# 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)?

##### 1 Answer
Feb 22, 2017

(a)
(1) 40 bulbs of 100 W working 8 hrs a day$= \left(40 \times 100\right) \times 8 = 4 k W \times 8 = 32 k W h$
(2) 20 fans each drawing a current 0.8 A and working 15 HRS A DAY, Using $W = V A$$= \left(20 \times 200 \times 0.8\right) \times 15 = 3.2 k W \times 15 = 48 k W h$
(3) 2 tv sets each offering a resistance of 200 ohm and working 4 hrs a day, Using $P = {V}^{2} / R$
$= \left(2 \times {200}^{2} / 200\right) \times 4 = 0.4 k W \times 4 = 1.6 k W h$
(4) 2 electric motors of 1.5 H.P. and each working 4 hrs a day, Using $1 H P = 746 W$
$\left(2 \times 1.5 \times 746\right) \times 4 = 2.238 k W \times 4 = 8.952 k W h$

Monthly consumption, taking $30$ number of days $= 30 \left(32 + 48 + 1.6 + 8.952\right) = 30 \times 90.552 = 2716.56 k W h$
Monthly Bill$= 5.25 \times 2716.56 = R s 14261.94$

(b)
Total load$= 4 + 3.2 + 0.4 + 2.238 = 9.838 k W$
Using the expression $W = V A$
We get current $A = \frac{983.8}{200} = 49.19 A$
We must use minimum fuse of $50 A$, assuming all appliances are used simultaneously.