Namaste, dear students! Welcome to today's science class. I am so happy to see you all here, ready to learn about one of the most important and fascinating topics in physics – Electricity. This chapter is going to be extremely useful for you, not just for your exams, but also for understanding the world around you. We use electricity every single day – from switching on a light bulb to charging our phones, from using an electric fan to watching television. But have you ever wondered what electricity actually is? How does it flow? What controls how much current flows through a circuit? Today, we are going to answer all these questions and much more. So, let's begin our journey into the world of electricity!
In this chapter, we are going to study about electric current, how it flows, what controls its flow, and the heating effect of electric current. We will also learn about some very important concepts like potential difference, resistance, and electric power. I will explain everything step by step, and we will solve many numerical problems together. So, pay attention and don't worry if you find some concepts difficult at first – we will build our understanding slowly, from simple to complex.
Let us start with the very basics. What do we mean by electric current?
## 11.1 Electric Current and Circuit
Students, you are all familiar with water flowing in a river. That flowing water is called water current. Similarly, when electric charges flow through a conductor, we say there is an electric current. A conductor is simply a material that allows electricity to pass through it easily – like a metallic wire.
Now, think about a torch. You must have used a torch at some point. When you put the cells in a torch and switch it on, the bulb glows. But have you noticed that the bulb only glows when the switch is on? What does the switch do? The switch creates a continuous path for the electric charges to flow from the cell to the bulb and back. This continuous and closed path through which electric current flows is called an electric circuit. If you break this path anywhere – for example, by turning off the switch – the current stops flowing and the bulb stops glowing. Simple, isn't it?
Now, how do we express or measure electric current? Electric current is defined as the amount of charge flowing through a particular area in unit time. In other words, it is the rate of flow of electric charges. Let me write this mathematically for you.
If a net charge Q flows through any cross-section of a conductor in time t, then the current I through that cross-section is:
I = Q/t
This is equation 11.1 in your textbook. Here, I stands for current, Q stands for charge, and t stands for time.
Now, what is the unit of electric charge? The SI unit of electric charge is coulomb, denoted by the letter C. Did you know that one coulomb is equivalent to the charge contained in nearly 6 × 10^18 electrons? That's a huge number! And we also know that an electron possesses a negative charge of 1.6 × 10^-19 coulomb. So, if you divide 1 by 1.6 × 10^-19, you get approximately 6 × 10^18. This is how we arrived at that number.
The unit of electric current is called ampere, named after the French scientist Andre-Marie Ampere, who lived from 1775 to 1836. One ampere is constituted by the flow of one coulomb of charge per second. So, 1 A = 1 C/1 s. We often use smaller units like milliampere, where 1 mA = 10^-3 A, and microampere, where 1 μA = 10^-6 A.
An instrument called an ammeter measures electric current in a circuit. It is always connected in series in the circuit. Do you remember what 'in series' means? It means the ammeter is connected in such a way that the same current passes through it and the other components of the circuit.
Now, here's an important point about the direction of current. In circuits using metallic wires, electrons constitute the flow of charges. Electrons are negatively charged particles. However, when electricity was first discovered, electrons were not known. Scientists at that time thought that electric current was the flow of positive charges. So, they took the direction of flow of positive charges as the direction of electric current. Conventionally, even today, we take the direction of electric current as opposite to the direction of flow of electrons. So, in an electric circuit, the current flows from the positive terminal of the cell to the negative terminal. Keep this in mind – it's very important!
Now, let me give you an example to understand this better.
### Example 11.1
A current of 0.5 A is drawn by a filament of an electric bulb for 10 minutes. Find the amount of electric charge that flows through the circuit.
So students, let's solve this problem together. We are given: Current I = 0.5 A Time t = 10 minutes
First, we need to convert minutes into seconds because the SI unit of time is seconds. So, 10 minutes = 10 × 60 = 600 seconds.
Now, using the formula I = Q/t, we can write Q = I × t
So, Q = 0.5 A × 600 s = 300 C
Therefore, the amount of charge that flows through the circuit is 300 coulombs. Simple, right?
Now, let me ask you some questions to check your understanding:
Question 1: What does an electric circuit mean? Question 2: Define the unit of current. Question 3: Calculate the number of electrons constituting one coulomb of charge.
For question 3, let me help you. We know that charge on one electron is 1.6 × 10^-19 C. So, number of electrons in 1 coulomb = 1 / (1.6 × 10^-19) = 6.25 × 10^18 electrons. That's approximately 6 × 10^18 electrons, as I mentioned earlier.
Now, let's move on to the next concept. What makes the electric charge to flow?
## 11.2 Electric Potential and Potential Difference
Students, think about water flowing in a pipe. Water flows from a higher level to a lower level, right? This happens because there is a pressure difference between the two ends. Similarly, for electric charges to flow through a conductor, there must be a difference in electric pressure. This electric pressure is called potential difference.
