Chapter 10.4 - Solved examples on Electric power

In the previous section we completed a discussion on the basics of 'electric power'. We also saw some solved examples.  In this section, we will see a few more solved examples.

Solved example 10.8
A bulb of resistance 529 Ω works on 230 V. The filament was broken. It was joined together and made to work. Let the resistance of the filament now be 460 Ω. 
(a) What is the change in power of the bulb? 
(b) What is the current flowing through the bulb?
Solution:
Part (a):
1. Given: R₁ = 529 Ω, R₂ = 460 Ω, V = 230 volts.
2. When the bulb was manufactured, it had a certain resistance. It was 529 Ω.
• So the manufacturer knows how much energy it will consume in one second (if 230 V is supplied) .
• We can calculate it ourselves using the equation which relates power to [voltage, resistance] 
• We can use Eq.10.5: P = ^V2⁄_R
• Substituting the values, we get: P₁ = ²³⁰²⁄₅₂₉ = 100 W.
3. When the new resistance is 460, the new power = P₂ = ²³⁰²⁄₄₆₀ = 115 W.
So change in power = (115 - 100) = 15 W
4. The power has increased. That means, more energy is consumed, which is equivalent to saying: 'more energy is produced'.
• So more heat will be produced. But the filament may not be able to withstand this greater heat. Thus it will melt and break again 
5. Another explanation:
• We have seen in the previous section that, when the broken pieces of the filament are joined together, the current will increase. 
• According to Joule's law, when current increases, heat produced also increases
• But the filament may not be able to withstand this greater heat. Thus it will melt and break again 
Part (b):
1. We are required to find the currents.
We can use Eq.10.6: P = VI, which relates power to [Voltage, current] 
2. Voltage remains constant at 230 V
So we can write:
• P₁ = V × I₁ ⟹100 = 230 × I₁ ⟹ I₁ = ¹⁰⁰⁄₂₃₀ = 0.4348 A 
• P₂ = V × I₂ ⟹115 = 230 × I₂ ⟹ I₂ = ¹¹⁵⁄₂₃₀ = 0.5 A
3. So change in current = (I₂ - I₁) = (0.5 - 0.4348) = 0.0652 A

Solved example 10.9
A bulb is designed to work in 230 V. The resistance of the filament is 529 Ω. If the bulb works at 115 V, what will be it's power?
Solution:
1. Given: V₁ = 230 V, V₂ = 115 V, R = 529 Ω.
2. The manufacturer designed the bulb to work at 230 V. For that, a resistance of 529 Ω was given
• We can use Eq.10.5: P = ^V2⁄_R, which relates power to [Voltage, Resistance]
3. We can write: P₂ = ^(V₂)2⁄_R
• Substituting the values, we get: P₂ = ⁽¹¹⁵⁾²⁄₅₂₉ = 25 W
■ Initial power = P₁ = ⁽²³⁰⁾²⁄₅₂₉ = 100 W 
Note that, there is no change in resistance. In such a situation, if voltage is halved (ie., from 230 to 115V), the power becomes one fourth (ie., from 100 to 25)

Solved example 10.10
Two bulbs are marked as follows:
(i) 40 W, 240 V
(ii) 100 W, 240 V
Which one has greater resistance?
Solution:
■ We want an equation which relates power to: [voltage, resistance]  
• We can use Eq.10.5: P = ^V2⁄_R,
Case (i): Substituting the values, we get: 40 = ²⁴⁰²⁄_R ⟹ R = ²⁴⁰²⁄₄₀ = 1440 Ω 
Case (ii): Substituting the values, we get: 100 = ²⁴⁰²⁄_R ⟹ R = ²⁴⁰²⁄₁₀₀ = 576 Ω 

Solved example 10.11
An electric heater of power 920 W is working on a 230 V supply. If current flows for 5 minutes through it, Calculate the heat generated
Solution:
1. From Joule's law, we have: Heat generated (H) = I²Rt 
2. So we want 3 items:
• Current flowing through the heater (I)
• Resistance of the heater (R) 
• Time for which current flows through the heater (t)
3. In the question, we are given only 't'
• We have to calculate the other two items ourselves
4. To calculate them, we can use the given P and V
• To find I, we want a relation that will connect the power (P) to: [V, I]
• We will use Eq.10.6: P = VI
• Substituting the values, we get: 920 = 230 × I ⟹ I = ⁹²⁰⁄₂₃₀ = 4 A
5. To find R, we want a relation that will connect the power (P) to: [V, I]
• We will use Eq.10.5: P = ^V2⁄_R
• Substituting the values, we get:  920 = ²³⁰²⁄_R  ⟹ R = ²³⁰²⁄₉₂₀ = 57.5 Ω
6. Now we have the required items. Substituting the values in (1),we get:
H =  I²Rt =  4² × 57.5 × (5 × 60) = 276000 joules
7. Another method:
• After calculating I in step (4), we can directly use Eq.10.3: H = VIt (Details here)
• In this method, we need not calculate R
• Substituting the values, we get: H = 230 × 4 × (5 × 60) = 276000 joules 

