Physics Topics cover a broad range of concepts that are essential to understanding the natural world.

## How does Heating Effect of Current Work?

An electric current can produce three important effects. These are : (1) Heating effect, (2) Magnetic effect, and (3) Chemical effect. We will now discuss the heating effect of current. The magnetic effect of current will be discussed in the next Chapter whereas the chemical effect of current will be described in higher classes.

When an electric current is passed through a high resistance wire, like nichrome wire, the resistance wire becomes very hot and produces heat. This is called the heating effect of current. The heating effect of current is obtained by the transformation of electrical energy into heat energy.

Just as mechanical energy used to overcome friction is converted into heat, in the same way, electrical energy is converted into heat energy when an electric current flows through a resistance wire. Thus, the role of ‘resistance’ in electrical circuits is similar to the role of friction in mechanics. We will now derive a formula for calculating the heat produced when an electric current flows through a resistance wire.

Since a conductor, say a resistance wire, offers resistance to the flow of current, so work must be done by the current continuously to keep itself flowing. We will calculate the work done by a current I when it flows through a resistance R for time t. Now, when an electric charge Q moves against a potential difference V, the amount of work done is given by :

W = Q × V …………….. (1)

From the definition of current we know that :

Current, I = \(\frac{Q}{t}\)

So, Q = I × t …………… (2)

And from Ohm’s law, we have :

\(\frac{V}{I}\) = R

or Potential difference, V = I × R ………….. (3)

Now, putting Q = I × t and V = I × R in equation (1), we get :

W = I × t × I × R

So, Work done, W = I^{2} × R × t

Assuming that all the electrical work done or all the electrical energy consumed is converted into heat energy, we can write ‘Heat produced’ in place of ‘Work done’ in the above equation. Thus, Heat produced,

H = I^{2} × R × t joules

This formula gives us the heat produced in joules when a current of I amperes flows in a wire of resistance R ohms for time t seconds. This is known as Joule’s law of heating. According to Joule’s law of heating given by the formula H = I^{2} × R × t, it is clear that the heat produced in a wire is directly proportional to :

- square of current (I
^{2}) - resistance of wire (R)
- time (t), for which current is passed

(a) Since the heat produced is directly proportional to the square of current:

H ∝ I^{2}

so, if we double the current, then the heat produced will become four times. And if we halve the current, then heat generated will become one-fourth.

(b) Since the heat produced in a wire is directly proportional to the resistance :

H ∝ R

so, if we double the resistance, then heat produced will also get doubled. And if we halve the resistance, then the heat produced will also be halved. This means that a given current will produce more heat in a high resistance wire than in a low resistance wire.

We know that when two similar resistance wires are connected in series, then their combined resistance gets doubled but when they are connected in parallel then their combined resistance gets halved. So, a given current will produce more heat per unit time if the two resistances are connected in series than when they are connected in parallel.

(c) Since the heat produced in a wire is directly proportional to the time for which current flows :

H ∝ t

so, if the current is passed through a wire for double the time, then the heat produced is doubled. And if the time is halved, the heat produced is also halved.

We will now solve some problems based on the heating effect of current. Please note that the formula : H = I^{2} × R × t for calculating the heat produced can be used only if the current I, resistance R and time t are known to us. In some cases, however, they give us the power P and time t only. In that case the heat energy is to be calculated by using the formula : E = P × t.

It should be noted that all the appliances which run on electricity do not convert all the electric energy into heat energy. Only the electrical heating appliances convert most of the electric energy into heat energy. For example, when electric current is passed through an electric appliance such as a fan, then most of the electric energy is used up in running the fan (or turning the fan), only a very small amount of electric energy is converted into heat energy by a fan. Due to this, an electric fan becomes slightly warm when run continuously for a long time.

On the other hand, when electric current is passed through an electrical heating appliance such as an electric heater, electric kettle, hair dryer, immersion rod or a geyser, then most of the electrical energy is converted into heat. All the electrical heating appliances have a ‘heating element’ or ‘heating coil’ made of high resistance wire (like nichrome wire) which helps in converting most of the electric energy into heat energy. We will now solve some problems based on the heating effect of current.

**Example Problem 1.**

A potential difference of 250 volts is applied across a resistance of 500 ohms in an electric iron. Calculate (i) current, and (it) heat energy produced in joules in 10 seconds.

**Solution:**

(i) Calculation of Current. The current can be calculated by using Ohm’s law equation :

\(\frac{V}{I}\) = R

Here, Potential difference, V = 250 volts

Current, I = ? (To be calculated)

Resistance, R = 500 ohms

Putting these values in the above formula, we get :

\(\frac{250}{I}\) = 500

So, I = \(\frac{250}{500}\)

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

= 0.5 ampere

Thus, the current flowing in the electric iron is 0.5 A.

(ii) Calculation of Heat Energy. The heat energy in joules can be calculated by using the formula :

H = I^{2} × R × t

Here, Current, I = 0.5 A

Resistance, R = 500 Ω

And, Time, t = 10 s

Putting these values in the above formula, we get:

H = (0.5)^{2} × 500 × 10

= 1250 joules

**Example Problem 2.**

Calculate the heat produced when 96,000 coulombs of charge is transferred in 1 hour through a potential difference of 50 volts.

