Two bulbs of `100 W` each and two coolers of `250 W` each, work on an average `6` hours a day. If the energy costs Rs. `1.75 per kWh`, calculate the monthly bill and the minimum fuse rating when power is supplied at `250 V`.

Answer

`Rs.220.50, 2.8 A`

Explanation

Total power, `P = (2 xx 100 + 2 xx 250) W = 700 W`
Electric energy consumed per month `= P xx t = (700 W)(6 h)(30)`
=`126000 Wh = 126 kWh`
Monthly bill `= (126 kWh) (Rs. 1.75//kWh) = Rs. 220.50`
As `P = VI, I` (minimum fuse rating) `= (P)/(V) = (700 W)/(250 V) = 2.8 A`.