High school Physics Lessons: conductor

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Thursday, May 10, 2018

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

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

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

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⁄_(I_₂_)²Rt = ^(I₁^)²⁄_(I_₂_)² = [^(I₁⁾⁄_(I_₂₎]²
But given that I₂ = 2I₁
• So we get: ^H₁⁄_H₂= [^(I₁⁾⁄_(2I_₁₎]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⁄_(I_₂_)²Rt = ^(I₁^)²⁄_(I_₂_)² = [^(I₁⁾⁄_(I_₂₎]²
But given that I₂ = 0.5I₁
• So we get: ^H₁⁄_H₂= [^(I₁⁾⁄_(0.5I_₁₎]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

• 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 = (^V^²⁄_R_²)Rt = (^V^²⁄_R)t = ^V^²t⁄_R
So we can write:
Eq.10.2: H = ^V^²t⁄_R

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

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 = ^V^²t⁄_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

■ 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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Thursday, April 19, 2018

Chapter 8.2 - Relation between Voltage and current

In the previous section we saw how to connect voltmeter and ammeter to circuits. In this section, we will see the relation between voltage and current.

Let us do an activity. The steps are written below:
1. Make a circuit with the following components:
An ammeter, switch, a cell and a bulb.
• The circuit diagram is shown in fig.8.16 below:

Fig.8.16

2. Turn on the switch
• Note down the ammeter reading
• Note down the intensity of light from the bulb
• Turn off the switch. This completes one trial.
3. Add one more cell in the circuit. The two cells should be connected in series
4. Turn on the switch
• Note down the ammeter reading
• Note down the intensity of light from the bulb
• Turn off the switch. This completes the second trial.
5. Add one more cell in the circuit. The three cells should be connected in series
6. Turn on the switch
• Note down the ammeter reading
• Note down the intensity of light from the bulb
• Turn off the switch. This completes the third and final trial.

The trials are complete. The observations are tabulated below:

Table.8.1

From the table we can make the following two conclusions:
1. When the number of cells (connected in series) increase, ammeter reading increases
• The increase in ammeter reading indicates increase in current
■ So we can write:
When the number of cells (connected in series) increase, the current in the circuit increases
2. When the number of cells (connected in series) increase, intensity of light increases
• That means, when current increases, intensity of light also increases
■ What is the reason?
We will write the answer in steps:
(i) We know that, light is produced as a result of the heating of the filament of the bulb
(ii) More current passing through the filament means that more electrons passing through it per second
(iii) So the filament will glow with more intensity
• We can write this in another way also:
When current increase, the heat produced also increases

Another activity:
This activity is performed to find the relation between current and potential difference. We will write the steps:
1. Make a circuit with the following components:
An ammeter, a voltmeter, switch, a 1.5 V cell and a 30 cm long nichrome wire.
• The circuit diagram is shown in fig.8.17 below:

Fig.8.17

2. Turn on the switch
• Note down the voltmeter reading. This reading should be entered in the table in the 'V' column
• Note down the ammeter reading. This reading should be entered in the table in the 'I' column
• Turn off the switch. This completes one trial.
3. Add one more 1.5 V cell in the circuit. The two cells should be connected in series
4. Turn on the switch
• Note down the voltmeter reading. This reading should be entered in the table in the 'V' column
• Note down the ammeter reading. This reading should be entered in the table in the 'I' column
• Turn off the switch. This completes the second trial.
5. Add one more 1.5 V cell in the circuit. The three cells should be connected in series
6. Turn on the switch
• Note down the voltmeter reading. This reading should be entered in the table in the 'V' column
• Note down the ammeter reading. This reading should be entered in the table in the 'I' column
• Turn off the switch. This completes the third and final trial.