Let me explain this with an example. Suppose you have a water tank at a higher level connected to a pipe. Water will flow out of the other end of the pipe because there is a pressure difference. Similarly, for charges to move in a conducting wire, there must be a potential difference along the conductor. This potential difference is produced by a battery, which consists of one or more electric cells.
Now, what exactly is potential difference? We define electric potential difference between two points in an electric circuit carrying some current as the work done to move a unit charge from one point to the other. Let me write this mathematically:
Potential difference (V) = Work done (W) / Charge (Q)
So, V = W/Q
This is equation 11.2 in your textbook.
The SI unit of potential difference is volt, named after Alessandro Volta, an Italian physicist who lived from 1745 to 1827. One volt is the potential difference between two points in a current-carrying conductor when 1 joule of work is done to move a charge of 1 coulomb from one point to the other. So, 1 volt = 1 joule/1 coulomb, or 1 V = 1 J C^-1.
The potential difference is measured by an instrument called a voltmeter. Unlike the ammeter, the voltmeter is always connected in parallel across the points between which the potential difference is to be measured. Remember this – ammeter is connected in series, voltmeter is connected in parallel.
Now, let's solve an example to understand this better.
### Example 11.2
How much work is done in moving a charge of 2 C across two points having a potential difference 12 V?
We are given: Charge Q = 2 C Potential difference V = 12 V
Using the formula V = W/Q, we can write W = V × Q
So, W = 12 V × 2 C = 24 J
Therefore, the work done is 24 joules.
Now, let me ask you some questions:
Question 1: Name a device that helps to maintain a potential difference across a conductor. Question 2: What is meant by saying that the potential difference between two points is 1 V? Question 3: How much energy is given to each coulomb of charge passing through a 6 V battery?
For question 3, the answer is 6 joules. Because energy given to each coulomb of charge is equal to the potential difference, which is 6 V, so 6 joules per coulomb.
Now, let's move on to the next section. We are going to learn about circuit diagrams.
## 11.3 Circuit Diagram
Students, you have learned that an electric circuit consists of a cell or battery, a plug key, electrical components, and connecting wires. Drawing the actual picture of a circuit every time would be very complicated and time-consuming. That's why we use symbols to represent different components of a circuit. These symbols make it much easier to draw and understand circuits. This is called a circuit diagram.
Your textbook provides a table (Table 11.1) showing the conventional symbols for some commonly used electrical components. Let me describe these to you:
An electric cell is represented by a long vertical line and a short vertical line. The long line represents the positive terminal, and the short line represents the negative terminal.
A battery or a combination of cells is represented by multiple cells connected in series.
A plug key or switch when open is represented by a break in the circuit with a diagonal line. When closed, it is represented as a continuous line.
A wire joint is represented by a dot at the point where two wires meet.
Wires crossing without joining are represented by one wire going over another without a dot.
An electric bulb is represented by a circle with a cross inside.
A resistor of resistance R is represented by a zigzag line.
A variable resistance or rheostat is represented by a zigzag line with an arrow pointing diagonally.
An ammeter is represented by a circle with the letter A inside.
A voltmeter is represented by a circle with the letter V inside.
These symbols are used universally, so it's important that you remember them. You will be using these symbols frequently in your exams and while solving problems.
Now, let's move on to one of the most important laws in electricity – Ohm's Law.
## 11.4 Ohm's Law
Students, this is a very important section. Ohm's law is fundamental to understanding electricity. Let me explain what Ohm's law states.
Is there a relationship between the potential difference across a conductor and the current through it? Let's find out through an activity.
### Activity 11.1
We set up a circuit consisting of a nichrome wire XY of length 0.5 m, an ammeter, a voltmeter, and four cells of 1.5 V each. First, we use only one cell and note the ammeter reading (current I) and voltmeter reading (potential difference V). Then we use two cells, three cells, and four cells, and note the readings each time. We calculate the ratio V/I for each case.
What do you think we will find? You will find that approximately the same value for V/I is obtained in each case. This means that V/I is constant. When we plot a graph between V and I, we get a straight line passing through the origin. This shows that V is directly proportional to I.
In 1827, a German physicist Georg Simon Ohm (1787–1854) discovered this relationship. According to Ohm's law, the potential difference V across the ends of a given metallic wire in an electric circuit is directly proportional to the current flowing through it, provided its temperature remains the same. So we can write:
V ∝ I
or V/I = constant = R
or V = IR
Here, R is a constant for the given metallic wire at a given temperature and is called its resistance. Resistance is the property of a conductor to resist the flow of charges through it. Its SI unit is ohm, represented by the Greek letter Ω.
According to Ohm's law, if the potential difference across the two ends of a conductor is 1 V and the current through it is 1 A, then the resistance of the conductor is 1 Ω. So, 1 ohm = 1 volt/1 ampere.
From the equation V = IR, we can also write I = V/R. This shows that the current through a resistor is inversely proportional to its resistance. If the resistance is doubled, the current gets halved. This is very important!
Sometimes, we need to increase or decrease the current in an electric circuit. A component used to regulate current without changing the voltage source is called variable resistance. In an electric circuit, a device called a rheostat is often used to change the resistance in the circuit.
