`color{red} ♦` HEATING EFFECT OF ELECTRIC CURRENT

`color{red} ♦`ELECTRIC POWER

`color{red} ♦`ELECTRIC POWER

We know that a battery or a cell is a source of electrical energy.

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.

We have also seen, in Section 12.2, that to maintain the current, the source has to keep expending its energy. Where does this energy go?

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.

We often observe this in our everyday life. For example, an electric fan becomes warm if used continuously for longer time etc. On the other hand, 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. This effect is utilised in devices such as electric heater, electric iron etc. Consider a current I flowing through a resistor of resistance R. Let the potential difference across it be V (Fig. 12.13).

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 is VQ. Therefore, the source must supply energy equal to VQ in time t. Hence the power input to the circuit by the source is

` P = V Q/t = V I` .........(12.19)

Or the energy supplied to the circuit by the source in time t is `P xx t`, that is, VIt. What happens to this energy expended by the source? 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 = V I t` ..........(12.20)

Applying Ohm’s law [Eq. (12.5)], we get

`H = I^2 Rt` .............(12.21)

This is known as Joule’s law of heating. The law implies that heat produced in a resistor is (i) directly proportional to the square of current for a given resistance, (ii) directly proportional to resistance for a given current, and (iii) directly proportional to the time for which the current flows through the resistor.

In practical situations, when an electric appliance is connected to a known voltage source, Eq. (12.21) is used after calculating the current through it, using the relation `I = V//R`.

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.

We have also seen, in Section 12.2, that to maintain the current, the source has to keep expending its energy. Where does this energy go?

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.

We often observe this in our everyday life. For example, an electric fan becomes warm if used continuously for longer time etc. On the other hand, 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. This effect is utilised in devices such as electric heater, electric iron etc. Consider a current I flowing through a resistor of resistance R. Let the potential difference across it be V (Fig. 12.13).

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 is VQ. Therefore, the source must supply energy equal to VQ in time t. Hence the power input to the circuit by the source is

` P = V Q/t = V I` .........(12.19)

Or the energy supplied to the circuit by the source in time t is `P xx t`, that is, VIt. What happens to this energy expended by the source? 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 = V I t` ..........(12.20)

Applying Ohm’s law [Eq. (12.5)], we get

`H = I^2 Rt` .............(12.21)

This is known as Joule’s law of heating. The law implies that heat produced in a resistor is (i) directly proportional to the square of current for a given resistance, (ii) directly proportional to resistance for a given current, and (iii) directly proportional to the time for which the current flows through the resistor.

In practical situations, when an electric appliance is connected to a known voltage source, Eq. (12.21) is used after calculating the current through it, using the relation `I = V//R`.

Q 3284591457

An electric iron consumes energy at a rate of `840 W` when heating is at the maximum rate and `360 W` when the heating is at the minimum. The voltage is `220 V`. What are the current and the resistance in each case?

Class 10 Chapter 12 Example 10

Class 10 Chapter 12 Example 10

From Eq. (12.19), we know that the power input is

`P = V I`

Thus the current `I = P//V`

(a) When heating is at the maximum rate,

`I = 840 W//220 V = 3.82 A`;

and the resistance of the electric iron is

`R = V//I = 220 V//3.82 A = 57.60 Ω`.

(b) When heating is at the minimum rate,

`I = 360 W//220 V = 1.64 A;`

and the resistance of the electric iron is

`R = V//I = 220 V//1.64 A = 134.15 Ω`

Q 3204591458

`100 J` of heat are produced each second in a `4 Ω` resistance. Find the potential difference across the resistor.

Class 10 Chapter 12 Example 11

Class 10 Chapter 12 Example 11

`H = 100 J, R = 4 Ω, t = 1 s, V = ?`

From Eq. (12.21) we have the current through the resistor as

`I = √(H//Rt)`

`= √[100 J/(4 Ω xx 1 s)]`

`= 5 A`

Thus the potential difference across the resistor, V [from Eq. (12.5)] is

`V = I R`

`= 5 A xx 4 Ω`

`= 20 V`.

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. However, heating effect of electric current has many useful applications.

The electric laundry iron, electric toaster, electric oven, electric kettle and electric heater are some of the familiar devices based on Joule’s heating. 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 bulb filaments.

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. Another common application of Joule’s heating is the fuse used in electric circuits. It protects circuits and appliances by stopping the flow of any unduly high electric current.

The fuse is placed in series with the device. It consists of a piece of wire made of a metal or an alloy of appropriate melting point, for example aluminium, 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.

In electric circuits, the unavoidable heating can increase the temperature of the components and alter their properties. However, heating effect of electric current has many useful applications.

The electric laundry iron, electric toaster, electric oven, electric kettle and electric heater are some of the familiar devices based on Joule’s heating. 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 bulb filaments.

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. Another common application of Joule’s heating is the fuse used in electric circuits. It protects circuits and appliances by stopping the flow of any unduly high electric current.

The fuse is placed in series with the device. It consists of a piece of wire made of a metal or an alloy of appropriate melting point, for example aluminium, 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.

You have studied in your earlier Class that the rate of doing work is power. This is also the rate of consumption of energy.

Equation (12.21) 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 = V^2//R`..........(12.22)

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`. Thus,

`1 W = 1 `volt `xx 1` ampere `= 1 V A` ..........(12.23)

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 `xx 3600 ` second

`= 3.6 xx 10^6` watt second

`= 3.6 xx 10^6` joule (J)

Equation (12.21) 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 = V^2//R`..........(12.22)

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`. Thus,

`1 W = 1 `volt `xx 1` ampere `= 1 V A` ..........(12.23)

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 `xx 3600 ` second

`= 3.6 xx 10^6` watt second

`= 3.6 xx 10^6` joule (J)

Q 3214591459

An electric bulb is connected to a `220 V` generator. The current is `0.50 A`. What is the power of the bulb?

Class 10 Chapter 12 Example 12

Class 10 Chapter 12 Example 12

`P = VI`

`= 220 V xx 0.50 A`

`= 110 J//s`

`= 110 W`.

Q 3214691550

An electric refrigerator rated `400 W` operates `8` hour/day. What is the cost of the energy to operate it for `30` days at Rs `3.00` per kW h?

Class 10 Chapter 12 Example 13

Class 10 Chapter 12 Example 13

The total energy consumed by the refrigerator in `30` days would be

`400 W xx 8.0` hour/day `xx 30` days `= 96000` W h

`= 96` kW h

Thus the cost of energy to operate the refrigerator for `30 ` days is

`96 kW \ \ h xx R s 3.00` per `kW \ \ h = Rs 288.00`