What is the Commercial Unit of Electrical Energy?

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 × 10⁶ J)

From this discussion we conclude that 1 kilowatt-hour is equal to 3.6 × 10⁶ 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 × 10⁶ J
So, 3 kWh = 3.6 × 10⁶ × 3 J
= 10.8 × 10⁶ J (or 10.8 × 10⁶ 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 = I² × 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

Watt‘hour (W ᐧ h) : Work done or electrical energy consumed in 1 hour by an electrical appliance having power 1W is called 1W h.
∴ 1W ᐧ h = 1W × 1h = 1W × 3600s = 3600J
It is obvious that, W ᐧ h = W × h = V × A × h
i. e., if 1A current flows for lh under the potential difference of 1V, the amount of electrical energy spent will be 1W ᐧ h.

kilowatt ᐧ hour (kW ᐧ h): Work done or electrical energy spent in 1 h by an electrical appliance having power 1kW is called 1kW ᐧ h or 1BOT unit.
Obviously, this is a bigger unit of electrical energy. The electric-ity supplied to us for our use is measured by this unit. Many times, kW ᐧ h or BOT unit is simply called unit.
1kW ᐧ h = 1000W ᐧ h = 1000W × 1h
= 1000W × 3600s = 3.6 × 10⁶W ᐧ s
= 3.6 × 10⁶J
kW ᐧ h = \(\frac{W \times h}{1000}\) = \(\frac{V \times A \times h}{1000}\)
For example, ¡fan electric bulb marked 100 W glows for 20h then,
amount of electrical energy spent = \(\frac{100 \times 20}{1000}\) = 2kW ᐧ h
= 2 unit
So, in this case the value of electric meter wilL rise up by 2 units.

Numerical Examples

Example 1.
In a house there are 10 lamps of 40W each, 5 fans of 80W each and a TV set of 80 W.Theyrun for 6 hours a day. Find the consumption of electrical energy in a month of 30 days. What is its value in BOT unit? [HS ‘05]
Solution:
Totalpower = (10 × 40) + (5 × 80) + 80 = 880W.
Total time = 6 × 30 = 180 h = 180 × 3600 s
∴ Energy consumed in a month = 880 × 180 × 3600
= 5.7 × 10⁸J
Its value in BOT unit = \(\frac{\mathrm{W} \times \mathrm{h}}{1000}\) = \(\frac{880 \times 180}{1000}\) = 158.4

Example 2.
In an evening college there are 100 bulbs of 60 W each, 80 bulbs of 100 W each and 70 fans of 100 W each. They run for 5h, 4h and 4h respectively per day. If each kW ᐧ h costs ₹ 0.50, calculate the electric bill of the college for a month.
Solution:
Electrical energy consumed by 60W bulbs
= \(\frac{(60 \times 100) \times 5 \times 30}{1000}\) = 900 kW ᐧ h
Electrical energy consumed by 1oo W bulbs
= \(\frac{(100 \times 80) \times 4 \times 30}{1000}\) = 960kW ᐧ h
Electrical energy consumed by fans
= \(\frac{(100 \times 70) \times 4 \times 30}{1000}\) = 840 kWh
∴ Cost of electrical energy = (900 + 960 + 840) × 0.50 = ₹ 1350

Example 3.
In a house there are 20 lamps of 60W each, 10 fans’ which operate in 0.5A current. If main power supply is 220V, expense per kW ᐧ h is 50 paise and each appliance runs 6 h per day then calculate the electric bill of November. [HS ‘07]
Solution:
November month has 30 days.
∴ Total time = 30 × 6h = 180h; 50 paise = ₹0.5
Power of each fan = 220 × 0.5 = 110W
∴ Electrical energy,
= \(\frac{(60 \times 20+110 \times 10) \times(30 \times 6)}{1000}\)
= \(\frac{2300 \times 30 \times 6}{1000}\) = ₹(23 × 18) × 0.5 = ₹ 207

Example 4.
There are six 40 W and two 100 W lamps, four 40 W fans and a 1000 W electric heater in a house. If in April, each lamp runs for 5 hours a day, each fan 15 hours a day and the heater 2 hours a day, what will be the electric bill for that month? It may be supposed that the main supply voltage is 200 V and cost of each BOT unit = ₹ 1.50.
Which one of the given three fuse wires of ratIng 5 A, 10 A and 15 A will be appropriate for connection in the main switch? [HS 2000]
Solution:
Number of days in April = 30
Electrical energy consumed by the lamps
= \(\frac{(6 \times 40+2 \times 100) \times 5 \times 30}{1000}\) = 66 BOT unit
Electrical energy consumed by the fans
= \(\frac{(4 \times 40) \times 15 \times 30}{1000}\) = 72 BOT unit
Electrical energy consumed by the heater
= \(\frac{1000 \times 2 \times 30}{1000}\) = 60 BOT unit
∴ Monthly expenditure of electricity
=(66 + 72 + 60) × 1.50 = ₹ 297
Again, total power of the appliances
= (6 × 40) + (2 × 100) + (4 × 40) + 1000 = 1600W
If all the appliances run simultaneously, according to the relation, P = V × I,
we have, I = \(\frac{P}{V}\) = \(\frac{1600}{200}\) = 8 A
So, 10 A fuse will be appropriate for connection in the main switch because, 5 A fuse will melt and there is no need of 15 A fuse.

