Question

a bulb is rated at 220 V -100 W what is its resistance 5 such bulbs are situated switched on for 4 hours each day for a month of March what is the total energy consumed in the month calculate the cost if the rate is 50 paise per unit

Answer 1

Given,
voltage:220V
power:100w
A/Q,
we know that ,
P=VI
100=220×I
I=220/100
current (I)=2.2A
Now,
From ohm's law,
V=IR
220=2.2×R
R=220×10/22
Resistance for 1 bulb=100ohm
similarly,Resistance for 5 bulbs=500ohm
Now,
t=4hrs.
For 1 bulb,
P=100w=0.1kWh
t=14400sec.
energy consumed in 1 day=4×0.1=0.4kWh.
Therefore,energy consumed by 5bulbs in a day=5×0.4=2kWh.
Now,
we know that there are 31 days in month of March
Therefore, energy consumed in March by 5 bulbs of 100w = 2×31=62kWh or unit
If the cost of 1 unit = 50 paise = 0.5 rupees
Therefore,the cost of 62 unit = 62×0.5=31rupees
Hence,the cost of electricity using in month of March= Rs.31.00