Class 10 Science Electricity: Important Concepts for Board Exams

Class 10 Science Electricity: Master Class 10 Science Chapter Electricity with easy explanations, key formulas, NCERT solutions, and exam-focused notes. Perfect for quick revision and board exam preparation.

Here is a comprehensive blog post about electricity, drawing extensively from the provided YouTube transcript for Class 10 Science.

Class 10 Science Electricity: A Comprehensive Guide to Key Concepts and Applications

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Welcome to this comprehensive guide on Electricity, a fundamental chapter in Class 10 Science that often appears in mid-term and board examinations. This blog post aims to serve as a one-stop solution for both students seeking a quick revision of concepts they’ve already learned and those encountering the topic for the first time. We’ll explore all the essential concepts of electricity, delve into numerical problems, and highlight important, often confusing questions to ensure this chapter no longer feels challenging. After completing this guide, you should feel confident enough to tackle NCERT questions on your own.

So, have you ever seen someone look shocked or angry, or observed an “attitude” that seemed to spark something? Or perhaps wondered what causes the jolt of electricity? The answer often lies with charge. Let’s begin our journey by understanding electric charge and its pivotal role in electricity.

1. Electric Charge: The Foundation of Electricity

We don’t jump straight into current when starting this chapter; instead, we first grasp the concept of charge.

Definition of Electric Charge

Electric charge is defined as a property associated with matter. Just as matter has properties like mass and volume, it also possesses the property of acquiring charge. This property enables matter to produce and experience electrical or magnetic effects. For instance, the repulsion between like poles (North-North, South-South) or attraction between opposite poles (North-South) in magnets also originates from charge, though the detailed understanding of magnetism’s origin comes later. In this chapter, our focus will be on the motion of charge and the phenomena it exhibits.Class 10 Science Electricity

SI Unit and Other Units of Charge

Since charge is a measurable quantity, it has an SI unit.Class 10 Science Electricity

The SI unit of charge is the Coulomb, denoted by C.
There are also smaller units that are frequently used in problems:
- 1 milliCoulomb (mC) = 10⁻³ Coulomb.
- 1 microCoulomb (µC) = 10⁻⁶ Coulomb. It’s crucial to note these smaller units alongside the SI unit to avoid problems when solving numerical questions.

Denotion of Charge

Charge is typically denoted by the symbol Q. Both a capital ‘Q’ or a lowercase ‘q’ can be used, but in Class 10, the capital ‘Q’ is generally preferred. For example, Q = 3 milliCoulomb means Q = 3 × 10⁻³ Coulomb.

Where is Charge Present in the Universe?

Fundamentally, charge is present on elementary particles. These are the building blocks of matter, and since charge is a property associated with matter, it naturally resides on these tiny constituents. In the context of Class 10, the elementary particles we consider are:

Electron: Carries a negative charge.
Proton: Carries a positive charge.
Neutron: Is neutral in nature.

It’s important to clarify the term “neutral” for neutrons. It doesn’t mean neutrons have no charge; rather, they contain an equal amount of positive and negative charge internally, which cancels out, resulting in a net charge of zero. Since neutrons are also matter, they must possess charge, but it is balanced.

All these sub-atomic particles (electrons, protons, neutrons) reside within an atom. Atoms combine to form molecules, and molecules make up all the substances in our universe, including ourselves, pens, and boards. Therefore, it is this fundamental presence of charge on elementary particles that ultimately explains why bodies attract or repel each other.Class 10 Science Electricity

How Does Charge Appear on a Body?

Consider a common experiment: rubbing a balloon on your hair, causing your hair to be attracted to the balloon. How does this phenomenon, seemingly generating charge, occur?

When two bodies are rubbed together (e.g., your hands or a balloon and hair), electrons transfer from one body to the other.

The body that loses electrons becomes positively charged.
The body that gains electrons becomes negatively charged.

This happens because only electrons are typically mobile; protons, being inside the nucleus, do not easily transfer. The attraction between the oppositely charged balloon and hair is an example of this electric attraction. This effect is often more visibly seen with lighter objects.

Essentially, charge appears on a body by disturbing its electrical neutrality. An atom is normally neutral, with an equal number of protons (positive charges) and electrons (negative charges).Class 10 Science Electricity

If an atom loses an electron, it becomes a positively charged ion (cation), as the positive protons now outnumber the negative electrons. For example, a neutral sodium atom with 11 protons and 11 electrons becomes Na⁺ (a cation) if it loses one electron, leaving 11 protons and 10 electrons.
If an atom gains an electron, it becomes a negatively charged ion (anion). For example, if the lost electron goes to a chlorine atom, it becomes Cl⁻.