Example 5.
The power of a small electric motor is \(\frac{1}{8}\)HP. If it is connected to 220 V supply line, how much current will it draw? If the motor runs for 80 h, what will be the cost? (Each BOT unit costs 70 palse)
Solution:
Power P = \(\frac{1}{8}\)HP = \(\frac{1}{8}\) × 746W
According to the relation, P = V × I,
we have, I = \(\frac{P}{V}\) = \(\frac{1}{8}\) × 746 × \(\frac{1}{220}\) = 0.424 A
Number of BOT units = \(\frac{1}{8}\) × 746 × 80 × \(\frac{1}{1000}\)
∴ Cost = \(\frac{1}{8}\) × 746 × 80 × \(\frac{1}{1000}\) × 70 paise = ₹ 5,22(approx.)

Example 6.
A heating coil of resistance 100 Ω is connected for 30 min to 220 V. By this time, determine
(i) amount of charge flowing,
(ii) amount of electrical energy consumed and
(iii) amount ofheat generated. Determine the cost of consumed electrical energy If 1 kW ᐧ h costs ₹ 1.
Solution:
i) Amount of charging flowing,
Q = It = \(\frac{V}{R} t\) = \(\frac{220}{100}\) × 30 × 60 = 3960 C

ii) Amount of electrical energy consumed,
W = Q ᐧ V = 3960 × 220 = 8.712 × 10⁵J

iii) Amount of heat generated,
H = \(\frac{W}{J}\) = \(\frac{8.712 \times 10^5}{4.2}\) = 2.074 × 10⁵J

iv) Cost of consumed electrical energy = \(\frac{8.712 \times 10^5}{3600 \times 1000}\) × 1
= ₹ 0.24 = 24 paise

Electric Fuse

Every metal, such as copper, aluminium etc. used as the element of connecting wires in electrical circuits and electrical appliances has a definite melting point. Due to short circuit or any other reason, if an excessively high current flows in the circuit, the temperature of the wire ihay rise to its melting point. As a result, the whole circuit may be damaged and may cause fire. A fuse inserted in the circuit may prevent the accidental fire. It is prepared from an alloy (e.g., 3 parts lead and 1 part tin) having a comparatively low melting point.

This fuse is placed in an insu-lating box and is connected in series with the main circuit. Before the current of the main circuit reaches the danger point, the temperature of the fuse reaches its melting point. As a result, the fuse melts down and the main circuit is disconnected. Hence, there is no chance of damage to the main circuit.

Circuit breaker: This is an automatic switch system which disconnects the circuit before being damaged due to short-circuit or overload. At the very moment the circuit is disconnected, the switch of the breaker moves to the ‘off’ position. After replacing the defective instrument for which the circuit was broken, as the switch is made ‘on! current flows as usual through the circuit. In different types of circuit breakers different physical phenomena are used.

For example, in the circuit through which high current flows, a bimetallic strip bends due to heating and disconnects the circuit; or if a high current flows through an electromagnet, it attracts a soft-iron plate of the circuit and thus the circuit is cut off.

The main difference between a fuse wire and a circuit breaker is that, the fuse wire is burnt off and needs to be replaced with a new wire every time the fuse melts. On the other hand, a circuit breaker is not at all damaged and hence there is no need to replace it. Once the fault has been repaired, the contacts must again be closed to restore power to the interrupted circuit.

The circuit breaker which is used in low voltage line (0 -1000 V) is generally called Mechanical or Miniature Circuit Breaker (MCB). Nowadays, in place of fuse wire MCB is widely used. Switch gear is generally used as circuit breaker in the supply line of high voltage (11000 V or more) from the generating station.

Highest safe current: If l be the length of the fuse wire, r be its radius and ρ its specific resistance, then resistance of the wire is given by,
R = \(\rho \frac{l}{\pi r^2}\)
So, heat generated in the wire per second is,
H = \(\frac{I^2 R}{J}\) = \(\frac{I^2 \rho l}{\mathrm{~J} \pi r^2}\)

This heat is used in two ways:

1. One part increases the temperature of the wire,
2. the other part radiates to the surroundings from the outer surface of the wire.

Thus, a situation arises when the total heat generated in the wire, radiates to the surroundings. So, temperature does not increase further and reaches a fixed value. If this temperature lies just below the melting point of the material of the fuse wire, then the current flowing in the wire at this stage is called the highest safe current.
Now, the outer surface area of the wire = 2πrl.
If h be the amount of radiant heat from unit area of the wire per second, then the total radiant heat from the outer surface of the wire per second is, H’ = 2π rlh.
So, for highest safe current I,
H = H’ or, \(\frac{I^2 \rho l}{J \pi r^2}\) = 2πrlh or, I² = \(\frac{2 \pi^2 J h}{\rho} \cdot r^3\)
i.e., I² ∝ r³ or, I ∝ r^3/2
It is to be noted that, the expression for highest safe current l is independent of the length l of the fuse wire.