Advertisements
Advertisements
Question
A 100-watt electric bulb is used for 5 hours daily and four 60 watt bulbs are used for 5 hours daily. Calculate the energy consumed (in kWh) in the month of January.
Solution
Energy used by 100 W bulb is E = P × t
= 100 × 5 = 500 Wh
Energy used by four 60 W bulbs E = 4 × 60 × 5 = 1200 Wh
Total energy per day = 500 + 1200 = 1700 Wh
= 1.7 kWh
Number of days in January = 31 days.
Energy consumed in January = 31 × 1.7 = 52.7 kWh.
APPEARS IN
RELATED QUESTIONS
An electric bulb is rated at 220 V, 100 W. What is its resistance?
Which quantity has the unit of watt?
An electric heater of resistance 8 Ω takes a current of 15 A from the mains supply line. Calculate the rate at which heat is developed in the heater.
Electrical power p is given by the expression P = (Q × V) ÷ time. Express the power P in terms of current and resistance explaining the meaning of symbols used there in.
What would happen and why, to an electric bulb when it is connected across a supply of voltage (i) lower (ii) higher than its proper rating?
You are required to connect a bulb, a fan and a socket outlet to the mains in one set A and an air-conditioner and a refrigerator to the mains in other set B. Will you recommend the wire of same thickness and same insulation in both? explain your answer.
Find the cost of operating an electric toaster for two hours if it draws 8 A current on a 110 volt circuit. The cost of electrical energy is Rs. 2.50 per kWh.
The following table gives the electrical appliances used, their power and the average time for which they are used each day in a home. Estimate the monthly electricity bill if the rate is 60 paise per unit.
| Sr.no. | Name | Nos. | Power rating | Time/day |
| 1 | Bulb | 4 | 100 W | 7.5 hr |
| 2 | Fans | 2 | 50 W | 10 hr |
| 3 | T.V. | 1 | 100 W | 2 hr |
| 4 | Iron | 1 | 500 W | 1 hr |
| 5 | Electric stove | 1 | 750 W | 2 hr |
Unit of electric power may also be expressed as:
A family uses 8 kW of power. Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square meter. If 20% of this energy can be converted to useful electrical energy, how large an area is needed to supply 8 kW?