Now, let's learn about what factors affect the resistance of a conductor.
## 11.5 Factors on Which the Resistance of a Conductor Depends
### Activity 11.3
Let's perform another activity to understand what factors affect the resistance of a conductor.
We set up a circuit with a cell, an ammeter, a nichrome wire of length l, and a plug key. We note the current in the ammeter. Then we replace the nichrome wire with another nichrome wire of the same thickness but twice the length (2l). We note the ammeter reading again. Then we replace it with a thicker nichrome wire of the same length l. And finally, we replace it with a copper wire of the same length and same cross-sectional area as the first nichrome wire.
What do we observe? The ammeter reading decreases to one-half when the length of the wire is doubled. The ammeter reading increases when a thicker wire of the same material and same length is used. And we observe a change in ammeter reading when a wire of different material is used, even if the length and cross-sectional area are the same.
This shows that the resistance of a conductor depends on three factors: 1. Its length 2. Its area of cross-section 3. The nature of its material
Precise measurements have shown that the resistance of a uniform metallic conductor is directly proportional to its length (l) and inversely proportional to the area of cross-section (A). So we can write:
R ∝ l R ∝ 1/A
Combining these, we get: R ∝ l/A
or R = ρ l/A
This is equation 11.10. Here, ρ (rho) is a constant of proportionality called the electrical resistivity of the material of the conductor. The SI unit of resistivity is Ω m. It is a characteristic property of the material.
Metals and alloys have very low resistivity, in the range of 10^-8 Ω m to 10^-6 Ω m. They are good conductors of electricity. Insulators like rubber and glass have resistivity of the order of 10^12 to 10^17 Ω m.
Table 11.2 in your textbook shows the electrical resistivity of some substances at 20°C. Let me highlight some important points from this table:
Among conductors, silver has the lowest resistivity, followed by copper, then aluminium. Tungsten has a higher resistivity than copper, which is why it gets heated up and glows when current passes through it – that's why we use tungsten for light bulb filaments.
Among alloys, nichrome has a very high resistivity. Alloys generally have higher resistivity than pure metals. Also, alloys do not oxidize readily at high temperatures, which is why they are used in electrical heating devices like electric irons and toasters.
Now, let's solve some examples to understand these concepts better.
### Example 11.3
(a) How much current will an electric bulb draw from a 220 V source if the resistance of the bulb filament is 1200 Ω? (b) How much current will an electric heater coil draw from a 220 V source if the resistance of the heater coil is 100 Ω?
For part (a): Given: V = 220 V, R = 1200 Ω Using Ohm's law, I = V/R = 220/1200 = 0.18 A
For part (b): Given: V = 220 V, R = 100 Ω Using Ohm's law, I = V/R = 220/100 = 2.2 A
Notice the difference! The heater draws much more current than the bulb because its resistance is much lower. This is why you should never connect high-power appliances to the same socket – they draw too much current!
### Example 11.4
The potential difference between the terminals of an electric heater is 60 V when it draws a current of 4 A from the source. What current will the heater draw if the potential difference is increased to 120 V?
Given: V1 = 60 V, I1 = 4 A First, find the resistance: R = V1/I1 = 60/4 = 15 Ω Now, when V2 = 120 V, I2 = V2/R = 120/15 = 8 A
So the current becomes 8 A. This makes sense – when voltage is doubled, current also doubles if resistance remains constant.
### Example 11.5
Resistance of a metal wire of length 1 m is 26 Ω at 20°C. If the diameter of the wire is 0.3 mm, what will be the resistivity of the metal at that temperature? Using Table 11.2, predict the material of the wire.
Given: R = 26 Ω, length l = 1 m, diameter d = 0.3 mm = 3 × 10^-4 m We know that R = ρ l/A, so ρ = RA/l Area A = πd²/4 = π × (3 × 10^-4)²/4 = π × 9 × 10^-8 / 4 = 7.07 × 10^-8 m² (approximately)
So, ρ = R × A / l = 26 × 7.07 × 10^-8 / 1 = 1.84 × 10^-6 Ω m
From Table 11.2, this is the resistivity of manganese. So the wire is made of manganese!
### Example 11.6
A wire of given material having length l and area of cross-section A has a resistance of 4 Ω. What would be the resistance of another wire of the same material having length l/2 and area of cross-section 2A?
For the first wire: R1 = ρ l/A = 4 Ω For the second wire: R2 = ρ (l/2) / (2A) = ρ l / (4A) = (1/4) × (ρ l/A) = (1/4) × R1 = 1 Ω
So the resistance of the new wire is 1 Ω. Notice that when length is halved and area is doubled, the resistance becomes one-fourth.
Now, let's answer some questions:
Question 1: On what factors does the resistance of a conductor depend? Answer: Length, area of cross-section, and nature of the material.
Question 2: Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why? Answer: Current will flow more easily through a thick wire because resistance is inversely proportional to area of cross-section. A thicker wire has larger cross-sectional area, hence lower resistance.
