天同師兄請賜教!circuit問題

用戶2457

2010-12-09 10:21 am

以下有5張圖:
http://i1091.photobucket.com/albums/i400/eddierr/R2.jpg?t=1291747994 fig.1
http://i1091.photobucket.com/albums/i400/eddierr/R3.jpg?t=1291747994 fig.2
http://i1091.photobucket.com/albums/i400/eddierr/R4.jpg?t=1291747994 fig.3
http://i1091.photobucket.com/albums/i400/eddierr/R5.jpg?t=1291747994 fig.4
http://i1091.photobucket.com/albums/i400/eddierr/R6.jpg?t=1291747994 fig.5

q1)先睇fig.1,如果要計equivalent resistance我地會好似fig.2咁睇因為Vc=Vd,冇current,最後應該eq r=5ohm係唔係?
q2)再睇fig3,eq r應該係2ohm而current across c 同d同Vc,Vd就好似上圖所示,then我地用一條conducting wire連住佢好似fig.4咁,問題就黎喇,
i)Vc係唔係等於Vd?
因為以我所知係wire上既any pt都應該有same potential,咁refer番q1)既解度,
ii)基於我假設Vc=Vd,係唔係冇current through pt.c&d?
iii)如果係咁,點解current會由c去y而唔會由c去d再去y?
iv)再基於我假設Vc=Vd,咁i1+i2應該=i3+i4架嘛,我計到Vc=Vd=2.67,i1=3.33,i2=0.67,i3=1.33,i4=2.67,咁i1不等於i2and i3不等於i4,咪代表有current through pt.c&d?咁究竟係有current定冇?同fig1,2計eq r果個example有咩唔同?
q3)到到last fig.5,其實係我終極唔明既問題,呢個係potentiometer,跟住length1既potential係=emf of E2,咁冇current through galvanometer啦,
i)請問點解?我唔明既係我就咁將E2同一個同length1有same resistance既resistor駁埋都有current,但點解釋呢個situation?
ii)如果Va>Vb,係唔係有current由a去b?但果度係E2既negative terminal黎架喎,咁請問current係E1整出黎既定係E2?
iii)相反,Vb>Va又會係點?

如果天同師兄可以答到我真係十分感激,因為我對"same potential,no current"呢個concept唔係好掌握到,就好似由cell到resistor果條path一路到係same potentail架啦,但係又有current,所以希望您可以幫忙!

回答 (1)

天同

2010-12-10 4:44 am

✔ 最佳答案

Q1: this question is trivial. Connecting C and D has no effect on the currents.

Q2(i): After you connect C and D by a wire, you could regard C and D join together to become a point. The circuit now reduces to a 1-ohm and a 5-ohm resistors in parallel, which is connected in series to a 2-ohm and a 1-ohm resistor in parallel. The potnetial at point C (or D) is now 2.67 volt (referenced to the potential at the -ve cell terminal).

(ii) As said above, C and D now becomes a point, there is no meaning to say about current flowing within a point.

(iii) When C and D are joined together, the currents in the whole circuit will have completely changed. The main current now becomes 4 A (contrast to 3 A before). This 4 A current divides itself into 3.33 A (flow through the 1-ohm resistor) and 0,67 A (through the 5-ohm resistor). These two currents combine together after passing through the resistors and re-divide again to form current of 1.33 A (through the 2-ohm resistor) and 2.67 A (through the 1-ohm resistor).

(iv) As said above, the currents combine together at the end of the first parallel resistor arrangement and re-divide again when flow through the second parallel resistor arrangemnt. I think what confuse you is the presence of the "wire connecting C and D". It may be more realistic to clear your doubt if we assume the wire CD has a finite but very small resistance, say 0.001 ohm. Analysis (using either Kirchoff's Law or Thevenin's Theorem) shows that there is a small voltage of (slightly less than) 2 mV across the wire, which causes a current of (slight less than) 2 A to flow from C to D.

Generally, what we are interested in the calculation of circuit of this kind is the currents flowing through the resistors. By assuming the wire joining CD as one with no resistance (hence points C and D could be seen as meshed together to form a point) would indeed simplify the matter much.

Q3: Pease refer to my answer given earlier

http://hk.knowledge.yahoo.com/question/question?qid=7010110901499

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