Question

How many $60{\text{W}}$ bulbs may be safely run on ${\text{220V}}$ using ${\text{5A}}$ fuse?
(A) $18$
(B) $16$
(C) $14$
(D) $12$

Answer 1

To solve this question, we need to use the formula for power to calculate the net power dissipated in the circuit. Then, on dividing this value with the power dissipated on each of the bulbs, we will get the required number of bulbs.

Formula used: The formula used to solve this question is given by
$P = VI$ , here $P$ is the power across a resistor through which a current of $I$ flows, and $V$ is the potential difference applied across it.

Complete step by step solution:
Let there be $n$ bulbs connected safely across the ${\text{220V}}$ source.
Since the fuse of ${\text{5A}}$ is used in the circuit, so the maximum current in the circuit cannot exceed ${\text{5A}}$ , as beyond this current, the fuse will melt and hence the circuit will be incomplete.
So when the maximum number of bulbs is connected in the circuit, maximum current will flow in the circuit. So the current in the circuit is equal to ${\text{5A}}$ .
Now, the net potential applied in the circuit is equal to ${\text{220V}}$ . We know that the power is given by
$P = VI$
So the total power dissipated through all of the bulbs is given by
$P = 220 \times 5$
$\Rightarrow P = 1100{\text{W}}$ ..................(1)
Now, according to the question, the power rating of each bulb is equal to $60{\text{W}}$ , so each bulb will dissipate a power of $60{\text{W}}$ . Since there are a total of $n$ bulbs connected, so the total power dissipated is given by
$P = 60n{\text{W}}$ ..................(2)
Equating (1) and (2) we get
$60n = 1100$
$\Rightarrow n = \dfrac{{1100}}{{60}}$
On solving we get
$n = 18.33$
Considering the maximum natural number, we get the maximum number of bulbs that can be safely run across the ${\text{220V}}$ source is equal to $18$ .
Hence, the correct answer is option A.

Note
Do not worry about the type of connection of the bulbs with the voltage source. This is because each bulb has to have a power of $60{\text{W}}$ to be run. By the way, all the bulbs must be connected in a parallel combination across the voltage source, so that the potential difference across each bulb is equal to ${\text{220V}}$ .