A 120-V power line is protected by a 15-A fuse.What is the maximum number of 120 V, 500 W light bulbs that can be operated at full brightness from this line?
A)1
B)2
C)3
D)4
E)5

Question

Quizplus · Accepted Answer

Using the formula P = IV, we can find the current needed for a 500 W bulb:

500 W = (120V)(I)

I = 4.17 A

Since the fuse is rated for 15 A, we can divide the maximum current allowed by the current needed for one bulb to find the maximum number of bulbs:

15 A / 4.17 A ≈ 3.6 bulbs

Therefore, the maximum number of bulbs that can be operated at full brightness is 3. We cannot operate a fraction of a bulb, so we take the integer part of 3.6, which is 3.

Therefore, the answer is choice C.

A 120-V Power Line Is Protected by a 15-A Fuse