Question 3: Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it? Answer: If potential difference is halved and resistance remains constant, the current will also halve, according to Ohm's law (I = V/R).
Question 4: Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal? Answer: Alloys have higher resistivity than pure metals, so they produce more heat for the same current. Also, alloys do not oxidize readily at high temperatures, making them more durable.
Question 5: Use the data in Table 11.2 to answer: (a) Which among iron and mercury is a better conductor? (b) Which material is the best conductor?
For (a): Iron has resistivity 10.0 × 10^-8 Ω m, while mercury has resistivity 94.0 × 10^-8 Ω m. Lower resistivity means better conductor. So iron is a better conductor than mercury.
For (b): Among all materials listed, silver has the lowest resistivity (1.60 × 10^-8 Ω m), so silver is the best conductor.
Now, let's move on to the next important topic – resistance of a system of resistors.
## 11.6 Resistance of a System of Resistors
In practical circuits, we often use multiple resistors connected together. There are two ways to connect resistors: in series and in parallel. Let's study both.
### 11.6.1 Resistors in Series
When resistors are connected end to end in a chain, they are said to be connected in series. Let me explain what happens to the current and voltage in a series combination.
#### Activity 11.4
We join three resistors of different values in series, connect them with a battery, an ammeter, and a plug key. We note the ammeter reading. Then we change the position of the ammeter to anywhere in between the resistors and note the reading each time.
What do we find? The value of current in the ammeter is the same, regardless of its position! This means that in a series combination, the current is the same through every part of the circuit. This is a very important point to remember.
#### Activity 11.5
Now, we connect a voltmeter across the ends of the series combination and note the reading (V). Then we measure the potential difference across each resistor separately (V1, V2, V3).
What do we find? The total potential difference V is equal to the sum of potential differences across individual resistors: V = V1 + V2 + V3.
Now, if we want to replace the three resistors in series by a single equivalent resistor R, such that the potential difference V across it and the current I through the circuit remain the same, we can use Ohm's law.
We have: V = IR Also, V1 = IR1, V2 = IR2, V3 = IR3
Since V = V1 + V2 + V3, IR = IR1 + IR2 + IR3 So, R = R1 + R2 + R3
This is equation 11.14. So, when several resistors are connected in series, the equivalent resistance equals the sum of their individual resistances, and is thus greater than any individual resistance.
Let me solve an example for you.
### Example 11.7
An electric lamp, whose resistance is 20 Ω, and a conductor of 4 Ω resistance are connected to a 6 V battery. Calculate (a) the total resistance of the circuit, (b) the current through the circuit, and (c) the potential difference across the electric lamp and conductor.
Given: R1 (lamp) = 20 Ω, R2 (conductor) = 4 Ω, V = 6 V
(a) Total resistance R = R1 + R2 = 20 + 4 = 24 Ω
(b) Current I = V/R = 6/24 = 0.25 A
(c) Potential difference across lamp, V1 = I × R1 = 0.25 × 20 = 5 V Potential difference across conductor, V2 = I × R2 = 0.25 × 4 = 1 V
Notice that V1 + V2 = 5 + 1 = 6 V, which is equal to the total voltage. This confirms our understanding!
Now, let's answer the questions:
Question 1: Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Ω resistor, an 8 Ω resistor, and a 12 Ω resistor, and a plug key, all connected in series.
For this, you need to draw a circuit with three cells (total 6 V) connected in series, then a 5 Ω resistor, an 8 Ω resistor, a 12 Ω resistor, and a plug key, all connected in series one after another.
Question 2: Redraw the circuit of Question 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 Ω resistor. What would be the readings in the ammeter and the voltmeter?
Total resistance = 5 + 8 + 12 = 25 Ω Voltage = 6 V Current = 6/25 = 0.24 A (this would be the ammeter reading) Voltage across 12 Ω = I × 12 = 0.24 × 12 = 2.88 V (this would be the voltmeter reading)
Now, let's study about resistors in parallel.
### 11.6.2 Resistors in Parallel
When resistors are connected such that each resistor has its own separate path connected to the same two points, they are said to be connected in parallel.
### Activity 11.6
We make a parallel combination of three resistors R1, R2, and R3, and connect it with a battery, a plug key, and an ammeter. We also connect a voltmeter in parallel with the combination. We note the ammeter reading (total current I) and voltmeter reading (potential difference V).
Then we measure the current through each resistor separately by connecting the ammeter in series with each resistor. Let these currents be I1, I2, and I3.
What do we find? The total current I = I1 + I2 + I3. This makes sense because the current divides into different branches.
Now, if we find the equivalent resistance Rp of the parallel combination, we can use Ohm's law:
I = V/Rp Also, I1 = V/R1, I2 = V/R2, I3 = V/R3
So, V/Rp = V/R1 + V/R2 + V/R3 Or 1/Rp = 1/R1 + 1/R2 + 1/R3
This is equation 11.18. So, the reciprocal of the equivalent resistance of a group of resistances joined in parallel is equal to the sum of the reciprocals of the individual resistances.
