Chapter 10 - Effects of Electric Current

In the previous section we completed a discussion on the 'relation between electricity and magnetism'. In this chapter, we will see the 'effects of electric current'.

Let us first see how an electric current can produce a 'heating effect'. We will do an activity:
1. In the fig.10.1 below, A and B are two beakers of 200 mL capacity.

Fig.10.1

• Each beaker contains 100 mL of water.
• The conductor PQ in the beaker A is a nichrome wire
• The conductor RS in the beaker B is a copper wire
• Both the conductors have the same length and diameter
2. Trial 1:
Measure the temperature of the water in both the beakers and note it down. Now we can start the trials:
• Turn the switch on
• Note down the current shown by the ammeter reading
• Allow the current to flow through the circuit for three or four minutes 
• Turn off the switch and immediately measure the temperature of water in each beaker. 
    ♦ Let them be t_A and t_B
• We will see that t_A is larger than t_B
3. So the observations are over. Let us analyse them:
• Beakers A and B are connected in series. So the same current (indicated by the ammeter) will be flowing through both nichrome and copper
• Both nichrome and copper pieces are of the same size. Because they have the same length and diameter
• Even when current, lengths and diameters are the same, nichrome produced greater heat. Why is that so?
Ans: Nichrome offers greater resistance to the flow of electricity. That is., electricity cannot flow easily through nichrome. So most of the electrical energy will be converted into heat energy.
4. But the opposite happens in the case of copper
• Copper offers lesser resistance to the flow of electricity. That is., electricity can flow easily through copper. So only a lesser electrical energy will be converted into heat energy.
• Since in this trial, current, lengths and diameters of both nichrome and copper are the same, we can say that, resistance to flow of electricity depends on the material with which the conductor is made. We discussed it in a previous chapter here.
■ So we can write: Greater the resistance (R), greater is the heat energy produced

5. Trial 2:
• Allow the water in the beakers to cool down to room temperature. 

• Remove the beaker B from the circuit. Now there is only the nichrome conductor in beaker A in the circuit. This is shown in fig.10.1(b) above

• Measure the temperature of water in beaker A and note it down.
• Turn on the switch and pass current for five minutes
• Note down the current. Let it be I₁
• Turn off the switch and immediately measure the temperature of water. Note it down. Let it be t₁
6. Allow the water to cool down back to room temperature
• Turn on the switch and pass current for seven minutes
• The current must be the same I₁ as before. Ensure this by the ammeter reading
• Turn off the switch and immediately measure the temperature of water. Note it down. Let it be t₂
• We will see that, t₂ is greater than t₁
■ So we can write: Greater the 'time duration (t) for which current is passed through a conductor', greater is the heat energy produced

7. Trial 3:
• Allow the water to cool down to room temperature. Measure this temperature and note it down
• Turn the switch on and allow the current to pass for three minutes
• Note down the current I₁
• Turn off the switch and immediately measure the temperature of water. Note it down. Let it be t₁
8. Allow the water to cool down back to room temperature
• Turn on the switch and pass current for the same three minutes 
• The current I₁ this time must be greater than I₁. 
    ♦ Ensure this by the ammeter reading. The current can be increased by adjusting the rheostat
• Turn off the switch and immediately measure the temperature of water. Note it down. Let it be t₂
We will see that t₂ is greater than t₁
■ So we can write: Greater the 'current (I) which is passed through a conductor', greater is the heat energy produced

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From the above activity, we can write:
The heat energy produced in a conductor due to the flow of current through it, depends on three items:
(i) The resistance (R) of the conductor
(ii) The time duration (t) for which the current is passed through the conductor
(iii) The intensity of the current (I)
• The English scientist James Prescott Joule discovered the relation connecting those three factors. The relation is known as Joule's Law:
■ The heat generated in a current carrying conductor is the product of the square of the current (I) in the conductor, the resistance (R) of the conductor and the time (t) of the flow of current

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In simple terms, we can say:
The heat will be equal to the product of three items:
(i) Square of the current (I²)
(ii) Resistance (R)
(iii) time (t)
• So if H is the heat generated, then:
Eq.10.1: H = I²Rt 
• I is measured in ampere, R in ohm and t in seconds

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Let us see some interesting cases:

■ Heat of H₁ joules is generated in a conductor when a current I₁ passes through it for t seconds. How much heat will be generated if the current is doubled and is passed for the same t seconds?
Solution:
• We have: H = I²Rt 
• Case 1: H₁ = (I₁)²Rt 
• Case 2: Let H₂ be the heat generated when a current I₂ passes for the same t seconds
• Then we can write:
H₂ = (I₂)²Rt
• Taking ratios, we get: ^H₁⁄_H2= ^(I₁)2Rt⁄_(I2)²Rt = ^(I₁)2⁄_(I2)² = [^(I₁)⁄_(I2)]² 
But given that  I₂ = 2I₁
• So we get: ^H₁⁄_H2 = [^(I₁)⁄_(2I1)]²  = [⁽¹⁾⁄₍₂₎]²  = [¹⁄₄]
⟹ ^H₁⁄_H2 = ¹⁄₄ ⟹ H₂ = 4H₁
• So we can write: The new heat will be four times the original heat

■ Heat of H₁ joules is generated in a conductor when a current I₁ passes through it for t seconds. How much heat will be generated if the current is halved and is passed for the same t seconds?
Solution:
• We have: H = I²Rt 
• Case 1: H₁ = (I₁)²Rt 
• Case 2: Let H₂ be the heat generated when a current I₂ passes for t seconds
• Then we can write:
H₂ = (I₂)²Rt
• Taking ratios, we get: ^H₁⁄_H2= ^(I₁)2Rt⁄_(I2)²Rt = ^(I₁)2⁄_(I2)² = [^(I₁)⁄_(I2)]² 
But given that  I₂ = 0.5I₁
• So we get: ^H₁⁄_H2 = [^(I₁)⁄_(0.5I1)]²  = [⁽¹⁾⁄_(0.5)]²  = [⁽¹⁾⁄_(5/10)]²  = [⁽¹⁰⁾⁄₍₅₎]²  = [2]²  = 4
⟹ ^H₁⁄_H2 = 4 ⟹ H₂ = ^H₁⁄₄ 
• So we can write: The new heat will be one fourth of the original heat

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• The equation H = I²Rt gives us the relation between heat and the quantities: [current, resistance, time]
• Can we relate heat to voltage? Let us try:
• From ohm's law, we have: V = IR (Details here)
• From this we get: I = ^V⁄_R
• So we can use (^V⁄_R) instead of I
• Thus we get: H = (^V⁄_R)²Rt ⟹ H = (^V2⁄_R²)Rt =  (^V2⁄_R)t = ^V2t⁄_R
So we can write:
Eq.10.2: H = ^V2t⁄_R 

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Another derivation:
• From ohm's law, we have: V = IR
• From this we get: R = ^V⁄_I
• So we can use (^V⁄_I) instead of R
• Thus we get: H = I²(^V⁄_I)t ⟹ H = VIt
So we can write:
Eq.10.3: H = VIt

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Now we will see a solved example:
Solved example 10.1:
A bulb of resistance 920 Ω works on 230 V supply. Calculate the quantity of heat generated in 3 minutes
Solution:
1. Given: R = 920 Ω, V = 230 volts, t = 3 minutes = (3 × 60) = 180 seconds 
2. We want a relation that will connect the heat (H) to: [R, V and t]
• So we will use Eq.10.2: H = ^V2t⁄_R 
3. Substituting the values, we get: H = ^(230)2×180⁄₉₂₀ = 10350 J
Another method:
1. We have the basic equation: H = I²Rt 
2. But the current I is not given. We can calculate it using the Ohm's law: V = IR
• So we get: I = ^V⁄_R = ²³⁰⁄₉₂₀ = ¹⁄₄ = 0.25 A   
3. Thus we get: H = I²Rt = H = (0.25)² × 920 × 180 = 10350 J
One more method:
1. Here we will try to use Eq.10.3: H = VIt
• We have calculated I in the previous method. So we can use it
2. Substituting the values, we get: H = 230 × 0.25 × 180 = 10350 J

Solved example 10.2
An electric iron works on 230 V. A current of 3 A flows through it for half an hour. Calculate the amount of heat energy generated
Solution:
1. Given: V = 230 volts, I = 3 A, t = 30 minutes = (30 × 60) = 1800 seconds
2. We want a relation that will connect the heat (H) to: [V, I and t]
• So we will use Eq.10.3: H = VIt
3. Substituting the values, we get: H = 230 × 3 × 1800 = 1242000 J = 1242 kJ

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■ So we have seen a special property of electricity:
• It will produce heat when it passes through a conductor. 
• We also saw how to calculate the 'quantity of heat' produced. 
• In the next section we will see some practical applications of this property.

