Three lamps loo W, 60W and 40W are connected in.
parallel to a power supply of 220V. Calculate -
1) Total energy consumed 2) Total resistance 3)
Calculate Cost of keeping them lighted for 6 hours
daily for 30 days if the cost of electricity per
unit is Rs 2.80.
Answers
Answer:
(1) Power consumption = 100 + 60 + 40 = 200 watt.
Energy consumption for 6 Hrs for 30 days = 200 * 6 * 30 =36000 WH= 36 KWH
So the cost @ Rs 2.80 = 36 * 2.80 = Rs. 100.80
(2) Resistance = volt^2 / Watt = 220^2 / 200 = 242 Ohm.
Explanation:
Total resistance = 242 Ω , 200W energy consumed and Cost = Rs 100.8
Explanation:
Three lamps in Parallel Hence total consumption
= 100 + 60 + 40
= 200 W
Energy consumed in 30 days 6 hrs = 200 * 6 * 30 = 36000 Wh
= 36 kwh
1 kwh = 1 unit
= 36 units
cost of electricity per unit is Rs 2.80.
=> Total Cost = 36 * 2.8 = Rs 100.8
Power = V²/R₁
100 = 220²/R₁ => R₁ = 484 Ω
60 = 220²/R₂ => R₂ = 2420/3 Ω
40 = 220²/R₃ => R₃ = 1210 Ω
Total resistance = R
=> 1/R = 1/484 + 1/(2420/3) + 1/(1210)
=> 1/R = (5 + 3 + 2)/2420
=> 1/R = 10/2420
=> R = 242
Total resistance = 242 Ω
Learn more:
two lamps rated 100 watt and 60 watt are connected in parallel to ...
https://brainly.in/question/7366007
50 bulbs are joined in parallel, each being a 100 V bulb taking a ...
https://brainly.in/question/1217989
two lamps one rated 100w at 220v and other 60w ĺat 220v are ...
https://brainly.in/question/7392178