A 120V power line is protected by a 15A fuse. What is the maximum number of 500W lamps that can be simultaneously operated in parallel on this line without “blowing” the fuse because of an excess of current?

1. Voltage of the power line, V=120V
2. Current of the fuse, i= 15A
3. Power of the lamp, P=500W

The fuse in a circuit is used to protect the circuit from over-current flow. Here, the lamps are to be connected in parallel and have an individual contribution to the current in the fuse. Thus, the maximum power of the given power line that it holds divided by the power of each lamp will give us the number of lamps to be connected to prevent the breakdown of the fuse.

The power of a device, P=VI .....(i)

The maximum output power of the power line can be given using the given data in equation (i) as follows:

P=(120V)(15A)

= 1800W

Thus, the maximum number of lamps that can be operated in parallel connection is given as follows:

$$\begin{gathered} N=\frac{MaximumPower}{LampPower} \\ =\frac{1800W}{500W} \\ =3.6 \\ =3 \end{gathered}$$

Hence, the maximum number of lamps is 3.