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Chapter 6 - Heat

In the previous section we completed a discussion on wave motion. In this section we will discuss some basic details about Heat.

• Matter exists mainly in three states: Solids, Liquids and Gases. 

• The building blocks of all material are molecules. It is interesting to know how these molecules are placed inside the materials. Let us analyse:

1. In solids, the molecules are rigidly held in positions. That means they cannot move from their positions. 

• This is because, the inter molecular attractive forces are very high in solids. 

• Also, the molecules in solids are very closely packed. That means, the distances between molecules in solids are very small. 

2. In liquids, the molecules are not rigidly held in positions. So they can move from their positions. 

• This is because, the inter molecular attractive forces are low in liquids. 

• Also, the molecules in liquids are not very closely packed. The distances between molecules in liquids are greater than that in solids. 

3. In gases also, the molecules are not rigidly held in positions. So they can move from their positions.

• This is because, the inter molecular attractive forces are low in gases. These forces are so low that, the molecules of gases have greater freedom than the molecules of liquids

• Also, the molecules in gases are not very closely packed. The distances between molecules in gases are greater than that in solids and liquids.

■ The molecules of all matter are always in a state of motion.

• Even the molecules of solids are always in a state of motion. 

• Hence all the molecules possess kinetic energy. 

• Let us consider the three states of water. 

1. In the solid state, it is ice. When the water becomes ice, it is a solid. 

• The molecules of ice are rigidly held in position. The freedom of motion is very low. 

• Even in such a condition, they possess kinetic energy. But this kinetic energy will be very low.

2. In the liquid state, water molecules have a greater freedom of motion. 

• So they possess a little more kinetic energy than in solids

3. In the gaseous state, water is water vapour. The molecules have very large freedom of motion. 

• So the kinetic energy will also be very large.

Let us do an experiment to understand the relation between heat and kinetic energy. The steps are given below:

1. Wrap some potassium permanganate and a small piece of stone using plastic coated paper. 
• Make one more such packet. 

2. Put small holes in both the packets using a needle. 

3. Take some hot water in one beaker and cold water in another beaker.

• The quatities of water must be the same in both the beakers. 

4. Put the prepared packets into the beakers at the same time. 

• We can see that the colour of the potassium permanganate spreads in the hot water quickly. See fig.6.1 below:

Fig.6.1

• How can we explain this observation?

The explanation can be given in steps: 

(i) In hot water, the molecules have greater kinetic energy. So they have greater speeds. 

(ii) Because of the greater speeds that they possess, they can reach greater distances in lesser time. 

(iii) So the molecules of the potassium permanganate spreads out to a larger area with in a short span of time. 

4. This is not possible in cold water. In it, the molecules will travel only slowly. 

Conclusion:

When any substance is heated, the speed of motion of molecules in that substance increases. So kinetic energy of those molecules increases.

Another experiment:

1. Take equal amount of water in two beakers. 

2. Heat one of them for some time. 

3. Now touch the water in both beakers. 

• We can see that water in the heated beaker is at a higher temperature. 

• This is because, the water in the heated beaker absorbed heat energy. 

• When heat energy is absorbed, the kinetic energy of the water molecules increases 

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• But all the molecules in the hot water beaker will not be having the same kinetic energy.

    ♦ The molecules which are near the source of heat (the spirit lamp or burner) will be having a greater kinetic energy
    ♦ The molecules which are away from the source of heat will be having a lesser kinetic energy
    ♦ The molecules which are at an intermediate distance from the source of heat will be having an intermediate kinetic energy.
• Thus comes the need for mentioning an 'average kinetic energy' 

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So we will see Total kinetic energy and Average kinetic energy:

1. Let the total kinetic energy of the cold water be k1

2. Let the total kinetic energy of the hot water be k2

3. It is clear that k2 is greater than k1

4. The number of molecules in both the beakers are the same. Because we took equal amount of water in both beakers. Let this number be 'n'. 

5. So average kinetic energy in cold water = ^k1⁄_n

• average kinetic energy in hot water = ^k2⁄_n

6. The denominators are the same. And k2 > k1. So we get:

■ Average kinetic energy in hot water is greater than average kinetic energy in cold water.