The trials are complete. The observations are tabulated below:

Table.8.2

• Note that, the last column is filled up by calculating ^V⁄_I for each trial
• From the table we can make the following conclusions:
Conclusion 1: When cells are connected in series, the 'available potential difference' increases.
■ Since the potential difference is measured in volts, we say:
When cells are connected in series, voltage increases.
• When a single 1.5 V cell is connected in series, the voltmeter reading = 1.5 V
• When two 1.5 V cells are connected in series, the voltmeter reading = (1.5 × 2) = 3 V
• When three 1.5 V cells are connected in series, the voltmeter reading = (1.5 × 3) = 4.5 V
Conclusion 2:
We will arrive at this conclusion by writing the required steps:
(i) From the third and fourth columns we get:
• When voltage increases, current also increases.
(ii) This can be written mathematically as: V ∝ I
• That is., V is proportional to I
(iii) We can avoid the '∝' symbol by introducing a 'constant of proportionality'. See details here.
That is., V = (a constant) × I
⟹ ^V⁄_I = a constant
(iv) That means:
• We can do any number of trials we like.
• In each of those trials, we can calculate ^V⁄_I using the V and I obtained in that trial.
• Then we can compare those ^V⁄_I values. All those values will be the same.
• This is indeed true as can be seen from the last column. All values are '10'
■ So the conclusion is:
^V⁄_Iis a constant

• This property was first discovered by the German scientist George Simon Ohm
■ He formulated the Ohm's law. It states that:
When temperature remains constant, the current through a conductor is directly proportional to the potential difference between it's ends.
• Note that, the temperature should remain constant.
• This is important because, if we increase or decrease the temperature, the internal molecular and ionic properties of a conductor will change. So the resistance given (against the flow of current) by the conductor will change. We cannot do the calculations if the temperature changes.

• So we have: ^V⁄_I = a constant
• This constant is given a special name: 'Resistance of the conductor' or simply 'Resistance'
• It is denoted by the letter 'R'
• So we can write: ^V⁄_I= R

How can we apply this law to a practical situation?
We will write the answer in steps:
1. Consider the activity that we saw just above.
• We have: Current = 0.15 A (when 1.5 V cell is connected)
2. What if we want a higher current with the same 1.5 V?
• We have: ^V⁄_I = R = 10
• That is: R = ^1.5⁄_I= 10
3. The current 'I' is in the denominator. So if we decrease 10, I will increase.
4. How can we decrease 10?
• We can decrease it by decreasing the length of the nichrome wire.
• That is., if we decrease the length of the nichrome wire, the 'resistance to the flow of current' will decrease.
• So the current will increase even without any increase in the number of cells.
■ The reverse is also applicable. That is., if we want to reduce the current without decreasing the voltage, we can increase the length of the nichrome wire.

■ So 'resistance' is a very convenient way to increase or decrease current. The convenience is that:
Using resistance, we can change the current without changing the voltage
• Nichrome wire is often used to provide resistance in circuits.
♦ A longer nichrome wire will give a higher resistance
♦ A shorter nichrome wire will give a lower resistance.
• Some images of nichrome wire can be seen here.
• The components in a circuit whose function is to 'provide a resistance to the flow of current' are called resistors
■ The official definition is:
Resistors are conductors used to include a particular resistance in a circuit
• Some resistors available in the market can be seen here.
• In circuit diagrams, they are shown using the symbol:

Unit of resistance

1. In the definition, note the words: 'particular resistance'
• It means that, we must know 'how much resistance' is to be provided in a circuit.
• Then only we can purchase a resistor
2. So we must be able to 'measure resistance'.
• For 'measuring resistance', we must need an appropriate unit.
3. Let us try to establish a unit:
• We know that R = ^V⁄_I.
♦ Unit of voltage 'V' is volt
♦ Unit of current 'I' is ampere
• So unit of R is ^volt⁄_ampere
• This '^volt⁄_ampere' is given a special name: Ohm
• It's symbol is 'Ω'. It is the Greek letter 'omega'

■ So what can we say about '1 Ω'?
• That is., we want to know the peculiarity about a resistor, whose resistance is '1 Ω'
We will write the steps:
1. Consider the circuit shown in fig.8.18 below:

Fig.8.18

• A voltmeter is connected to know the 'potential difference across the two ends of a resistor'
• Recall that 'potential difference across the two ends of a resistor' is same as 'voltage across the two ends of a resistor'
• An ammeter is connected to know the current flowing through the circuit
• Let the voltmeter reading be 1 V
♦ Then we can say: The voltage between the ends of the resistor is 1 volt.
• Let the ammeter reading be 1 A
♦ Then we can say: A current of 1 ampere is flowing through the circuit.
♦ That is., a current of 1 ampere is flowing through the resistor
• So the resistance 'R' of the resistor shown in that circuit is: ^{1 V}⁄_{1 A}. = 1 Ω
■ The official definition is:
If a conductor connected to a voltage of one volt, passes a current 1 A, then the conductor will have '1 Ω' resistance

Using basic algebra, the equation R = ^V⁄_I can be written in two other forms also:
• I = ^V⁄_R
• V = IR
• We can use any one of the three equations. The choice depends on the requirements in the problem
• We can use the following triangle to remember the equations:

■ This is called the VIR triangle
• If we want to calculate V, then we put a finger over V.
♦ That leaves I and R
♦ I and R are on the same level
♦ So we get V = IR
• If we want to calculate I, then we put a finger over I
♦ That leaves V and R
♦ V is at top and R is at bottom
♦ So we get I = ^V⁄_R
• If we want to calculate R, then we put a finger over R
♦ That leaves V and I
♦ V is at top and I is at bottom
♦ So we get R = ^V⁄_I

Solved example 8.1
(a) In a circuit, the voltmeter connected across a 4 Ω resistor showed a reading of 12 V. What was the current flowing through that resistor at that time?
(b) In a circuit, a voltmeter is connected to a 3 Ω resistor. The ammeter reading shows that 2 A current is flowing through it. What is the potential difference across the resistor?
(c) In a circuit, a voltmeter is connected to a resistor. The voltmeter reading shows that, the potential difference across the resistor is 6 V. The ammeter reading shows that 3 A current is flowing through it. What is the resistance of the resistor?
Solution:
Part (a):
1. The given values are: V = 12 V, R = 4 Ω
We have to calculate I
2. We have: I = ^V⁄_R
Substituting the known values, we get: I = ¹²⁄₄= 3 A
Part (b):
1. The given values are: I = 2 A, R = 3 Ω
We have to calculate V
2. We have: V = IR
Substituting the known values, we get: V = 2 × 3 = 6 V
Part (c):
1. The given values are: V = 6 V, I = 3 A
We have to calculate R
2. We have: R = ^V⁄_I
Substituting the known values, we get: R = ⁶⁄₃= 2 Ω

• We have seen the relation between voltage and current.
• Let us draw a graph connecting the two
• We can use the values recorded in table 8.2 above. The resulting graph is shown below:

Let us see the features of the graph:
• The bold yellow line is our required graph.
• We can see that, it is a straight line. Why is it straight?
Let us analyse:
1. We have the relation: V= IR
• In this relation, R is a constant. Both V and I are variables. That is:
♦ V can take different values like 1.5, 3, 4.5 etc.,
♦ I can take different values like 0.15, 0.3, 0.45 etc.,
♦ But R will always remain a constant
2. So this is similar to the equation y = kx
• Where x and y are variables and k is a constant
3. The graph of 'y = kx' will always be a straight line
• Further more, this graph will always pass through the origin of the graph
• We can see that this is indeed true for our case also.
♦ If we extend our graph down wards, it will pass through the origin.
♦ This is indicated by the dashed yellow line.
4. Also note that, 'k' is multiplied with 'x'
• So the constant is multiplied with the 'variable which is plotted along the x axis'
• In our case, the constant R is multiplied with the variable I. So I is plotted along the x axis
• We cannot plot I along y axis

In the next section, we will see more details about resistors.