Question

In a residence $4$ tube lights, each of them of \[40\;W\] is operated daily for $5$ hours, and $3$ fans each of \[120\;W\] are operated daily for $4$ hours. Calculate the electricity bill at Rs. $5$ per unit for September month.

Answer 1

Hint: The amount of electrical energy transmitted to an appliance is determined by its power and the span of time it is switched on. The total mains electrical energy transferred is calculated using kilowatt-hours \[kWh\]. One unit is \[1{\text{ }}kWh\].

Complete step by step solution:
The expression indicating electric energy is given by \[E = P \times t\] where \[E\] is the energy transmitted in kilowatt-hours \[kWh\], \[P\] is the power in kilowatts \[kW\], and $t$ is the time in hours \[h\]. Note that power is calculated using kilowatts here as an alternative for the usual watts. To change from \[W\]to \[kW\]you must divide by \[1,000\].
Here the given Power of one tube light \[ = 40W\]
Power of $4$ tube lights \[ = 4 \times 40 = 160W = 0.16kW\]
Energy consumed by $4$ tube lights every day \[ = 0.16 \times 5 = 0.8kWh\]
Power of one fan \[\; = 120W\]
Power of $3$ fans \[ = {\text{ }}3 \times 120{\text{ }} = {\text{ }}360W = 0.36kW\]
Energy spent by $3$ fans each day \[ = 0.36 \times 4 = 1.44kWh\]
Total electrical energy spent per day \[ = 0.8 + 1.44 = 2.24kWh\]
Therefore the total number of units spent per day \[ = 2.24\,units\]
Cost of one unit \[ = 5\]
Number of days in September \[ = 30\;days\]
Therefore Electricity bill for September\[ = 2.24 \times 30 \times 5 = 336rs\]

Note:
For a given time interval, the energy spent (or delivered, depending on your viewpoint) is given by the equation \[PE = qV\], where \[E\] is the electric energy,\[\;V\] is the voltage, and \[q\] is the quantity of charge relocated in the time interval under. We can see the relationship between the total energy consumed and the power by integrating over time. Positive energy matches the consumed energy and negative energy matches the energy production. If the power is the same over the time interval then the energy can be given simply as \[E = Pt\].