Question

Physics question below ?

a boys hostel has following appliances when energy is supplied at 200 V and costs Rs 5.25 per kWh. (I) 40 bulbs of 100 W working 8 hrs a day. (2) 20 fans each drawing a current 0.8 A and working 15 HRS A DAY (3) 2 tv sets each offering a resistance of 200 ohm and working 4 hrs a day (4) 2 electric motors of 1.5 H.P. and each working 4 hrs a day .CALCULATE : (a) monthly bill (b) among the fuse of 48 A and 50 A, which one you will use and why ? ( PLS ANS URGENTLY DEAR EXPERTS , I REQUEST YOU)?

Answer 1

(a)
(1) 40 bulbs of 100 W working 8 hrs a day`=(40xx100)xx8=4kWxx8=32kWh`
(2) 20 fans each drawing a current 0.8 A and working 15 HRS A DAY, Using `W=VA` `=(20xx200xx0.8)xx15=3.2kWxx15=48kWh`
(3) 2 tv sets each offering a resistance of 200 ohm and working 4 hrs a day, Using `P=V^2/R`
`=(2xx200^2/200)xx4=0.4kWxx4=1.6kWh`
(4) 2 electric motors of 1.5 H.P. and each working 4 hrs a day, Using `1HP=746W`
`(2xx1.5xx746)xx4=2.238kWxx4=8.952kWh`

Monthly consumption, taking `30` number of days `=30(32+48+1.6+8.952)=30xx90.552=2716.56kWh`
Monthly Bill`=5.25xx2716.56=Rs 14261.94`

(b)
Total load`=4+3.2+0.4+2.238=9.838kW`
Using the expression `W=VA`
We get current `A=983.8/200=49.19A`
We must use minimum fuse of `50A`, assuming all appliances are used simultaneously.