Contents

By learning Physics Topics, we can gain a deeper appreciation for the natural world and our place in it.

## Formula for Calculating Electrical Energy

We will now derive a formula for calculating electrical energy in terms of power and time. We have already studied that :

Electric power = \(\frac{\text { Work done by electric current }}{\text { Time taken }}\)

Now, according to the law of conservation of energy,

Work done by electric current = Electric energy consumed

So, we can now write down the above relation as :

Power = \(\frac{\text { Electrical energy }}{\text { Time }}\)

or Electrical energy = Power × Time

or E = P × t

It is obvious that the electrical energy consumed by an electrical appliance is given by the product of its power rating and the time for which it is used. From this we conclude that the electrical energy consumed by an electrical appliance depends on two factors : (i) power rating of the appliance, and (ii) time for which the appliance is used. We should memorize the above formula for calculating electrical energy because it will be used in solving numerical problems.

In the formula : Electrical energy = Power × Time, if we take the power in ‘watts’and time in ‘hours’ then the unit of electrical energy becomes ‘Watt-hour’ (Wh). One watt-hour is the amount of electrical energy consumed when an electrical appliance of 1 watt power is used for 1 hour. We will now describe the commercial unit (or trade unit) of electrical energy called kilowatt-hour.

### Unit of Electrical Energy : Kilowatt-Hour

The SI unit of electrical energy is joule and we know that “1 joule is the amount of electrical energy consumed when an appliance of 1 watt power is used for 1 second”. Actually, joule represents a very small quantity of energy and, therefore, it is inconvenient to use where a large quantity of energy is involved.

So, for commercial purposes we use a bigger unit of electrical energy which is called “kilowatt-hour”. One kilowatt- hour is the amount of electrical energy consumed-when an electrical appliance having a power rating of 1 kilowatt is used for 1 hour. Since a kilowatt means 1000 watts, so we can also say that one kilowatt-hour is the amount of electrical energy consumed when an electrical appliance of 1000 watts is used for 1 hour.

In other words, one kilowatt-hour is the energy dissipated by a current at the rate of 1000 watts for 1 hour. From this discussion we conclude that the commercial unit of electrical energy is kilowatt-hour which is written in short form as kWh.

### Relation between kilowatt-hour and joule

1 kilowatt-hour is the amount of energy consumed at the rate of 1 kilowatt for 1 hour. That is,

1 kilowatt-hour = 1 kilowatt for 1 hour

or 1 kilowatt-hour = 1000 watts for 1 hour ………… (1)

But : 1 watt = \(\frac{1 \text { joule }}{1 \text { second }}\)

So, equation (1) can be rewritten as :

1 kilowatt-hour = 1000 \(\frac{1 \text { joule }}{1 \text { second }}\) for 1 hour

And, 1 hour = 60 × 60 seconds

So, 1 kilowatt-hour = 1000 \(\frac{1 \text { joule }}{1 \text { second }}\) × 60 × 60 seconds

or 1 kilowatt-hour = 36,00,000 joules (or 3.6 × 10^{6} J)

From this discussion we conclude that 1 kilowatt-hour is equal to 3.6 × 10^{6} joules of electrical energy. It should be noted that watt or kilowatt is the unit of electrical power but kilowatt-hour is the unit of electrical energy. Let us solve some problems now.

**Example Problem 1.**

A radio set of 60 watts runs for 50 hours. How much electrical energy is consumed ?

**Solution:**

We know that :

Electrical energy = Power × Time

or E = P × t ……………… (1)

We want to calculate the electrical energy in kilowatt-hours, so first we should convert the power of 60 watts into kilowatts by dividing it by 1000. That is :

Power, P = 60 watts

= \(\frac{60}{1000}\) kilowatt

= 0.06 kilowatt

And, Time, t = 50 hours

Now, putting P = 0.06 kW and, t = 50 hours in equation (1), we get :

Electrical energy, E = 0.06 × 50

= 3 kilowatt-hours (or 3 kWh)

Thus, electrical energy consumed is 3 kilowatt-hours.

Note. In the above problem we have calculated the electrical energy consumed in the commercial unit of energy ‘kilowatt-hour’ (kWh). We can also convert this electrical energy into SI unit of energy called joule by using the relation between kilowatt-hour and joule. Now,

1 kWh = 3.6 × 10^{6} J

So, 3 kWh = 3.6 × 10^{6} × 3 J

= 10.8 × 10^{6} J (or 10.8 × 10^{6} joules)

**Example Problem 2.**

A current of 4 A flows through a 12 V car headlight bulb for 10 minutes. How much energy transfer occurs during this time ?

**Solution:**

Energy = Power × Time .

or E = P × t ………….. (1)

First of all we should calculate power P by using the current of 4 A and voltage of 12 V.

Now, P = V × I

So, P = 12 × 4

or, Power, P = 48 watts

= \(\frac{48}{1000}\) kilowatts

Thus, Power, P = 0.048 kW

And, Time, t = 10 minutes

= \(\frac{10}{600}\) hours

= \(\frac{1}{6}\) hours

Now, putting P = 0.048 kW and, t = \(\frac{1}{6}\) hours in equation (1), we get :

E = 0.048 × \(\frac{1}{6}\)

= 0.008 kWh

Thus, the energy transferred is 0.008 kilowatt-hour.

**Example Problem 3.**

Calculate the energy transferred by a 5 A current flowing through a resistor of 2 ohms for 30 minutes.

