I 2 V V. = 0 write 1 loop equation for each loop with a voltage not in the current set of equations. or I using Ohm s Law V 1 5.

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1 Krchoff s Laws Drect: KL, KL, Ohm s Law G G Ohm s Law: 6 (always get equaton/esor) Ω 5 Ω 6Ω 4 KL: : 5 : 5 eq. are dependent (n general, get n ndep. for nodes) KL: 4 wrte loop equaton for each loop wth a voltage not n the current set of equatons. Elmnate ether or usng Ohm s Law eq: : KL: 4 v v ramer s ule: [ ] X X.... N Det[ ] Det [ ]

2 We can always wrte n terms of only, or varables usng ohm s law: Ω KL: : KL: M : 4 M : Ω 4Ω M M 8 urrent Dependent oltage Source: Note: G f G

3 ules for nodes: ) onvert all voltage to current sources eg: Ω 5 Ω. ) Determne a reference node and dentfy unknown relatve voltages ) Use KL at each unknown node: at node : G G... GNN : G G... G N: G G... G N N N N NN N N where: G Σ all conductances connected to node G j Σ all conductances between node and node j Σ all current sources connected to node Defne N equatons n N unknowns all other voltages and currents by Ohm s Law from and S. Eq: Ω Ω 5 Ω Ω Ω Ω 4 ef

4 ( ) : : 5. ( 5... ). : () ( ) One can also solve for N ndependent currents n N meshes onsder: We know that KL voltages around a loop. urrent flows n a smple mesh. : : We can as usual rearrange to a form that can be wrtten by nspecton: : : ( ) ( ) * Note that KL s always solved mplctly snce at each node we have a sum of currents n meshes: each mesh enters and leaves wth the same current.

5 n general: ) onnect each current source wth parallel res. to voltage source wth seres. ) Select a current varable and mesh for each smple loop (usually we traverse each loop n same drecton, e, clockwse. ) Use KL for each loop n terms of the mesh current varable. ff no dependent sources: :... NN :... NN N: N N... NN N N sum of all resance n mesh j sum of all common resance to meshes,j sum of voltage rses n mesh, n drecton of current Eq: Wheatone rdge Ω 4 8 Ω 4 Ω Ω Ω ( ) ( 8 4 6) ( )

6 How to solve ths syem? Gaussan Elmnaton to Trangular form: : : : (dvde both sdes by ) [ 9 ][ ] ' ' ' [ ] [ ] ' ' ' [ 9 ] " Then ' '

7 Dependent oltage & urrent Sources: We model the actvty of many actve components by use of a programmable voltage or current source whose rength s a functon of the voltage or currents elsewhere n the crcut: Eq:.8v 5.v.v 4 f we choose as the reference node ( ) : : (. ) (.. ) 5 4 ( ) now.. s. s

8 Superpoon for rcuts onsder: 6Ω Ω : : 4 Ω 5Ω We could solve ths by ether node or mesh analyss, but there may be a smpler approach: ' ' f we suppress source # (.e. make a short crcut) we can fnd,. Smlarly,, could be wrtten wthout source #. Total currents and voltages superpose suppress one at a tme and then superpose the results: Suppress : 6Ω Ω Ω Ω 4 5Ω 4 ( 6 ( 5) ) ' 7 ' So ' volt 9 ' 8 ' 8 8 8

9 Now: Suppress, run 6Ω Ω '' Ω Ω '' 5Ω '' ( 6 () 8 )

10 Exponental Exctaton of crcuts: dmttance & mpedance dea: Exponental functon s easy to analyze smple to add/multply/ntegrates, etc. lso s a common case for crcut exctaton onsder: L L where e for complex s,, real t. We know all currents are same n crcut, L KL or t () t () () t tdt () () t d t L() dt L L () L d t dt () t L t dt () for our crcut t () e t e ( ) () ', () t '' e where: e ' '' ''' e KL we have t e () ' () t '' se L e ''' sl

11 () Note: for ths knd of exctaton, L t '' se s () t () t s () t L L L () t t () So: s has same unts/behavor as : conductance. Eg. : Ω F Parallel admttances s ref sv s () for t e t s, e t e volts

12 Snewave (Snusodal) exctaton: We know: e o coso sn o e coso o e o e sn o o o e () ( φ) We wsh to udy crcuts wth exctaton: vt cos wt ( e wt ϕ e wt ϕ) ve wt v e wt v note s complex. e ϕ v e ϕ We defne: dmttance mpedance L s sl s sl for s w ( snusod case) w wl w wl

13 dmttances compose lke conductances, mpedances compose lke resances. d() t dt () () () So for our crcut: vt t L tdt for t () e ( ) vt () L s e s or Ls zs () s we wrte mpedances as Zs (), admttance as Y(s) () so t e eq: f e t Ls s e (@ t decreasng) Ω L H F t () e e s s t (mps) asc Trck: Extend the crcut technques for node and mesh analyss to also handle mpedence and dmttances generalze crcuts whch can be analyzed.

14 L t () coswt v v v v wt e e wt note: we can apply superpoon to solve ths: suppress v or v and solve for other. wl w wl w f we wrte, n exponental complex form: we get: e σ e σ wl w ( ) o tan wl w σ () σ t e e wt... wt wt ( ( σ ) ) ( σ ) () t e e ( w) wl ( ) cos wt σ, are called Phasors