Question

An electric heater or a motor of 750W and 1kW are used for domestic purpose for \[2\,hours\] and \[1\dfrac{1}{2}\,hours\] per day in the month of January respectively. Calculate the cost of energy used if the unit costs \[5.50\] .

Answer 1

Hint: Units are power consumed in \[kW\,H\] . Therefore convert the given powers into units and calculate for each day. Then use the units consumed per day to calculate for each month. Using the unitary method, calculate the total cost of units consumed.

Complete step by step answer:
Power is the rate of energy supplied or used. Its SI unit is watt ( \[W\] ).
The units of energy supplied are in \[kW\,H\] (kilowatt hour)
Energy used by electric heater in \[kW\,H\] = \[750\times {{10}^{-3}}\times 2=1.5\,units\] ---------(1)
 Similarly energy used by a motor is-
\[1000\times {{10}^{-3}}\times \dfrac{3}{2}=1.5\,units\] ---------- (2)
Total units used in a day will be-
 \[1.5+1.5=3\,units\] [From eq (1) and eq (2)]
Cost of units per day=
\[3\times 5.5=16.5\] ------ (3)
Total cost of units for the month of January can be calculated using eq (3),
\[16.5\times 31=511.5\]

The total cost of units for the month of January will be Rs \[511.5\] .

Additional Information: Power is a scalar quantity. It is also called the rate of doing work. Power can be calculated as \[P=\dfrac{E}{\Delta t}\] ( \[E\] is energy consumed while \[\Delta t\] is the time interval), \[P=Fv\] ( \[F\] is the force applied while \[v\] is the velocity).

Note: When you use \[1kW\text{ }or\text{ }1000W\] in \[1\,hour\] then you consume 1 unit of energy or a power of \[1kW\,H\] . The electricity consumed in real life is in units. Other units of power are calories per hour, ergs per second, foot-pounds per minute etc. The dimensions of power are- [ \[M{{L}^{2}}{{T}^{-2}}\] ].