Solved example 10.12
The marking on an electrical appliance is 800 W, 200 V
(a) If it works on 100 V, what is the power consumed?
(b) What is the power when it works on 50 V?
(c) What happens if 500 V is applied to it?
Solution:
• The marked values are: 800 W and 200 V
• The manufacturer is telling us that, if we supply 200 V to that appliance, it will do a work of 800 joules every second
• Another way of saying it is:
If we supply 200 V to that appliance, it will consume 800 joules of energy every second
• Yet another way of saying it is:
If we supply 200 V to that appliance, it will consume a power of 800 W
• The appliance has a fixed resistance R. We have to find this R first.
• So we want a relation that will connect the power (P) to: [R, V]
• We will use Eq.10.5: P = ^V2⁄_R
3. Substituting the values, we get: 800 = ²⁰⁰²⁄_R ⟹ R = ²⁰⁰²⁄₈₀₀ = 50 Ω.
Part (a):
1. If the voltage applied is 100 V (instead of 200 V), we will not get the 800 W power. It will be less
2. We have to find this new power. So we want a relation that will connect the power (P) to: [R, V]
• We will use Eq.10.5: P = ^V2⁄_R
3. Substituting the values, we get: P = ¹⁰⁰²⁄₅₀ = 200 W
■ Note: 
• The voltage decreased from 200 to 100 V. That is., new voltage is half of original
    ♦ Then the power decreased from 800 to 200 W. That is., power becomes one fourth of the original
Part (b):
1. If the voltage applied is 50 V (instead of 200 V), we will not get the 800 W power. It will be less
2. We have to find this new power. So we want a relation that will connect the power (P) to: [R, V]
• We will use Eq.10.5: P = ^V2⁄_R
3. Substituting the values, we get: P = ⁵⁰²⁄₅₀ = 50 W
■ Note: 
• The voltage decreased from 200 to 50 V. That is., new voltage is one fourth of original
    ♦ Then the power decreased from 800 to 50 W.
    ♦ But 50 = (800 × ¹⁄₁₆) 
    ♦ That is., power becomes one sixteenth of the original
Part (c):
1. 500 V is greater than the marked value of 200 V. 
• A greater voltage will cause damage to the appliance. Let us see how:
2. From Ohm's law, we have: V = IR
• Substituting the values, we get: 500 = I × 50 ⟹ I = 10 A
3. Using the same method, we can calculate the current expected by the manufacturer:
V = IR ⟹ 200 = I × 50 ⟹ I = 4 A
4. So the manufacturer is expecting a current of only 4 A
• We would be supplying 10 A
• The conductor in that appliance will not be able to carry 10 A. It will melt, and thus the appliance will be damaged.

Solved example 10.13
A potential difference of 200 V is applied to a 200 Ω resistor for 5 minutes
(a) Calculate the amount of heat generated
(b) If 200 Ω is replaced by 100 Ω, how much heat will be generated in 5 minutes?
(c) If 400 Ω resistance is used in this place for 5 minutes, how much heat is generated?
Solution:
Part (a):
1. From Joule's law, we have: Heat generated (H) = I²Rt 
2. So we want 3 items:
• Current flowing through the resistor (I)
• Resistance (R) 
• Time for which current flows through the resistor (t)
3. In the question, we are given 'R' and 't'
• We have to calculate 'I' ourselves
4. From Ohm's law, we have: V = IR
• Substituting the values, we get: 200 = I × 200 ⟹ I = 1 A
5. So now we have all the items. Substituting in (1), we get:
H =  I²Rt =  1² × 200 × (5 × 60) = 60000 joules
Part (b):
1. From Joule's law, we have: Heat generated (H) = I²Rt 
2. So we want 3 items:
• Current flowing through the resistor (I)
• Resistance (R) 
• Time for which current flows through the resistor (t)
3. In the question, we are given 'R' and 't'
• We have to calculate 'I' ourselves
4. From Ohm's law, we have: V = IR
Substituting the values, we get: 200 = I × 100 ⟹ I = 2 A
5. So now we have all the items. Substituting in (1), we get:
H =  I²Rt =  2² × 100 × (5 × 60) = 120000 joules

Part (c):
1. From Joule's law, we have: Heat generated (H) = I²Rt 
2. So we want 3 items:
• Current flowing through the resistor (I)
• Resistance (R) 
• Time for which current flows through the resistor (t)
3. In the question, we are given 'R' and 't'
• We have to calculate 'I' ourselves
4. From Ohm's law, we have: V = IR
Substituting the values, we get: 200 = I × 400 ⟹ I = 0.5 A
5. So now we have all the items. Substituting in (1), we get:
H =  I²Rt =  (0.5)² × 400 × (5 × 60) = 30000 joules

Solved example 10.14
Correct the wrong equations:
H =  I²Rt, H =  V²Rt, H = Pt, H = IVt
Solution:
• 'H =  V²Rt' is wrong. 
• The correct form is: H =  I²Rt

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We have completed this discussion on 'effects of electric current' and also 'electric power'. 
• In a heater, electrical energy is converted into heat energy 
• In an incandescent bulb, electrical energy is converted into both heat energy and light energy
• In an electric motor, electrical energy is converted into mechanical energy
■ A great advantage of electrical energy is that, it can be easily converted into other forms of energy
• But we must use appropriate appliances for the conversion.

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In the next section we will see 'electromagnetic induction'.

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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.

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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

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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.

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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⁄_I is a constant

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• 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.

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• 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

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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.

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■ 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:

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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'

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■ 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

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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

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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

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• 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

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

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