Therefore, charge is not “created” or “destroyed”; it merely becomes visible or apparent when the balance of positive and negative charges is disrupted.

2. Quantization of Charge

One of the most fundamental properties of electric charge is its quantization.

The Meaning of “Quanta”

The term “quanta” literally means packets. In physics, particularly in quantum physics, this refers to small, discrete packets. So, quantization of charge means that charge also exists in discrete packets.Class 10 Science Electricity

Imagine buying a packet of bread slices. You can buy a whole packet, but the shopkeeper won’t sell you half a slice or two-and-a-half slices. Similarly, charge is transferred in complete, indivisible packets. You cannot transfer half an electron, 1.5 electrons, or 10.2 electrons. Charge transfer always occurs in whole numbers.

The Smallest Packet of Charge

The smallest possible packet of charge that exists in nature is the charge of a single electron. Therefore, when one electron is transferred, it’s equivalent to one packet of charge being transferred; if two electrons are transferred, it’s two packets, and so on.Class 10 Science Electricity

The value of the charge on one electron (and one proton, only differing in sign) is a constant that students generally memorize:

e = 1.6 × 10⁻¹⁹ Coulombs.
- For an electron, the charge is -1.6 × 10⁻¹⁹ C.
- For a proton, the charge is +1.6 × 10⁻¹⁹ C. The sign primarily indicates whether the charge is associated with an electron or a proton, while the magnitude remains the same.Class 10 Science Electricity

Formula for Quantization of Charge

If a body has ‘n’ electrons (or ‘n’ excess electrons), the total charge (Q) on that body is given by: Q = ± ne Where:

Q is the total charge on the body.Class 10 Science Electricity
n is an integer representing the number of electrons (or elementary charges) transferred. Since ‘n’ must be a natural number (1, 2, 3…), it signifies that charge is quantized.
e is the magnitude of the elementary charge (1.6 × 10⁻¹⁹ C).
The ± sign indicates whether the charge is positive (due to loss of electrons/excess protons) or negative (due to gain of electrons/excess electrons).

For example, if a body has 10 electrons, its total charge would be 10 × 1.6 × 10⁻¹⁹ C.

Example: Finding the Number of Electrons in One Coulomb of Charge

This is a famous problem often found in NCERT textbooks. Query: Find the number of electrons present in 1 Coulomb of charge. Given:

Total charge, Q = 1 C.
Elementary charge, e = 1.6 × 10⁻¹⁹ C (a known constant). To find: Number of electrons, n.

Using the quantization formula: Q = ne 1 = n × (1.6 × 10⁻¹⁹)

Solving for n: n = 1 / (1.6 × 10⁻¹⁹) n = (1 / 1.6) × 10¹⁹ To simplify, multiply numerator and denominator by 10 to remove the decimal: n = (10 / 16) × 10¹⁹ Further simplification by dividing 10 by 16: n = (0.625) × 10¹⁹ In scientific notation (standard form, where the decimal is placed after the first non-zero digit): n = 6.25 × 10¹⁸ electrons

This shows that 6.25 × 10¹⁸ electrons together constitute 1 Coulomb of charge. This is an incredibly large number, highlighting that 1 Coulomb is a very significant amount of charge. The decimal in 6.25 does not imply fractional electrons; it’s a part of scientific notation.Class 10 Science Electricity

3. Electrical Substances: Conductors, Semiconductors, and Insulators

Based on how easily they permit electrons to flow, materials are categorized into three main types. The fundamental difference lies in the number of free electrons they possess.

3.1 Conductors

Definition: Conductors are substances or materials that permit electrons to flow freely from particle to particle. They readily conduct electricity.
Reason: This property is due to the presence of a large number of loosely bound electrons, also known as free electrons. With many free electrons, charges can easily move throughout the material.
Examples: Common conductors include copper, aluminum, silver, and impure water.

3.2 Semiconductors

Definition: Semiconductors are materials whose electrical conductivity falls between that of a conductor and an insulator. They conduct electricity, but only to a limited extent, passing small currents (in milliampere or microampere range).
Reason: Semiconductors have fewer free electrons compared to conductors.Class 10 Science Electricity
Examples: Silicon, arsenic, gallium arsenide, and titanium dioxide.Class 10 Science Electricity

3.3 Insulators

Definition: Insulators are substances or materials that resist the free flow of electrons from atom to atom and molecule to molecule. They do not allow electricity to pass through them.
Reason: This is due to the absence or a very small number of free electrons.
Examples: Glass, wood, rubber, and cloth.

