**What is Heating Effect of Electric Current?** When an electric current flows through a conductor, electrical energy is expended in overcoming the frictional resistance between the electrons and the molecules of the wire.

If potential difference of V volts is applied across a conductor and current of I amperes flows through it for time oft seconds, then energy expended will be equal to VIt watt-seconds or joules.

If R is the resistance of the conductor through which a current of I amperes flows and V is the potential difference applied across its ends then by Ohm’s law

V = I R

and energy expended,

W = V It = IR*I*t = I^{2} R t joules …..(1)

Or also W = joules ……(2)

According to the *law of conservation of energy* this electrical energy expended must be converted in some other form of energy and that other form is heat i.e. electrical energy expended is converted into heat energy and conversion of electrical energy into heat energy is called the heating or *thermal effect* of *electric current*.

## Joule’s Law of Electric Heating

From Eq. (1) the energy expended or heat generated in joules when a current of I amperes flows through a resistance of R ohms for t seconds is given as

H = I^{2} R *t* joules

The above expression is known as ** Joule’s law**, which states that the amount of heat produced in an electric circuit is

- proportional to the square of the current i.e. H α I
- proportional to the resistance of the circuit i.e. H α R and
- proportional to the time duration for which the current flows through the circuit i.e. H α t

Heat produced in k calls,

[1 kcal = 4, 180J = 4,200 J]

## Practical Application of Heating Effect of Electric Current

The use of heating effect of electric current is made in various appliances, called the electric appliances, in industrial heating, welding, space heating and in electrical installation protection.

The most familiar example of practical electrical heating is the metal filament lamp through which, when current is passed its filament gets heated to such a high temperature that it emits the visible radiations called the light.

Familiar domestic applications are the electric kettles, electric stoves, electric irons, electric percolators, the electric cookers etc. Every heating appliance is built around a heating element (or elements) consisting essentially of a spiral high resistivity alloy such as nichrome, which generates heat on passing of electric current through it.

Electric heating is also preferred over other methods of heating in industry owing to its inherent advantages. Electric ovens are in common use in large bakeries, electric kilns bake earthen ware and china, while electric furnaces are used for several purposes, such as for the melting of metals and the annealing of tools. The heat of the electric arc in welding has brought a revolution to the various metal industries. An electric arc is an electric discharge in gases accompanied by high heat and a bright glow and is formed when two conductors, called the electrodes, are brought together to make an electric contact and then separated.

Recent applications are the heating and conditioning of the air in large buildings, and what is called ‘space heating’, in which open fires or radiators are replaced by tubular electric heaters of about 51 mm diameter and taking about 16,000 Watts per meter run.

Safety fuses and thermal relays, which are used for the protection of electrical installations against overheating, also operate on the heating effect of electric current. A fuse is a length of wire or of strip (copper, lead or silver) of low melting point placed in a circuit to get heated and melt when the current through it has reached a pre-determined value.

## Conversion Efficiency

Energy converting machines or devices cannot deliver all the energy that is supplied to them. If they could, their efficiency would be 100 percent. Some of the energy supplied to the machine is lost either in overcoming friction or in some other way. The resistance of electrical conductor causes the heat generation when current flows through it. This is lost energy when it occurs in a current source such as a generator or a battery, or in a convertor of energy such as an electric motor. Friction also accounts for energy loss in a motor.

Conversion efficiency may be defined as the ratio of output energy to total input energy required to produce it. This must occur during the same time interval. The Greek letter eta (η) is commonly used as the letter symbol for efficiency

It is usually more convenient to calculate efficiency as the ratio of output power to input power.

[lates size=2] \eta = \dfrac{P_{out}}{P_{in}}[/latex]

## Electrical unit of Work, Power and Energy

The unit of work done and of energy expended is joule. It is equal to the energy expended in passing 1 coulomb of charge through a resistance of 1 ohm i.e. the energy expended in passing one ampere current for 1 second through a resistance of one ohm is taken as one joule. It may also be expressed as lwatt-Second i.e. one watt of power consumed for one second.

i.e., 1 joule = 1 watt-second

The unit of energy, joule or watt-Second is too small for practical purposes, so a bigger unit mega joule (MJ) or kilowatt-hour (kWh) is used in electrical engineering.

