two bulbs if 100 w each and two coolers of 250w each work on an average 6 hours a day .if the energy costs r.s 1.75 per kWh , calculate the monthly bill and minimum fuse rating when power is supplied at 250v.?
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total power p =
2 × 100 + 2× 250
= 700 W
electric energy consumed per month =
p × t = ( 700 W ) (6h) (30)
12600 Wh = 126 kWh
monthly bill = 126 × 1.75
= R.s 220.50
As p = VI , I ( minimum fuse rating )
= p / v
700w / 250 V
= 2.8 A.
be brainly
2 × 100 + 2× 250
= 700 W
electric energy consumed per month =
p × t = ( 700 W ) (6h) (30)
12600 Wh = 126 kWh
monthly bill = 126 × 1.75
= R.s 220.50
As p = VI , I ( minimum fuse rating )
= p / v
700w / 250 V
= 2.8 A.
be brainly
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