 = 2V, i 2 = 1A ^ 1 = 9− 31.4 = 4.8V, i 1 = 3A;2 = ?, i 2 =2 / 0.8 ^ ^ ^ ^   = = − + + = − + + B k k k k B k k k k i i R i i i i R i i 3 1 1 2 2 3 1 1  2 2   ^ ^ ^ ^ ^ ^   = = − + + = − + + B k k k B k k k i i i i i i 3 1 1 2 2 3 1 1  2 2     ^ ^ ^ ^ ^ ^ −43+ 22 / 0.8 = −4.82+2 1 ^ ^ ^  2 =1.6V  R R1 R2  s − +  2 2 1 i i − +  1