1 kWh = 1,000 watt-hours

= 1,000*3,600 watt-seconds or joules = 3.6 MJ.

The kWh, also called the Board of Trade (BOT) unit, is the energy absorbed by supplying a load of 1 kW or 1,000 watts for the period of one hour. This is legal unit on which charges for electrical energy are made, and, therefore, it is called the Board of Trade (BOT) unit.

Watt: It is defined as the power expended when there is an unvarying current of one ampere between two points having a potential difference of one volt. As already stated the bigger unit of power is kW or megawatt.

1 kW = 1,000 Watts

1 MW = 1,000 kW = 1 x 10^{6} watts

## Conversion of Electrical Unit into Mechanical and Thermal Unit or Vice-Versa

1 Watt = 1 joule/second = 1 N-m/s

1 kW = 1,000 Watts or J/s or N-m/s

i.e. 1.36 hp (metric)

kWh = 1,000 Watt-hours

= 3,600,000 watt-seconds or joules

= 1.36 hp-hour (metric)

1 calorie = 4.18 J or Watt-Seconds

1 k cal. = 4, 180 J or Watt-Seconds =

1 kWh = 36*10^{5} watt-seconds = i.e. 860 k cal.

## Examples of Electrical heating of current

*Example 1. An electric motor takes a power of 2 kW. The efficiency of the motor is 80%. Calculate the output of motor and express in kg-m/s.*

**Solution:**

Motor input = 2 kW

Motor output = Motor efficiency motor input

= 0.8 2 = 1.6 kW

= 1,600 watts or Joules/s

= kg-m/s **Ans. [1 kg-m/s = 9.81 J/s]**

*Example 2. A 220 V dc motor develops 10 hp at 600 rpm. Calculate the torque developed in newton meter. If the efficiency of the motor is 85%, calculate the current taken from the mains. Also, find the units consumed/day for 6 hours working per day. Find also the monthly bill if cost of one unit is Rs 4.5/kWh.*

**Solution:**

Output of motor = 10hp

Speed of motor, N = 600 rpm

Torque developed in N-m,

= 117 N-m **Ans. **[1 hp(metric) = 735.5 N-m/s]

= 8,653 N-m/s or watts

Current drawn from 220 V main.

**Ans.**

Units consumed per hour

Units consumed per day (6 hours)

= 8.6536 = 52 units **Ans.**

Unit consumed per month

= 5230 = 1560

Monthly bill = Rs4.51560 = Rs 7,020 **Ans.**

*Example 3: How much heat is produced by 2 kW electric heater when it is operated for 30 minutes.*

**Solution**:

Heat Produced, H = Rating in kW times op operation

= 2 kW hour = 1 kWh

= 3610^{5} joules or 860 k cal. **Ans.**

*Example 4: An electric iron is marked 250 V, 500 W. What current does it take if connected to the correct voltage? What is hot resistance? If the iron is used for one hour daily for 30 days in a month, what will be the monthly bill at Rs 4.50 per unit? How does the cost get affected if the voltage is only 200 V instead of 250 V?*

**Solution:**

Wattage rating of an iron,

P = 500 watts

Voltage rating, V = 250 V

Current drawn when connected to rated voltage of 250 volts,

**Ans.**

Hot resistance, **Ans.**

Working hours/month,

*t _{ }*= Working hrs/day working days/month

= 130 = 30

Monthly energy consumptions

=

Monthly bill = Rs. 4.5015 = Rs. 67.50 **Ans.**

Power drawn by the electric iron when connected to 200 V,

Monthly bill =

= Rs. 43.20 **Ans.**

*Example 5: A current was sent through a resistance wire of 10 ohms, which is fully immersed in 2 kg of water. At the end of 15 minutes, the rise in temperature was observed to be 60 ^{0}C. Determine the value of current.*

**Solution:**

Water to be heated, m = 2 kg

Rise in temperature, (*t _{2} – t_{1}*) = 60

^{0}C

Specified heat of water, s = 4,200 J/kg/^{0}C

Energy utilized in rising the temperature of water

= ms(*t _{2} – t_{1}*) = 24,20060 = 504,000 J

= 560,000

Assume heating efficiency to be 90 percent

=

Current drawn, **Ans.**

[power developed, P = I^{2}R]

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