電路圖中A2既readings係大左定細左?

┌(A1)─ Batt ─ r ───┐
│　　　　　　　　　　│
┝──Ｒ２──(A2)──┩
│　　　　　　　　　　│
\ switch 　　　　　　　│
│　　　　　　　　　　│
┝──Ｒ３──(A3)──┩
│　　　　　　　　　　│
└──Ｒ４──(A4)──┛
A1, A2, A3, A4 are ammeters.
Batt is the battery for the circuit.
r is internal resistance of the battery.
R2, R3, R4 are loads for the circuit (in this case it is light bulbs).
For ideal battery, r = 0, and ideal ammeters, internal resistance of ammeters = 0.
Just before the switch is closed, reading of A1 = reading of A2
V = I * R
where v is the voltage of the battery, I is the reading of A1 & A2, R is the resistance of R2.
Just after the switch is closed, the overall resistance across the whole circuit is
= 1/ (1/R2 + 1/R3 + 1/R4)
If R2 = R3 = R4 = R
the overall resistance of the whole circuit is 1/ (3/R) = R/3
the (current) reading of A1 is V / overall resistance
= V / (R/3)
= 3 (V/R)
= 3I
so the reading of the ammeter A1 will be tripled.
As the voltage across R2 is unchanged, the current flow through R2 will be unchanged.
When the switch is closed, A1 will increase, and A2 will unchange.
For ordinary battery, r = finite value which is not zero, and ordinary ammeters, internal resistance of ammeters = finite value which is not zero.
Just before the switch is closed, reading of A1 = reading of A2
V = I * (R + r + rA1 + rA2)
If R2 = R3 = R4 = R & rA1 = rA2 = rA3 = rA4 = rA
V = I * (R + r + 2rA)
where v is the voltage of the battery, I is the reading of A1 & A2, R is the resistance of R2, r is the internal resistance of the battery, rA1, rA2, rA3, rA4 are the internal resistance of the ammeters.
Just after the switch is closed, the overall resistance across the whole circuit is
= (1/ (1/(R2+rA2) + 1/(R3+rA3) + 1/(R4+rA4))) + rA1 + r
If R2 = R3 = R4 = R & rA1 = rA2 = rA3 = rA4 = rA
the overall resistance of the whole circuit is 1/ (3/(R+rA)) + rA + r
= (R+rA)/3 + rA + r
= (R-2rA)/3 + 2rA + r
For any positive values of R & rA, R > (R-2rA)/3
Thus just after the switch is closed, the overall resistance across the whole circuit will decrease, and as V = I * R
for fixed V and when R decrease, I will increase.
so the reading of the ammeter A1 will increase.
Just before the switch is closed, reading of A1 = reading of A2
V = IA2(before) * (R + r + 2rA)
IA2(before) = V / (R + r + 2rA)
Just after the switch is closed, reading of A1 = 3 * reading of A2
V = IA1(after) * ((R-2rA)/3 + 2rA + r)
V = 3 * IA2(after) * ((R-2rA)/3 + 2rA + r)
V = IA2(after) * ((R-2rA) + 6rA + 3r)
V = IA2(after) * (R + 4rA + 3r)
IA2(after) = V / (R + 3r + 4rA)
For any positive values of r & rA, 1 / (R + r + 2rA) > 1 / (R + 3r + 4rA)
thus the reading of the ammeter A2 will decrease
When the switch is closed, A1 will increase, and A2 will decrease.
I hope this can help your understanding. :)

2007-10-12 15:27:40 補充：
For ideal battery,the overall resistance of the whole circuit accross the battery is= 1/ (1/R2 + 1/R3 + 1/R4) but 'If R2 = R3 = R4 = R'= 1/ (1/R + 1/R + 1/R)= 1/ (3/R)= R/3

2007-10-12 15:29:04 補充：
I at A1 = Voltage of battery / Resistancethe (current) reading of A1 is V / (R/3)= 3V / Rbut before the switch is closed, the reading of A1 is V / Rso the current rises by 3x, from V/R to 3V/R.

當close左開關之後.A2同A3&A4是parallel的關係

所以這三行的voltage一樣.

而A2這行的燈泡的resistance不會因為多了其他行而改變,

by V = I x R

因V & R不變, 故current不變.

即是A2 的reading不變