《电路》（英文版）17-1 Direct analysis methods

8 17-1 Direct analysis methods KCL: Ai=0 K比:B,D2=0 L (t0)=i i(t) Cuc(to)=vo () +D(t) Uc(t0)=0 i()L2()=0 D(t) Let the initial voltages on the capacitors be represented as voltage sources in series with the capacitors and the initial currents through the inductors be represented as current sources in parallel with the inductors

§17-1 Direct analysis methods • • • KCL: Ai b = 0 • • • KVL: Bf b = 0 0 0 i(t) C  C (t ) = +(t) − Let the initial voltages on the capacitors be represented as voltage sources in series with the capacitors and the initial currents through the inductors be represented as current sources in parallel with the inductors. i(t) C + (t) − + −  0 ( ) 0  C t 0 = i(t) +(t)− L 0 0 i (t ) i L = i(t) + (t) − 0 i ( ) 0 L i L t 0 =

General branch of a linear time-invariant network bk -branch current sk bk Ubk --branch voltage ek element current ek element voltage k sk current source 十 bk sk voLtage source k -element impedance KCL: ibk =iek -isk k=l, 2, . B KVL: Ubk=Uek -Usk k=1, 2, .. B VCR:(DUek=riek k=1, 2,., B (2)Uek iek (tdt or vek(s)=le(s) i k k or yek (S)=sLKlek(S) dt

General branch of a linear time-invariant network i branch current bk − − bk − −branch voltage i element current ek − − ek − −element voltage  sk − −voltage source i sk − −current source Zek − −element impedance KCL : i bk = i ek − i s k k = 1,2,  ,B KVL:  bk =  ek − s k k = 1,2,  ,B VCR: (1)ek = Rk i ek k = 1,2,  ,B ( ) 1 ( ) ( ) 1 (2) 0 I s sC i d or V s C ek k ek t t ek k ek = =     (3) or V (s) sL I (s) dt di L ek k ek ek ek = k = bk i ek i sk i +ek − + bk −  sk Zek

bk +U-i Usk KCL: i=i -i b k ek k ek k KVL: D bk k 十 bk Ler:i=[in1in2…ig (KCL) 1“s2 1e2 B bI b2 bB D.U.(KVL 2 eB sIs2 B

bk ek sk KCL: i = i − i KVL bk =ek − sk : T b b b bB [ ]  =  1  2  • T b b b bB Let : i [i i i ] = 1 2  • T s s s sB i [i i i ] = 1 2  • T e e e eB i [i i i ] = 1 2  • i i i (KCL) b e s • • • = − T e e e eB [ ]  =  1  2  • T s s s sB [ ]  =  1  2  • (KVL) b e s • • •  =  − bk i ek i sk i +ek − + bk −  sk Zek

Source transformation The voltage source transformation We remove this voltage source and add a voltage source Usk to all the other branches that are incident to one of the terminal of bk

Source transformation We remove this voltage source and add a voltage source to all the other branches that are incident to one of the terminal of  sk . bk The voltage source transformation  sk − + k b  sk  sk + + − − k b

The current source transformation k sk sk k We remove the branch b, altogether and a current source isk parallel to each branch of any one loop of which b, is a branch

We remove the branch altogether and a current source parallel to each branch of any one loop of which is a branch. sk i bk bk The current source transformation sk i sk i sk i sk i k b