HD
11:13

LearnNext Lesson Video

Heating effect of electricity is one of the widely used effects in the world. When electric current is passed through a conductor, it generates heat due to the resistance it offers to the current flow. The work done in overcoming the resistance is generated as heat.

This is studied by James Prescott Joule and he enunciated various factors that affect the heat generated. The heat produced by a heating element is directly proportional to the square of the electric current (I) passing through the conductor, directly proportional to the resistance (R) of the conductor, time (t) for which current passes through the conductor. It is given by the expression

H = I^{2}Rt and is well known as Joule’s Law.

Applications of the heating effect of electric current include appliances like electric immersion water heater, electric iron box, etc. All of these have a heating element in it. Heating elements are generally made of specific alloys like, nichrome, manganin, constantan etc.

A good heating element has high resistivity and high melting point. An electric fuse is an example for the application of heating effect of electric current. The rating of 3 A of an electric fuse implies the maximum current it can sustain is three ampere.

The chemical reaction within the cell generates the potential difference between its two terminals that sets the electrons in motion to flow the current through a resistor or a system of resistors connected to the battery. To maintain the current, the source has to keep expending its energy.

A part of the source energy in maintaining the current may be consumed into useful work (like in rotating the blades of an electric fan). Rest of the source energy may be expended in heat to raise the temperature of gadget. For example, an electric fan becomes warm if used continuously for longer time, etc.

If the electric circuit is purely resistive, that is, a configuration of resistors only connected to a battery, the source energy continually gets dissipated entirely in the form of heat. This is known as the heating effect of electric current. When a conductor offers resistance to the flow of current the work done by the electric current in overcoming this resistance is converted into heat energy.

The generation of heat in a conductor is an inevitable consequence of electric current. In many cases, it is undesirable as it converts useful electrical energy into heat. In electric circuits, the unavoidable heating can increase the temperature of the components and alter their properties.

The electric heating is also used to produce light, as in an electric bulb. Here, the filament must retain as much of the heat generated as is possible, so that it gets very hot and emits light. It must not melt at such high temperature. A strong metal with high melting point such as tungsten (melting point 3380°C) is used for making the filaments of the bulb. The filament should be thermally isolated as much as possible, using insulating support, etc. The bulbs are usually filled with chemically inactive nitrogen and argon gases to prolong the life of filament. Most of the power consumed by the filament appears as heat, but a small part of it is in the form of light radiated.

Devices which work on the heating effect of electric current have a heating element or filament. Good heating elements have high resistivity, high melting point and negligible variation in resistance due to temperature changes.

The three metal alloys most commonly used as heating elements are: Nichrome (80% Ni + 20% Cr); Manganin (86% Cu + 12% Mn + 2% Ni); Constantan (60% Cu + 40% Ni).

**Joule's law**

The **Joule's law** states that the quantity of heat produced in a resistor is directly proportional to: (i)the square of current for a given resistance, (ii) the resistance for a given current, and (iii) the time for which the current flows through the resistor, i.e., H = I^{2}Rt.

Consider a current I flowing through a resistor of resistance R.

Let the potential difference across it be V.

Let t be the time during which a charge Q flows across.

The work done in moving the charge Q through a potential difference, V =VQ.

⇒ The source must supply energy equal to VQ in time t.

⇒ The power input to the circuit by the source is

P = $\frac{\text{VQ}}{\text{t}}$ = VI.

Or the energy supplied to the circuit by the source in time t

⇒H= P × t,

⇒ H = VIt

This energy gets dissipated in the resistor as heat.

Thus for a steady current I, the amount of heat H produced in time t is H = VIt.

Applying Ohm’s law, H = I^{2} Rt.

In practical situations, when an electric appliance is connected to a known voltage source, current can be calculated using the relation I = $\frac{\text{V}}{\text{R}}$ . Using this value in H = I^{2}Rt, the heat produced can be calculated.

One of the common applications of Joule’s heating is the fuse used in electric circuits. An electric fuse is a safety device used to protect circuits and appliances by stopping the flow of any unduly high electric current. It works on the heating effect of electric current.

**Electric Fuse**

The fuse is placed in series with the device. An electric fuse consists of a piece of wire made of a metal or an alloy of appropriate melting point, for example aluminum, copper, iron, lead, etc. If a current larger than the specified value flows through the circuit, the temperature of the fuse wire increases. This melts the fuse wire and breaks the circuit. The fuse wire is usually encased in a cartridge of porcelain or similar material with metal ends.

The fuses used for domestic purposes are rated as 1 A, 2 A, 3 A, 5 A, 10 A, etc. For an electric iron which consumes 1 kW electric power when operated at 220 V, a current of (1000/220) A, that is, 4.54 A will flow in the circuit. In this case, a 5 A

fuse must be used.

**Electric energy and power**

The rate of doing work is called power. This is also the rate of consumption of energy.

The equation H = I^{2 }Rt gives the rate at which electric energy is dissipated

or consumed in an electric circuit. This is also termed as electric power. The power P is given by P = VI. Or P = I^{2}R = $\frac{{\text{V}}^{\text{2}}}{\text{R}}$ .

The SI unit of electric power is watt (W).

It is the power consumed by a device that carries 1 A of current when operated at a potential difference of 1 V.

1 W = 1 volt × 1 ampere = 1 V A.

The unit ‘watt’ is very small. Therefore, in actual practice we use a much larger unit called ‘kilowatt’. It is equal to 1000 watts.

Since electrical energy is the product of power and time, the unit of electric energy is, therefore, watt hour (W h). One watt hour is the energy consumed when

1 watt of power is used for 1 hour.

The commercial unit of electric energy is kilowatt hour (kW h), commonly known as ‘unit’. 1 kW h = 1000 watt × 3600 second = 3.6 × 10^{6} watt second = 3.6 × 10^{6} joule (J).

We pay the electricity board or electric company to provide energy to move electrons through the electric gadgets like electric bulb, fan and engines. We pay for the energy that we use and not for the electrons. Electrons are not consumed in a circuit, as many people think.

**Activity 1**

**Ualg.pt/~jlongras/aulas** has deveoped an interactive online simulation in which one can understand the heating effects of electricity and Joule's law.

**Activity 2**

**Tutorvista** has deveoped an interactive online simulation in which one can understand the heating effects of electricity and Joule's experiment to determine the mechanical equivalent of heat i.e., Joule's constant.

**Activity 3**

**Harcourtschool** has deveoped an interactive online simulation in which one can understand the heating effects of electricity by a simulation of glowing an electric bulb.

**What is the resistance of an ideal voltmeter and ammeter? How are they connected in a circuit?**

**Answer: </**...

The resistance of a conductor depends on:-1. Length 2. Nature of material 3. Area of cross section 4. Temperature resistance does not change...

P = V
^{2}/R

If V = 220 V we have

100 W = 220
^{2}/R

R = 220
^{2}/100 ? = 484 ?...

SI Unit of Heat Energy is Joule(J)

Maximum current that can be allowed to passthrough a bulb is

I
_{Max} = P/V

I
_{Max} = 2000/220

I
_{}...