點知邊個light blub 光 D?? 點計?

Question

Two light blubs A and B ratings ' 10W, 6V' and '5W , 6V' respectively are connected in series to a 6V battery.Which is brighter ?

Answer 1

Use P = V^2 / R to calculate the internal resistances of 2 light bulbs.
The one with higher resistance should be brighter as they are connected in series, i.e. they draw the same current.

2007-01-14 17:08:39 補充：
kaiyan0226, please also consider the current that two bulbs draw. The resistances of them are different.

2007-01-15 17:27:31 補充：
謝謝Chan Chi Wang的回應, 但我要補充一點, 我沒有說過答案不是B.用公式 P = V^2 / R,可得出A和B燈泡的電阻分別為3.6和7.2歐姆,因此B燈泡的電阻確是大於A燈泡的,而燈泡的電壓可用 P' = I^2 * R計出, (注意: 先前的P值跟這個P'值有不同意思)由於串聯緣故,通過兩個燈泡的電流一樣,電壓跟電阻成正比,又B燈泡的電阻大於A燈泡的,所以B燈泡輸出的電壓較A燈泡的為大.

2007-01-16 23:08:19 補充：
不好意思, 用錯了名詞, 以上的電壓應該改為功率(Power).

2007-01-16 23:10:21 補充：
方便起見, 我重答一次吧.用公式 P = V^2 / R,可得出A和B燈泡的電阻分別為3.6和7.2歐姆,因此B燈泡的電阻確是大於A燈泡的,而燈泡的功率可用 P' = I^2 * R計出, (注意: 先前的P值跟這個P'值有不同意思)由於串聯緣故,通過兩個燈泡的電流一樣,功率跟電阻成正比,又B燈泡的電阻大於A燈泡的,所以B燈泡輸出的功率較A燈泡的為大.

Answer 2

Use R=V^2/P to calculate the resistance of each bulb
(where R is the resistance, V is the voltage and P is the power of the bulb)

For bulb A, Ra=6^2/10 ohms = 3.6 ohms
For bulb B, Rb = 6^2/5 ohms = 7.2 ohms

Since the two bulbs are connected in series, the voltage across bulb A is,
Va = 6 x[3.6/(3.6+7.2)] v = 2 v

Power consumed by bulb A, Pa = 2^2/3.6 W = 1.11 W

Similarly, Vb = 6x[7.2/(3.6+7.2)] v = 4 v
Pb = 4^2/7.2 W = 2.22 W

Clearly, bulb B is twice as bright as bulb A.

Answer 3

琴生兄所說有道理，但不正確(這是一條iq題)。

正確答案是B，因為A及B燈泡是串聯，電源為6V，所以每個燈泡所用之電壓並不相同，總電壓為6V，A燈泡比B燈泡阻值大一倍，電阻越大，反而所用之電壓越小，輸出power亦越小。

Answer 4

10w,6v即係話你俾6v電佢佢就可以俾10w power你
俾一樣咁多既電, a俾出既power大d
咁即係a係光d

2007-01-15 21:55:29 補充：
sor...答錯野添

2007-01-15 21:55:39 補充：
sor...答錯野添