Let me solve some examples for you.
### Example 11.8
In a circuit, resistors R1, R2, and R3 have values 5 Ω, 10 Ω, and 30 Ω respectively, connected to a battery of 12 V. Calculate (a) the current through each resistor, (b) the total current in the circuit, and (c) the total circuit resistance.
Given: R1 = 5 Ω, R2 = 10 Ω, R3 = 30 Ω, V = 12 V
(a) Current through R1: I1 = V/R1 = 12/5 = 2.4 A Current through R2: I2 = V/R2 = 12/10 = 1.2 A Current through R3: I3 = V/R3 = 12/30 = 0.4 A
(b) Total current I = I1 + I2 + I3 = 2.4 + 1.2 + 0.4 = 4 A
(c) Total resistance Rp: 1/Rp = 1/5 + 1/10 + 1/30 = 6/30 + 3/30 + 1/30 = 10/30 = 1/3 So, Rp = 3 Ω
### Example 11.9
If R1 = 10 Ω, R2 = 40 Ω, R3 = 30 Ω, R4 = 20 Ω, R5 = 60 Ω, and a 12 V battery is connected, calculate (a) the total resistance in the circuit, and (b) the total current flowing in the circuit.
This is a more complex circuit where some resistors are in parallel and others in series. Let's solve it step by step.
First, R1 and R2 are in parallel. Their equivalent R' = (10 × 40)/(10 + 40) = 400/50 = 8 Ω
Similarly, R3, R4, and R5 are in parallel. Their equivalent R'' = 1/(1/30 + 1/20 + 1/60) = 1/(2/60 + 3/60 + 1/60) = 1/(6/60) = 60/6 = 10 Ω
Now, R' and R'' are connected in series. So total resistance R = R' + R'' = 8 + 10 = 18 Ω
Current I = V/R = 12/18 = 0.67 A
Now, let me tell you about some important advantages and disadvantages of series and parallel connections.
In a series circuit, the current is constant throughout. This is why it is impractical to connect an electric bulb and an electric heater in series – they need different amounts of current to operate properly. Another major disadvantage of a series circuit is that when one component fails, the circuit is broken and none of the components work. You might have seen this with fairy lights – when one bulb fuses, the entire string stops working!
In a parallel circuit, the current divides through the different branches. The total resistance in a parallel circuit is less than any individual resistance. This is helpful when each gadget has different resistance and requires different current to operate properly. That's why domestic circuits are always connected in parallel!
Now, let's answer some questions:
Question 1: Judge the equivalent resistance when the following are connected in parallel: (a) 1 Ω and 10^6 Ω (b) 1 Ω and 10^3 Ω
For (a): 1/Rp = 1/1 + 1/10^6 = 1 + 0.000001 = 1.000001 So Rp ≈ 1 Ω (very close to 1 Ω)
For (b): 1/Rp = 1/1 + 1/10^3 = 1 + 0.001 = 1.001 So Rp ≈ 0.999 Ω (very close to 1 Ω)
Question 2: An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?
First, find current through each appliance: Lamp: I1 = 220/100 = 2.2 A Toaster: I2 = 220/50 = 4.4 A Filter: I3 = 220/500 = 0.44 A Total current = 2.2 + 4.4 + 0.44 = 7.04 A
For the iron to take the same current (7.04 A) from 220 V: R = V/I = 220/7.04 = 31.25 Ω
Question 3: What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
Advantages: - Each device gets the full voltage of the battery - If one device fails, others continue to work - We can control each device independently - Different devices can draw different amounts of current as needed
Question 4: How can three resistors of resistances 2 Ω, 3 Ω, and 6 Ω be connected to give a total resistance of (a) 4 Ω, (b) 1 Ω?
For (a) 4 Ω: Connect 2 Ω and 6 Ω in parallel: (2×6)/(2+6) = 12/8 = 1.5 Ω. Then connect this in series with 3 Ω: 1.5 + 3 = 4.5 Ω. That's not 4 Ω.
Alternatively, connect 2 Ω and 3 Ω in parallel: (2×3)/(2+3) = 6/5 = 1.2 Ω. Then connect in series with 6 Ω: 1.2 + 6 = 7.2 Ω.
Actually, let's try another combination: Connect 2 Ω and 3 Ω in series = 5 Ω. Then connect this in parallel with 6 Ω: (5×6)/(5+6) = 30/11 = 2.73 Ω.
Hmm, let me think differently. For 4 Ω: Connect 2 Ω in series with the parallel combination of 3 Ω and 6 Ω. The parallel combination = (3×6)/(3+6) = 18/9 = 2 Ω. So total = 2 + 2 = 4 Ω. Yes!
For (b) 1 Ω: Connect all three in parallel. 1/Rp = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1. So Rp = 1 Ω.
Question 5: What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?
