A domestic lighting circuit has a fuse of 5 A. If the mains supply is at 230 V, calculate the maximum number of 36 W tube-lights that can be safely used in this circuit.

Given,

Main supply voltage, V = 230 V
Maximum current that the fuse can hold, I = 5 A

We know that-
$P=V\times I$
Substituting the given values we get-
$P=230\times 5$
$P=1150W$
So, the maximum power that can be delivered, is 1150W.

Now,
Let the maximum number tube-lights that can be used safely in this circuit be y.
Power of each tube-lights = 36 W (given)

Therefore, 
$P=36\times y$
$1150=36\times y$              $(putting\ the\ value\ of\ power)$
$y=\frac{1150}{36}$
$y=31.9$

Thus, the required number of tube-lights are 31.

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Updated on: 10-Oct-2022

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