The core idea is that a higher number of free electrons leads to better conductivity, while fewer or no free electrons result in poor or no conductivity.

4. Electric Current

With a clear understanding of charge and materials, we can now define electric current.

What is Electric Current?

Electric current is are negatively charged) that flow through the wire, moving from the negative terminal of a battery to the positive terminal (repelled by negative, attracted by positive).Class 10 Science Electricity

Conventional Current: By convention, however, current is considered to flow in the direction opposite to the flow of electrons. This means conventional current flows from the positive terminal to the negative terminal. This convention was established by earlier scientists and, as current is a scalar quantity, its assigned direction doesn’t significantly impact energy calculations.Class 10 Science Electricity

Definition of Electric Current

Electric current is defined as the rate of flow of charge through a cross-section of a conductor per unit time. This means that for current to exist, not only does the quantity of charge matter, but also its speed of flow.

Formula for Electric Current

The formula for electric current (I) is: I = Q / T Where:

I is the electric current.
Q is the amount of charge flowing.
T is the time taken for the charge to flow.

SI Unit of Electric Current

Based on its formula:

The unit of charge is Coulomb (C).
The unit of time is Second (s).
So, the unit of current is Coulomb per Second (C/s).
However, the SI unit of current is Ampere, denoted by A. Therefore, 1 Ampere = 1 Coulomb per Second (or C s⁻¹).

What Constitutes Electric Current Flowing in a Conductor?

A bare wire, by itself, does not produce current. For charge to flow and constitute an electric current, an external agency is required.

This external agency is known as potential difference (which we will explore further).Class 10 Science Electricity
When a battery (which provides potential difference) is connected across the ends of a conductor, it “pushes” the free electrons present in the wire.
This push causes the electrons to flow, thereby constituting an electric current. So, a battery provides the necessary external force to drive the charges.

Example: Calculating Charge Given Current and Time

This is another typical NCERT problem. Query: A current of 0.5 Ampere is drawn by a filament of an electric bulb for 10 minutes. Find the amount of electric charge that flows through the circuit. Given:

Current, I = 0.5 A.
Time, T = 10 minutes. To find: Amount of electric charge, Q.

Step 1: Convert time to SI units (seconds). T = 10 minutes × 60 seconds/minute = 600 seconds.

Step 2: Use the formula I = Q / T. 0.5 A = Q / 600 s

Step 3: Solve for Q. Q = 0.5 A × 600 s Q = 300 C Therefore, 300 Coulombs of charge flow through the circuit.

5. Voltage and Potential Difference

The “external agency” we mentioned that drives current is called voltage or potential difference.

Why Does Electric Charge Flow?

Electric charge in a conductor flows due to the presence of a potential difference across the ends of the conductor. This potential difference is also known as voltage. If there is no potential difference (e.g., when a switch is off or a battery is removed), current will not flow, and devices like bulbs or TVs won’t operate. The establishment of potential difference is a prerequisite for current to flow.

What Entity Provides Potential Difference?

The entity that provides and maintains potential difference in a circuit is a battery. Batteries have two ends, a positive (anode) and a negative (cathode), which are chemically created.

How Does a Battery Drive Current?

Let’s understand the mechanism.

When a battery is connected to a wire, its negative end is attached to one end of the wire, and its positive end to the other.
The negative terminal of the battery repels the free electrons in the wire (like charges repel).
Simultaneously, the positive terminal of the battery attracts these negatively charged electrons (opposite charges attract).
This combined pushing and pulling action causes the electrons to move continuously in one direction (from negative to positive terminal within the external circuit).
As established earlier, the conventional current then flows in the opposite direction (from positive to negative terminal within the external circuit).

So, the battery’s role is to “push” the charges, making it possible for current to flow.Class 10 Science Electricity

Definition of Electric Potential Difference (Voltage)

Potential difference (or voltage) is the amount of work done on a unit positive charge to move it from infinity to a specific point, or from one point to another, against an external electric force. In simpler terms, it’s the work done by the battery to move charge.Class 10 Science Electricity

Formula for Potential Difference

The formula for potential difference (V) is: V = W / Q Where:

V is the potential difference or voltage.
W is the work done.
Q is the amount of charge moved.