For highest resistance: Connect all in series: 4 + 8 + 12 + 24 = 48 Ω
For lowest resistance: Connect all in parallel: 1/Rp = 1/4 + 1/8 + 1/12 + 1/24 = 6/24 + 3/24 + 2/24 + 1/24 = 12/24 = 1/2 So Rp = 2 Ω
Now, let's move on to the next topic – the heating effect of electric current.
## 11.7 Heating Effect of Electric Current
Students, you must have noticed that when you use an electric fan for a long time, it becomes warm. Similarly, the filament of an electric bulb becomes extremely hot and glows. This is because of the heating effect of electric current.
When electric current flows through a conductor, the electrical energy is converted into heat energy. This is known as the heating effect of electric current. This effect is utilized in devices such as electric heater, electric iron, etc.
Let me explain the science behind this. Consider a current I flowing through a resistor of resistance R. Let the potential difference across it be V. 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 = VQ/t = VI
Or the energy supplied to the circuit by the source in time t is P × t, that is, VIt.
What happens to this energy? 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 = VIt
Applying Ohm's law (V = IR), we get:
H = I²Rt
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 (iii) directly proportional to the time for which the current flows through the resistor
Let me solve some examples for you.
### Example 11.10
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?
We know that P = VI, so I = P/V
(a) Maximum heating: I = 840/220 = 3.82 A R = V/I = 220/3.82 = 57.60 Ω
(b) Minimum heating: I = 360/220 = 1.64 A R = V/I = 220/1.64 = 134.15 Ω
Notice that when the heating is at minimum, the resistance is higher. This is because the iron has a thermostat that increases resistance to reduce heat when it gets too hot.
### Example 11.11
100 J of heat is produced each second in a 4 Ω resistance. Find the potential difference across the resistor.
Given: H = 100 J, R = 4 Ω, t = 1 s
From H = I²Rt, we get I² = H/(Rt) = 100/(4×1) = 25 So I = 5 A
Now, V = IR = 5 × 4 = 20 V
Now, let's answer some questions:
Question 1: Why does the cord of an electric heater not glow while the heating element does?
Answer: The cord is made of a material with low resistance, so it doesn't heat up much. The heating element is made of a material with high resistance (like nichrome), so it heats up and glows.
Question 2: Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.
Given: Q = 96000 C, V = 50 V, t = 1 hour = 3600 s First, find current: I = Q/t = 96000/3600 = 26.67 A Now, H = VIt = 50 × 26.67 × 3600 = 4,800,000 J = 4.8 MJ
Alternatively, using H = VQ = 50 × 96000 = 4,800,000 J
Question 3: An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s.
Given: R = 20 Ω, I = 5 A, t = 30 s H = I²Rt = 5² × 20 × 30 = 25 × 20 × 30 = 15,000 J
### 11.7.1 Practical Applications of Heating Effect of Electric Current
The heating effect of electric current has many useful applications. Electric laundry iron, electric toaster, electric oven, electric kettle, and electric heater are some 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 get very hot to emit light. It must not melt at such high temperature. That's why we use tungsten, which has a very high melting point of 3380°C. The bulbs are usually filled with chemically inactive nitrogen and argon gases to prolong the life of the filament.
Another common application is the fuse. A fuse protects circuits and appliances by stopping the flow of unduly high electric current. It consists of a piece of wire made of a metal or alloy of appropriate melting point. If a current larger than the specified value flows through the circuit, the temperature of the fuse wire increases, it melts, and breaks the circuit. The fuses used for domestic purposes are rated as 1 A, 2 A, 3 A, 5 A, 10 A, etc.
For example, for an electric iron which consumes 1 kW power when operated at 220 V, the current is (1000/220) A = 4.54 A. In this case, a 5 A fuse must be used.
Now, let's move on to the final section of this chapter – electric power.
## 11.8 Electric Power
Students, you have already learned in your earlier classes that power is the rate of doing work. Similarly, electric power is the rate at which electrical energy is consumed or dissipated.
From our earlier discussion, we know that: P = VI
And using Ohm's law, we can also write: P = I²R = V²/R
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. So, 1 W = 1 V × 1 A.
The unit 'watt' is very small. Therefore, we often use kilowatt, which is equal to 1000 watts.
Since electrical energy is the product of power and time, the unit of electric energy is 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 (kWh), commonly known as 'unit'.
1 kW h = 1000 watt × 3600 second = 3.6 × 10^6 watt second = 3.6 × 10^6 joule
Let me solve some examples.
### Example 11.12
An electric bulb is connected to a 220 V generator. The current is 0.50 A. What is the power of the bulb?
P = VI = 220 × 0.50 = 110 W
### Example 11.13
An electric refrigerator rated 400 W operates 8 hours per day. What is the cost of the energy to operate it for 30 days at Rs 3.00 per kW h?
Energy consumed in one day = 400 W × 8 h = 3200 Wh = 3.2 kWh Energy consumed in 30 days = 3.2 × 30 = 96 kWh Cost = 96 × 3 = Rs 288
Now, let's answer the questions:
Question 1: What determines the rate at which energy is delivered by a current?
