有題物理問題不慬

max current is 15A for each socket, otherwise circuit overload activated.

current of 1100W and 550W:
P=VI
(1100+550)=220I
I=7.5A

max remaining current for a set of lamps:
15-7.5
=7.5A

each lamp takes current:
P=VI
100=200I
I=0.5A

7.5A allows 15 lamps.

(16 lamps must blow the circuit, danger!)

2009-04-17 16:46:56 補充：
each lamp takes current:
P=VI
100=220I
I=0.46A

number of lamps:
7.5/0.46
=16.3

so, 16 lamps.

2009-04-17 16:50:20 補充：
one can consider the solution by power, or by resistance, or me by current.

2009-04-17 17:01:21 補充：
another method:

max power:
P=VI
=220x15
=3300W

each lamp 100W, so max number of lamps:
3300/100
=33 (lamps)

now in 3300W, 1100 and 550W is used, power remains:
3300-1100-550
=1650W

2009-04-17 17:04:31 補充：
half power remains:
so half of 33 lamps is allowed, it is 16.

2009-04-17 17:06:34 補充：
20-4-09 exam 成功!

普通數學題啫，(220x15-1100-550)/100=16.5，當然冇半個燈泡，最多咪喺16個。