SI Unit of Potential Difference

Based on its formula:

The unit of work done is Joule (J).
The unit of charge is Coulomb (C).
So, the unit of potential difference is Joule per Coulomb (J/C).
However, the SI unit of potential difference is Volt, denoted by V. Therefore, 1 Volt = 1 Joule per Coulomb (or J C⁻¹).

Definition of 1 Volt

One Volt is defined as the potential difference between two points when 1 Joule of work is done to move 1 Coulomb of charge from one point to the other. It signifies that a 1 Joule “push” is applied to 1 Coulomb of charge.

Example: Calculating Work Done Given Charge and Potential Difference

Another NCERT problem. Query: What is the work done required to move a charge of 2 Coulombs across a potential difference of 12 Volts?. Given:

Potential difference, V = 12 V.
Charge, Q = 2 C. To find: Work done, W.

Using the formula V = W / Q: 12 V = W / 2 C

Solving for W: W = 12 V × 2 C W = 24 J Thus, 24 Joules of work is required.

6. Electric Circuit

An electric circuit is the complete path for electric current.Class 10 Science Electricity

Definition of an Electric Circuit

An electric circuit is a continuous and closed path made of wires, along which electric current flows. It must be closed because current will not flow if there is any break or gap in the path.Class 10 Science Electricity

Basic Elements of an Electric Circuit

A complete electric circuit typically consists of four fundamental components:

Electric devices (appliances): Something that utilizes the electricity, like a bulb, fan, or TV.
Source of energy: The component that provides the potential difference, typically a battery.
Connecting wires: The medium through which the current flows.Class 10 Science Electricity
Switch: A device to control the flow of current (to turn the circuit on or off).Class 10 Science Electricity

Circuit Elements and Their Symbols

To represent circuits clearly, standard symbols are used. Here are some common ones:

Electric Cell: A long line (positive terminal) and a short, thick line (negative terminal).
- —| |—
Battery (Combination of Cells): Multiple cell symbols joined.Class 10 Science Electricity
- —| |—| |—
Plug Key or Switch (Open): A break in the circuit, indicating no current flow (zero current).
- —o o—
Plug Key or Switch (Closed): A dot between the lines, indicating a closed circuit and current flow (non-zero current).
- —o•o—
Wire Joint: A dot at the intersection of two wires.
- — —•— —
Wires Crossing Without Joining: One wire arches over another.
- —^—
Electric Bulb: A circle with a cross or a loop inside.
- —(x)— or —(~)—
Resistor (Resistance): A zigzag line.
- —/\/\/\—
Variable Resistor or Rheostat: A zigzag line with an arrow crossing it or a sliding contact.
- —/\/\/\>— or —/\/\/\—|—
Ammeter (Measures Current): A circle with ‘A’ inside.
- —Ⓐ—
Voltmeter (Measures Voltage): A circle with ‘V’ inside.
- —Ⓥ—

Real-life Circuit Elements

In a school lab, you might encounter physical components that correspond to these symbols:Class 10 Science Electricity

Battery Eliminator: Converts AC current from mains to DC current suitable for experiments.
Resistance Box: Contains various resistors, allowing different resistance values to be selected.
Voltmeter: An instrument with positive and negative terminals to measure potential difference.
Ammeter: An instrument to measure current, also with positive and negative terminals.Class 10 Science Electricity
Plug Key: A physical switch where you insert or remove a plug to open or close the circuit.
Rheostat: A variable resistor, typically a long wire wound on a cylinder with a sliding contact to change resistance.
Connecting Wires: Wires used to connect different components.Class 10 Science Electricity

7. Ohm’s Law

Ohm’s Law establishes a fundamental relationship between potential difference, current, and resistance.

Statement of Ohm’s Law

Ohm’s Law states that the current flowing through a conductor is directly proportional to the potential difference (voltage) applied across its ends, provided physical conditions (like temperature) remain constant.

This means that if you increase the voltage across a conductor, the current flowing through it will also increase proportionally. An analogy is water flowing from a higher to a lower level: the greater the height difference (voltage), the stronger the water flow (current).Class 10 Science Electricity

Mathematically, this relationship is expressed as: V ∝ I Where:

V is the potential difference.
I is the current.

To convert this proportionality into an equality, a constant is introduced: V = IR Here, R is the constant of proportionality known as Resistance.