Answer: The rate at which energy is delivered (power) is determined by the current and the potential difference. P = VI.
Question 2: An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.
Power P = VI = 220 × 5 = 1100 W = 1.1 kW Energy consumed in 2 h = 1.1 × 2 = 2.2 kWh
Now, students, we have covered all the topics in this chapter. Let me now solve all the exercise questions for you.
## Exercises
Let me solve each exercise question one by one.
### Question 1
A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R', then the ratio R/R' is – (a) 1/25 (b) 1/5 (c) 5 (d) 25
When the wire is cut into five equal parts, each part has resistance R/5. When these are connected in parallel, the equivalent resistance is: 1/R' = 5/(R/5) = 25/R So R' = R/25 Therefore, R/R' = R/(R/25) = 25 So the answer is (d) 25.
### Question 2
Which of the following terms does not represent electrical power in a circuit? (a) I²R (b) IR² (c) VI (d) V²/R
We know that electrical power can be represented as P = VI, P = I²R, and P = V²/R. But IR² is not a valid expression for power. So the answer is (b) IR².
### Question 3
An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be – (a) 100 W (b) 75 W (c) 50 W (d) 25 W
First, find the resistance of the bulb: R = V²/P = 220²/100 = 48400/100 = 480 Ω
When operated at 110 V, power = V²/R = 110²/480 = 12100/480 = 25.2 W ≈ 25 W
So the answer is (d) 25 W.
### Question 4
Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then 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
Let the resistance of each wire be R. In series, total resistance = 2R. In parallel, total resistance = R/2.
Heat produced H = V²t/R
For series: H1 = V²t/(2R) For parallel: H2 = V²t/(R/2) = 2V²t/R
So H1:H2 = (V²t/2R) : (2V²t/R) = 1/2 : 2 = 1:4
So the answer is (c) 1:4.
### Question 5
How is a voltmeter connected in the circuit to measure the potential difference between two points?
A voltmeter is always connected in parallel across the two points between which the potential difference is to be measured.
### Question 6
A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10^-8 Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?
Given: diameter d = 0.5 mm = 5 × 10^-4 m, ρ = 1.6 × 10^-8 Ω m, R = 10 Ω
Area A = πd²/4 = π × (5 × 10^-4)²/4 = π × 25 × 10^-8 / 4 = 19.63 × 10^-8 m² (approximately)
We know R = ρl/A, so l = RA/ρ = 10 × 19.63 × 10^-8 / (1.6 × 10^-8) = 10 × 19.63 / 1.6 = 122.7 m
If diameter is doubled, new diameter = 1 mm = 10^-3 m New area A' = π × (10^-3)²/4 = π × 10^-6 / 4 = 7.85 × 10^-7 m²
New resistance R' = ρl/A' = 1.6 × 10^-8 × 122.7 / (7.85 × 10^-7) = 1.6 × 122.7 × 10^-8 / 10^-7 / 7.85 = 196.32 × 10^-1 / 7.85 = 2.5 Ω
So the resistance decreases from 10 Ω to 2.5 Ω when diameter is doubled.
### Question 7
The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below:
I (amperes): 0.5, 1.0, 2.0, 3.0, 4.0 V (volts): 1.6, 3.4, 6.7, 10.2, 13.2
Plot a graph between V and I and calculate the resistance of that resistor.
To plot the graph, we take V on y-axis and I on x-axis. The graph will be approximately a straight line. The resistance is given by the slope of the graph, which is V/I.
For the first reading: V/I = 1.6/0.5 = 3.2 Ω For the second: 3.4/1.0 = 3.4 Ω For the third: 6.7/2.0 = 3.35 Ω For the fourth: 10.2/3.0 = 3.4 Ω For the fifth: 13.2/4.0 = 3.3 Ω
Average resistance ≈ 3.35 Ω
### Question 8
When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.
Given: V = 12 V, I = 2.5 mA = 2.5 × 10^-3 A = 0.0025 A R = V/I = 12/0.0025 = 4800 Ω = 4.8 kΩ
### Question 9
A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistor?
In a series circuit, the same current flows through all resistors. So we just need to find the total current.
Total resistance = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 Ω Current I = V/R = 9/13.4 = 0.67 A
So the current through the 12 Ω resistor is 0.67 A.
### Question 10
How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?
Let n resistors be connected in parallel. The equivalent resistance Rp = 176/n Ω We need current I = 5 A from V = 220 V So R required = V/I = 220/5 = 44 Ω
Therefore, 176/n = 44 n = 176/44 = 4
So 4 resistors are required.
### Question 11
Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω, (ii) 4 Ω.
For (i) 9 Ω: Connect two 6 Ω resistors in parallel: (6×6)/(6+6) = 36/12 = 3 Ω. Then connect this in series with the third 6 Ω resistor: 3 + 6 = 9 Ω.
For (ii) 4 Ω: Connect two 6 Ω resistors in series: 6 + 6 = 12 Ω. Then connect this in parallel with the third 6 Ω resistor: (12×6)/(12+6) = 72/18 = 4 Ω.
