Electricity Bill

At home we run many kinds of electrical appliances. A heater, an iron, a rice cooker and an electric jug all turn electrical energy into heat. Lights, televisions and fans use electrical energy too. They do not all use it at the same rate. Some appliances change a lot of energy every second, and others change very little.

The electrical energy an appliance changes in one second is its electric power, P, measured in watts (W). The maker prints this power on a sticker or plate on the back or underneath every appliance. The television plate below shows 145 W. That means the television uses 145 joules of electrical energy every second while it is on. Working out an electricity bill always begins by reading this printed power.

The amount of electrical energy that is changed into other forms while an appliance is used is called electricity consumption. A lamp gives light and heat, a fan gives motion, a heater gives heat. The electrical energy spent in each case is the consumption. The electricity meter in a house measures the consumption of all its appliances together.

The SI unit of energy is the joule (J). But the energy used in a home is so large that writing it in joules gives huge numbers. A heater uses millions of joules in just one hour. So a much bigger unit, the kilowatt hour (kWh), is used when electricity is bought and sold. It is called the commercial unit of electricity consumption.

To find how many joules make 1 kWh, change kilowatts to watts and hours to seconds. Remember 1 kW = 1000 W and 1 h = 3600 s.

\[ 1\ \text{kWh} = 1000\ \text{W} \times 1\ \text{h} \]

\[ 1\ \text{kWh} = 1000\ \text{W} \times 3600\ \text{s} \]

\[ = 3.6 \times 10^{6}\ \text{J} = 3.6\ \text{MJ} \]

Power is the energy used each second. So the longer an appliance runs, the more energy it uses. The energy an appliance changes is its power multiplied by the time it is used.

\[ E = P \times t \]

Here E is the energy changed, P is the electric power and t is the time. If P is written in kilowatts and t in hours, E comes out directly in kWh, which is units. For example, a 2 kW heater run for 3 hours uses \( 2 \times 3 = 6 \) kWh, or 6 units. The same heater run for only half an hour uses \( 2 \times 0.5 = 1 \) unit.

A house may have more than one appliance of the same kind, such as four bulbs or two fans. If one appliance has power P in kilowatts (kW), is used for time t in hours (h), and there are n such appliances, their electricity consumption is:

\[ E = P \times n \times t \quad (\text{kWh}) \]

For example, three 10 W LED bulbs lit for 4 hours a day: first change 10 W to 0.01 kW. The daily consumption is then \( 0.01 \times 3 \times 4 = 0.12 \) kWh.

Every house has an electricity meter (kWh meter) fitted where the supply line enters. All the energy used by every appliance in the house passes through it. The number on its display tells the total kWh used since the meter was fitted. The meter below shows 12345.6 kWh.

A meter reading never goes down; it only keeps rising. So to find one month's consumption, subtract last month's reading from this month's reading.

Example: a house meter read 12230.4 kWh last month and 12345.6 kWh this month. If the rate is Rs. 10 per unit, what is the bill?

Step 1: Subtracting last month's reading from this month's reading:

\[ 12345.6 - 12230.4 = 115.2\ \text{units} \]

Step 2: Multiplying the units by the rate per unit:

\[ 115.2 \times 10 = \text{Rs. } 1152 \]

So that month's electricity bill is Rs. 1152.

\[ \text{Bill} = \text{units} \times \text{rate} \]

Even without looking at the meter, the monthly bill can be worked out from a list of the household appliances. The work is done in this order:

• Write each appliance's power P in kilowatts.
• Write how many of it there are (n) and how many hours a day it is used (t).
• Find each appliance's daily consumption with \( P \times n \times t \).
• Add them all to get the total daily consumption of the house.
• Multiply the daily total by 30 to get the monthly consumption.
• Multiply the monthly consumption by the rate per unit to get the monthly bill.

A house uses six 15 W CFLs for 5 hours a day, four 7 W LED bulbs for 6 hours a day, two 1 kW heaters for 2 hours a day, a 900 W rice cooker for 50 minutes a day and an 800 W iron for 30 minutes a week. What is its monthly electricity consumption? If 1 unit costs Rs. 10, what is the monthly bill?

First put every figure into the right units in a table:

Step 1: Daily consumption of the CFLs, using P × n × t:

\[ \frac{15}{1000} \times 6 \times 5 = 0.45\ \text{kWh} \]

Step 2: Daily consumption of the LED bulbs:

\[ \frac{7}{1000} \times 4 \times 6 = 0.168\ \text{kWh} \]

Step 3: Daily consumption of the heaters:

\[ 1 \times 2 \times 2 = 4\ \text{kWh} \]

Step 4: Daily consumption of the rice cooker:

\[ \frac{9}{10} \times 1 \times \frac{5}{6} = 0.75\ \text{kWh} \]

Step 5: Daily consumption of the iron:

\[ \frac{8}{10} \times 1 \times \frac{0.5}{7} = 0.057\ \text{kWh} \]

Step 6: Adding the daily consumption of all the appliances:

\[ 0.45 + 0.168 + 4 + 0.75 + 0.057 = 5.425\ \text{kWh} \]

Step 7: Multiplying by 30 for the monthly consumption:

\[ 5.425 \times 30 = 162.75\ \text{units} \]

Step 8: Multiplying by the rate of Rs. 10 per unit for the monthly bill:

\[ 162.75 \times 10 = \text{Rs. } 1627.5 \]

So the house uses 162.75 units a month and its monthly electricity bill is Rs. 1627.5.

In the example, 4 of the house's 5.425 daily units were used by the two heaters alone. Six CFLs and four LEDs together used less than 1 unit. Heating appliances have high power, so they use many units even in a short time. If a 7 W LED gives as much light as a 15 W CFL, the LED uses less than half the energy. Because consumption is P × n × t, there are three ways to lower a bill: choose appliances of lower power, switch off the ones not needed, and use them for less time.