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Question: When we run a \(40\)watt bulb for \(5\)hours, if \(1\)unit \(\left( { = 1\,{\rm{kWh}}} \right)\) cos...

When we run a 4040watt bulb for 55hours, if 11unit (=1kWh)\left( { = 1\,{\rm{kWh}}} \right) costs 22rupees, find the bill we have to pay for this?

Explanation

Solution

Use the formula to find energy, E=P×tE = P \times t, where EE is energy, PP is power and tt is time duration in hours. Latter, find the cost by multiplying the number of units and the cost of one unit.

Complete step by step solution:
In the given problem,
The rated power of the bulb (P)\left( P \right) is 40watt.40\,{\rm{watt}}{\rm{.}}
Time duration (t)\left( t \right) for which the bulb is used is 5hours.5\,{\rm{hours}}{\rm{.}}
The energy consumed by the bulb is equivalent to the product of power of the bulb and the time duration for which the bulb is used. But the time duration of the bulb should be in hours, as the commercial unit of the energy say in kilowatt-hour.
Hence, energy consumed by the bulb (E)\left( E \right) is P×tP \times t
E=P×t =40watt×5hours=401000kilowatt×5hours=0.04×5kWh=0.2kWh\begin{array}{c}E = P \times t\\\ = 40\,{\rm{watt}} \times 5\,{\rm{hours}}\\\\{\rm{ = }}\dfrac{{40}}{{1000}}\,{\rm{kilo}}\,{\rm{watt}} \times 5\,{\rm{hours}}\\\\{\rm{ = 0}}{\rm{.04}} \times {\rm{5 }}\,{\rm{kWh}}\\\\{\rm{ = 0}}{\rm{.2 kWh}}\end{array}
Hence, the energy consumed by the bulb is 0.2kWh.{\rm{0}}{\rm{.2 kWh}}{\rm{.}}
It can also be said that the energy consumed is 0.2unit.{\rm{0}}{\rm{.2}}\,{\rm{unit}}{\rm{.}}
Cost of 1unit{\rm{1}}\,{\rm{unit}}is 22rupees.
Hence, the cost of 0.2unit{\rm{0}}{\rm{.2}}\,{\rm{unit}}is:
2×0.2=0.4rupees.2 \times 0.2 = 0.4\,{\rm{rupees}}{\rm{.}}

Additional information:
Measurement of energy is done in Kilowatt hour. It can also be written askWh{\rm{kWh}}. The measurement of power is horsepower and it can be written ashp{\rm{hp}}. One kilowatt is equal to 10001000watts. A kilowatt-hour or kWh{\rm{kWh}}means 10001000watt hour. Horsepower is that unit which is used to measure the efficiency of engines of vehicles and other machinery as well as motor parts. It is a mathematical approach to evaluate the power of engines and motors. The term horsepower was invented by James Watt who was the prime inventor of steam engines.
Kilowatt is the unit that is used to measure electricity. When a device is more powerful the number of watts also increases. A kilowatt-hour is an energy unit amounting to 36003600 kilojoules (3.63.6 Mega joules). The kilowatt-hour is widely used as the electricity billing unit supplied by electric utilities to customers.

Note: In the given problem, you are asked to find the cost of the total number of units consumed by the bulb. To do this you must convert the power into kilo-watts.