Advertisements
Advertisements
प्रश्न
In a house there are 6 bulbs of 100 W each used for 4 hours a day, 4 bulbs of 60 W each used for 8 hours a day, an immersion heater of 2.5 kW used for 1 hour per day, and an electric iron of 800 W used for 2 hours per day. Calculate the cost of monthly electric bill for the month of April at the rate of 80 p per unit.
Advertisements
उत्तर
Energy consumed by 6 bulbs per day = `6 xx 100/1000 xx 4 = 2.4 "kWh"`
Energy consumed by 4 bulbs per day = `4 xx 60/1000 xx 8 = 1.92 "kWh"`
Energy consumed by immersion heater per day = 2.5 x I= 2.5 kWh
Energy consumed by electric iron per day = `800/1000 xx 2 = 1.6 "kWh"`
Total energy consumed in I day= 2.4 + 1.92 + 2.5 + 1.6 = 8.42 kWh
Total energy consumed in the month of April (30 days) = 8.42 x 30 = 252.6 kWh
Electric bill for the month of April= 252.6 x 0.8 =Rs. 202.08
APPEARS IN
संबंधित प्रश्न
The atoms of copper contain electrons and the atoms of rubber also contain electrons. Then why does copper conduct electricity but rubber does not conduct electricity?
Which particles constitute the electric current in a metallic conductor?
Which of the two is connected in series : ammeter or voltmeter?
What is the conventional direction of the flow of electric current? How does it differ from the direction of flow electrons?
Which effect of current is utilised in the working of an electric fuse?
State expression for Cells connected in series.
Match the items in column-I to the items in column-II:
| Column - I | Column - II | ||
| (i) | electric current | (a) | volt |
| (ii) | potential difference | (b) | ohm meter |
| (iii) | specific resistance | (c) | watt |
| (iv) | electrical power | (d) | joule |
| (v) | electrical energy | (e) | ampere |