**Solution:**

First of all we will calculate the current by using the values of charge and time. We know that:

Current, I = \(\frac{Q}{t}\)

I = \(\frac{96,000}{60 \times 60}\) (Because 1 h = 60 × 60 s)

I = 26.67 A

We will now calculate the resistance by using Ohm’s law :

R = \(\frac{V}{I}\)

R = \(\frac{50}{26.67}\)

R = 1.87 Ω

Heat produced, H = I^{2} × R × t

= (26.67)^{2} × 1.87 × 60 × 60

= 4788400 J

= 4788.4 kJ

Thus, the heat produced is 4788.4 kilojoules.

**Example Problem 3.**

Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then in parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be :

(a) 1:2

(b) 2:1

(c) 1:4

(d) 4:1

**Solution:**

Suppose the resistance of each one of the two wires is x.

(i) When the two resistance wires, each having a resistance x, are connected in series, then :

Combined resistance, R_{1} = 2x

And, if the potential difference in the circuit is V, then applying Ohm’s law :

Current, I_{1} = \(\frac{V}{2 x}\)

Suppose the heat produced with the series combination of wires is H_{1}. Then :

H_{1} = I_{1}^{2} × R_{1} × t

or H_{1} = \(\left(\frac{V}{2 x}\right)^2\) × 2x × t = \(\frac{V^2}{4 x^2}\) × 2x × t

or H_{1} = \(\frac{V^2 \times t}{2 x}\)

(ii) When the two resistance wires, each of resistance x, are connected in parallel, then :

Combined resistance, R_{2} = \(\frac{x}{2}\)

And if the potential difference in the circuit is V, then applying Ohm’s law :

Current, I_{2} = \(\frac{V \times 2}{x}\)

Suppose the heat produced with the parallel combination of wires is H_{2}. Then :

H2 = if x R2 x t

Dividing equation (1) by equation (2), we get :

H_{2} = I_{2}^{2} × R_{2} × t

or H_{2} = \(\left(\frac{V \times 2}{x}\right)^2 \times \frac{x}{2}\) = \(\frac{V^2 \times 4 \times x \times t}{x^2 \times 2}\)

or H_{2} = \(\frac{V^2 \times 2 \times t}{x}\)

Dividing equation (1) by equation (2), we get :

\(\frac{H_1}{H_2}\) = \(\frac{V^2 \times t \times x}{2 x \times V^2 \times 2 \times t}\)

\(\frac{H_1}{H_2}\) = \(\frac{1}{4}\) or H_{1} : H_{2} = 1 : 4

Thus, the correct option is : (c) 1 : 4

### Applications of the Heating Effect of Current

The important applications of the heating effect of electric current are given below :

1. The heating effect of current is utilised in the working of electrical heating appliances such as electric iron, electric kettle, electric toaster, electric oven, room heaters, water heaters (geysers), etc. All these heating appliances contain coils of high resistance wire made of nichrome alloy. When these appliances are connected to power supply by insulated copper wires then a large amount of heat is produced in the heating coils (because they have high resistance), but a negligible heat is produced in the connecting wires of copper (because copper has very, very low resistance).

For example, the heating element (or coil) of an electric heater made of nichrome glows because it becomes red-hot due to the large amount of heat produced on passing current (because of its high resistance), but the cord or connecting cable of the electric heater made of copper does not glow because negligible heat is produced in it by passing current (because of its extremely low resistance). The temperature of the heating element (or heating coil) of an electrical heating device when it becomes red-hot and glows is about 900°C.

2. The heating effect of electric current is utilised in electric bulbs (electric lamps) for producing

light. When electric current passes through a very thin, high resistance tungsten filament of an electric bulb, the filament becomes white-hot and emits light. Please note that the same current flowing through the tungsten filament of an electric bulb produces enormous heat but almost negligible heat is produced in the connecting wires of copper. This is because of the fact that the fine tungsten filament has very high resistance whereas copper connecting wires have very low resistance.

Tungsten metal is used for making the filaments of electric bulbs because it has a very high melting point (of 3380°C). Due to its very high melting point, the tungsten filament can be kept white-hot without melting away. The other properties of tungsten which make it suitable for making filaments of electric bulbs are its high flexibility and low rate of evaporation at high temperature. Please note that when the tungsten filament of an electric bulb becomes white-hot and glows to emit light, then its temperature is about 2500°C !

If air is present in an electric bulb, then the extremely hot tungsten filament would burn up quickly in the oxygen of air. So, the electric bulb is filled with a chemically unreactive gas like argon or nitrogen (or a mixture of both). The gases like argon and nitrogen do not react with the hot tungsten filament and hence prolong the life of the filament of the electric bulb.

It should be noted that most of the electric power consumed by the filament of an electric bulb appears as heat (due to which the bulb becomes hot), only a small amount of electric power is converted into light. So, filament-type electric bulbs are not power efficient. On Metai caps for connecting the other hand, tube-lights are much more power efficient, because they have no filaments.

3. The heating effect of electric current is utilised in electric fuse for protecting household wiring and electrical appliances. A fuse is a short length of a thin tin plated copper wire having low melting point. The thin fuse wire has a higher resistance than the rest of the electric wiring in a house. So, when the current in a household electric circuit rises too much due to some reason, then the fuse wire gets heated too much, melts and breaks the circuit (due to which the current stops flowing).

This prevents the fire in house (due to over-heating of wiring) and also prevents damage to various electrical appliances in the house due to excessive current flowing through them. Thus, an electric fuse is a very important application of the heating effect of current.