Question

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.

Answer 1

Given: Fuse current limit (I) = 5 A

Main supply voltage (V) = 230 V

Power of one tube-light (P) = 30 W

By using the formula,

P_total = VI

⇒ P_total = 230 × 5

⇒ P_total = 1150 W

We know that,

n = `P_"total"/P`

⇒ n = `1150/36`

⇒ n = 31.94