### Question 12
Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?
Each lamp rated 10 W at 220 V draws current I = P/V = 10/220 = 0.045 A
If maximum current is 5 A, number of lamps = 5/0.045 = 111.1
So we can connect 111 lamps (approximately).
### Question 13
A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?
Case 1: Using only one coil (say A) R = 24 Ω, V = 220 V I = V/R = 220/24 = 9.17 A
Case 2: Using both coils in series R = 24 + 24 = 48 Ω I = 220/48 = 4.58 A
Case 3: Using both coils in parallel Equivalent resistance = (24×24)/(24+24) = 576/48 = 12 Ω I = 220/12 = 18.33 A
### Question 14
Compare the power used in the 2 Ω resistor in each of the following circuits: (i) a 6 V battery in series with 1 Ω and 2 Ω resistors (ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors
For (i): Total resistance = 1 + 2 = 3 Ω Current I = 6/3 = 2 A Power in 2 Ω resistor = I²R = 2² × 2 = 8 W
For (ii): In parallel, voltage across each resistor is 4 V (same as battery voltage) Power in 2 Ω resistor = V²/R = 4²/2 = 16/2 = 8 W
So the power is the same in both cases: 8 W.
### Question 15
Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?
Current through 100 W lamp = 100/220 = 0.455 A Current through 60 W lamp = 60/220 = 0.273 A Total current = 0.455 + 0.273 = 0.728 A
### Question 16
Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?
Energy used by TV = 250 W × 1 h = 250 Wh Energy used by toaster = 1200 W × (10/60) h = 1200 × 1/6 = 200 Wh
So the TV set uses more energy (250 Wh vs 200 Wh).
### Question 17
An electric heater of resistance 44 Ω draws 5 A from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater.
Rate of heat developed means power. P = I²R = 5² × 44 = 25 × 44 = 1100 W = 1.1 kW
So heat is developed at the rate of 1100 watts or 1.1 kilowatts.
### Question 18
Explain the following:
(a) Why is the tungsten used almost exclusively for filament of electric lamps? Answer: Tungsten has a very high melting point (3380°C), so it can be heated to very high temperatures without melting. When heated, it emits light.
(b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal? Answer: Alloys have higher resistivity than pure metals, so they produce more heat for the same current. Also, alloys do not oxidize readily at high temperatures, making them more durable.
(c) Why is the series arrangement not used for domestic circuits? Answer: In series arrangement, if one appliance fails, the entire circuit breaks and all appliances stop working. Also, all appliances would get the same voltage, but different appliances require different voltages. Moreover, we cannot control individual appliances independently.
(d) How does the resistance of a wire vary with its area of cross-section? Answer: Resistance is inversely proportional to the area of cross-section. If the area is doubled, resistance becomes half.
(e) Why are copper and aluminium wires usually employed for electricity transmission? Answer: Copper and aluminium have low resistivity, so they are good conductors of electricity. They are also relatively cheap and readily available.
Now, students, we have completed the entire chapter. Let me give you a comprehensive summary of everything we have learned.
## Summary
In this chapter on Electricity, we have learned the following important concepts:
1. **Electric Current**: It is the rate of flow of electric charges. Its SI unit is ampere (A). Conventionally, the direction of current is taken opposite to the direction of flow of electrons.
2. **Electric Potential Difference**: It is the work done to move a unit charge from one point to another. Its SI unit is volt (V).
3. **Ohm's Law**: According to this law, the potential difference across the ends of a conductor is directly proportional to the current flowing through it, provided temperature remains constant. Mathematically, V = IR.
4. **Resistance**: It is the property of a conductor to resist the flow of current. Its SI unit is ohm (Ω). Resistance depends on: - Length of the conductor (directly proportional) - Area of cross-section (inversely proportional) - Nature of the material
5. **Resistivity**: It is a characteristic property of the material. Metals have low resistivity, insulators have high resistivity.
6. **Resistors in Series**: When connected in series, the equivalent resistance is the sum of individual resistances. The current is the same through all resistors.
7. **Resistors in Parallel**: When connected in parallel, the reciprocal of equivalent resistance is the sum of reciprocals of individual resistances. The voltage is the same across all resistors.
8. **Heating Effect of Electric Current**: When current flows through a conductor, heat is produced. The heat produced is given by H = I²Rt. This is Joule's law of heating.
9. **Electric Power**: It is the rate of consumption of electrical energy. P = VI = I²R = V²/R. Its SI unit is watt (W). Commercial unit is kilowatt hour (kWh).
10. **Fuse**: It is a safety device that protects circuits from excessive current.
Students, electricity is a vast and important topic. Make sure you understand all the concepts clearly and practice numerical problems regularly. Pay special attention to the formulas and their applications. Remember to always draw circuit diagrams neatly and label all components.
Thank you for your attention in today's class. Keep studying hard and don't hesitate to ask questions if you have any doubts. Good luck with your studies!