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• For measuring the 'quantity of heat', we need a physical quantity. The physical quantity that we use for this purpose is temperature.
• 'Temperature' is related to 'quantity of heat' in the same way as 'volume' is related to 'quantity of space'. 

• Let us see an example:

1. Consider the box shown in fig.6.2(a) below:

Fig.6.2

2. A group of students walk upto the box and take it's measurements. They note down the measurements and do some calculations. 

3. Then they write down the result: Volume of the box is 0.0394 cubic meter.

4. Now, another group of students from another part of the world, walk upto the box and take it's measurements. They note down the measurements and do some calculations. 

5. Then they write down the result: Volume of the box is 1.39 cubic feet.

6. The volume of the box remains the same. 

• The two different values (0.0394 and 1.39) are obtained because they used two different system of units. 

    ♦ The first group of used the SI system of units. 

    ♦ The second group used the Imperial system.

7. What ever system is used, the 'quantity of space' or the 'volume' occupied by the box does not change. 

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1. Now consider the sphere shown in fig.6.2(b).
2. A group of scientists walk upto the sphere and measure it's temperature.
3. Then they write down the result: The temperature is 37 degrees Celsius.

4. Another group of scientists from another part of the world walk up to the sphere and measure it's temperature. 

5. Then they write down the result: The temperature of the sphere is 98.6 degrees Fahrenheit.

6. The 'quantity of heat energy' possessed by the sphere is the same. 

• The two different values (37 and 98.6) are obtained because they used two different system of units. 

    ♦ The first group of used the Celsius scale. The '37 degrees Celsius' can be abbreviated as: 37^o C

    ♦ The second group used the Fahrenheit scale. The '98.6 degrees Fahrenheit' can be abbreviated as: 98.6^o F

7. What ever system is used, the 'quantity of heat energy' possessed by the sphere does not change. 

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■ But we have to learn the relation ship between the two systems: 'Degrees Celsius' and 'Degrees Fahrenheit'. Let us analyse:

1. Consider a solid rod shown in red colour in fig.6.3(a) below:

Fig.6.3

• It is made of a suitable material so that, it can withstand extreme temperatures. 

• That is., it can withstand 'very low temperatures' and also 'very high temperatures'. 

2. The rod in fig.6.3(a) is experiencing such extremes. 

• It's left end is at a very low temperature. 

• As we move to the right, the temperature increases gradually. 

• The 'increases in temperature' is uniform. So if we draw a graph, it will be a single straight line. This is shown in fig.6.3(b). 

• The graph shown in fig.(c) shows a non uniform increase. 

3. For our present discussion, the 'increase in temperature' is uniform. 

4. If we consider any particle along the length of the rod, that particle will have more heat than the particles on it's left side.

■ Now we will make a closer study on the rod. 

5. A group of scientists walk upto the rod. They measure the temperatures at various points along the length of the rod. 

6. Then they mark a point 'A' on the rod. Also at that point, they write '0^o C'. It is shown in fig.6.4 below:

Fig.6.4

7. The onlookers asked them: 'Why mark 0^o C at that point?' 

• The scientists replied: 'Because the heat content at that point is 'just low enough' to make water freeze'. 

    ♦ 'Just low enough' means that if we place water any where to the right of 'A', it will not freeze.

    ♦ The word 'just' is used on many occasions in physics. It indicates a 'border'.

8. The temperature at which water freezes is a 'good base mark'. We can relate all other temperatures to it. 

• Now the reader may wonder: The point 'A' is not at the exact left end of the rod. It is at some distance away from the left end. 

• That means the particles to the left of 'A' are colder than the freezing point of water. Are such colder temperatures possible?

• Of course they are possible. For example, 'dry ice' is colder than 'ice made from water'.

9. Now another group of scientists from another part of the world, walk upto the rod. They measure the temperatures at various points along the length of the rod. 

10. Then at the same point 'A', they write '32^o F'. This is shown in fig.6.5 below:

Fig.6.5

11. The onlookers asked the same question as before: 'Why mark 32^o F at that point?' 

• The reply was also the same as before: 'Because the heat content at that point is 'just low enough' to make water freeze'.

12. The 'quantity of heat at point A is the same. But the two groups of scientists used different systems of measurements. That is why we have two different values: 0 and 32 at the point A 

• The first group used the Celsius scale. 

• The second group used the Fahrenheit scale.

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The scientists continued their work. We will see it in the next section.

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