**Solution:**

We will first calculate the power by using the given values of current and resistance. This can be done by using the formula :

P = I^{2} × R

Here, Current, I = 5 amperes

And, Resistance, R = 2 ohms

So, Power, P = (5)^{2} × 2

= 25 × 2 = 50 watts

= \(\frac{50}{1000}\) kilowatts

Thus, Power, P = 0.05 kW

And, Time, t = 30 minutes

= \(\frac{30}{60}\) hours

= \(\frac{1}{2}\) hours

= 0.5 hours ……………… (2)

Now, Energy, E = P × t

= 0.05 × 0.5

Energy, E = 0.025 kWh

### How to Calculate the Cost of Electrical Energy Consumed

Kilowatt-hour is the “unit” of electrical energy for which we pay to the Electricity Supply Department of our City. One unit of electricity costs anything from rupees 3 to rupees 5 (or even more). The rates vary from place to place and keep on changing from time to time. Now, by saying that 1 unit of electricity costs say, 3 rupees, we mean that 1 kilowatt-hour of electrical energy costs 3 rupees.

The electricity meter in our homes measures the electrical energy consumed by us in kilowatt-hours (see Figure). Now, we use different electrical appliances in our homes. We use electric bulbs, tube-lights, fans, electric iron, radio, T.V., and refrigerator, etc. All these household electrical appliances consume electrical energy at different rates.

Our electricity bill depends on the total electrical energy consumed by our appliances over a given period of time, say a month. We will now describe how the cost of electricity consumed is calculated. Since the electricity is sold in units of kilowatt-hour, so first we should convert the power consumed in watts into kilowatts by dividing the total watts by 1000.

The kilowatts are then converted into kilowatt-hours by multiplying the kilowatts by the number of hours for which the appliance has been used. This gives us the total electrical energy consumed in kilowatt-hours. In other words, this gives us the total number of “units” of electricity consumed. Knowing the cost of 1 unit of electricity, we can find out the total cost. This will become more clear from the following examples.

**Example Problem 1.**

A refrigerator having a power rating of 350 W operates for 10 hours a day. Calculate the cost of electrical energy to operate it for a month of 30 days. The rate of electrical energy is Rs. 3.40 per kWh.

**Solution:**

Electrical energy, E = P × t

Here, Power, P = 350 W

= \(\frac{350}{0.35}\) kW

= 0.35 kW

And, Time, t = 10 × 30 hours

= 300 h

Now, putting these values of P and t in the formula,

E = P × t

We get: E = 0.35 × 300 kWh

= 105 kWh

Thus, the electrical energy consumed by the refrigerator in a month of 30 days is 105 kilowatt-hours.

Now, Cost of 1 kWh of electricity = Rs. 3.40

So, Cost of 105 kWh of electricity = Rs. 3.40 ×105

= Rs. 357

**Example Problem 2.**

A bulb is rated at 200 V-100 W. What is its resistançe? Five such bulbs burn for 4 hours. What is the electrical energy consumed? Calculate the cost if the rate is ₹ 4.60 per unit.

**Solution:**

(a) Calculation of Resistance. Here we know the voltage and power of the bulb. So, the resistance can be calculated by using the formula :

P = \(\frac{V^2}{R}\)

Here, Power, P = 100 watts

Voltage, V = 200 volts

And, Resistance, R = ? (To be calculated)

Now, putting these values in the above formula, we get :

100 = \(\frac{(200)^2}{R}\)

100 R = 40000

And, R = \(\frac{40000}{100}\)

= 400 ohms

(b) Calculation of Electrical Energy Consumed. The electrical energy consumed in kilowatt-hours can be calculated by using the formula :

E = P × t

Here, Power, P = 100 watts

= \(\frac{100}{1000}\) kilowatt

= 0.1 kilowatt …………. (1)

And, Time, t = 4 hours ………….. (2)

So, Energy consumed by 1 bulb = 0.1 × 4

= 0.4 kilowatt-hours

And,Energy consumed by 5 bulbs = 0.4 × 5

= 2 kilowatt-hours (or 2 kWh)

Thus, the total electrical energy consumed is “2 kilowatt-hours” or “2 units”.

(c) Calculation of Cost of Electrical Energy. We have been given that : ,

Cost of 1 unit of electricity = ₹ 4.60

So, Cost of 2 units of electricity = ₹ 4.60 × 2

= ₹ 9.20

**Example Problem 3.**

An electric heater draws a current of 10 A from a 220 V supply. What is the cost of using the heater for 5 hours everyday for 30 days if the cost of 1 unit (1 kWh) is ₹ 5.20 ?

**Solution:**

In this problem, first of all we have to calculate the power of the heater by using the given values of current and voltage. This can be done by using the formula :

P = V × I

Here, Voltage (or p.d.), V = 200 V

And, Current, I = 10 A

So, Power, P = 220 × 10 W

= 2200 W

= \(\frac{2200}{1000}\) kW

= 2.2 kW …………. (1)

Now, Electric energy consumed, E = P × t

Here, Power, P = 2.2 kW

And, Time, t = 5 h

So, Electric energy consumed in 1 day = 2.2 × 5

= 11 kWh

And, Electric energy consumed in 30 days = 11 × 30

= 330 kWh (or 330 units) ………………… (2)

Now, Cost of 1 unit of electricity = ₹ 5.20

So, Cost of 330 units of electricity = ₹ 5.20 × 30

= ₹ 1716