Resistance (R)

Definition: Resistance is the opposition offered by the atoms and other particles of a conductor to the flow of electric current.
Reason: As electrons move through a conductor, they constantly collide with the atoms and other particles within the material. These collisions hinder the free flow of electrons, creating an “obstruction” or “opposition”. The more collisions, the higher the resistance.
Denotion: Resistance is denoted by R.
SI Unit: The SI unit of resistance is Ohm, represented by the Greek letter Ω (omega).
Circuit Symbol: In a circuit diagram, a resistor is represented by a zigzag line.Class 10 Science Electricity

VI Graph and Slope of Ohm’s Law

Since V is directly proportional to I, the graph plotting Voltage (V) on the y-axis against Current (I) on the x-axis for an ohmic conductor will always be a straight line passing through the origin, indicating a linear relationship.Class 10 Science Electricity

The slope of the VI graph gives the Resistance (R).

In a general y-x graph, slope = Δy / Δx.
For a VI graph, slope = ΔV / ΔI.
From Ohm’s Law (V = IR), if we rearrange, V/I = R.
Therefore, Slope (ΔV / ΔI) = R (Resistance). This is a very important concept for graphical problems.

Example: Calculating Current Using Ohm’s Law

Query: How much current will an electric bulb draw from a 220 V source if the resistance of the bulb filament is 1200 Ohm?. Given:

Voltage, V = 220 V.
Resistance, R = 1200 Ω. To find: Current, I.

Using Ohm’s Law, V = IR: 220 V = I × 1200 Ω

Solving for I: I = 220 / 1200 A I = 22 / 120 A (canceling a zero) I = 11 / 60 A (dividing by 2) The current is 11/60 Amperes.

Example: Plotting VI Graph and Calculating Resistance from Slope

Query: The values of current (I) flowing in a given resistor for corresponding values of potential difference (V) across the resistor are given below. Plot a graph between V and I and calculate the resistance.

Current (I) (A)	Voltage (V) (V)
0.5	1.6
1.0	3.2
1.5	4.8
2.0	6.4
2.5	8.0

Step 1: Plot the V-I Graph.

Draw axes, with V on the y-axis and I on the x-axis.
Mark the given current values (0.5, 1.0, 1.5, 2.0, 2.5 A) on the x-axis.
Mark the corresponding voltage values (1.6, 3.2, 4.8, 6.4, 8.0 V) on the y-axis.
Plot each (I, V) point (e.g., (0.5, 1.6), (1.0, 3.2), etc.).
Connect the plotted points. You will get a straight line passing through the origin, confirming the linear relationship as per Ohm’s Law.

Step 2: Calculate Resistance from the Slope. To calculate the slope (ΔV / ΔI) for a linear graph, pick any two points or, for average resistance, pick the extreme points.

ΔV = V₂ – V₁ = 8.0 V – 1.6 V = 6.4 V.
ΔI = I₂ – I₁ = 2.5 A – 0.5 A = 2.0 A.
Resistance, R = ΔV / ΔI = 6.4 V / 2.0 A.
R = 3.2 Ω. The resistance of the resistor is 3.2 Ohms.

8. Factors Affecting Resistance

While R is a constant of proportionality in Ohm’s Law, it is not a universal constant; its value depends on several physical factors. There are four main factors that affect the resistance of a conductor:Class 10 Science Electricity

Length (L) of the conductor.
Area of cross-section (A) of the conductor.
of Cross-section

Resistance is inversely proportional to the area of cross-section of the conductor (R ∝ 1/A).
A thicker wire (larger cross-sectional area) provides more space for electrons to flow, reducing the chances of collisions and making it easier for current to pass. Conversely, a thinner wire (smaller area) offers more hindrance.Class 10 Science Electricity

8.3 Dependence on Nature of Material (Resistivity, ρ)

The material a conductor is made of fundamentally determines its ability particles within the conductor vibrate more vigorously. This increased agitation leads to more frequent collisions with the flowing electrons, making it harder for them to pass through. Class 10 Science Electricity
This increased hindrance results in higher resistance. The exact formula for temperature dependence is complex and usually studied in higher classes.Class 10 Science Electricity

9. Resistance vs. Resistivity

Resistance and resistivity are related but distinct concepts, often causing confusion.

Resistance

Definition: Resistance is the opposition offered by the internal particles (atoms, inner electrons) of a conductor to the flow of current. It is an actual obstruction or hindrance to electron flow.
Denotion: Denoted by R.
Factors affecting: Depends on length (L), area (A), material (ρ), and temperature (T).
SI Unit: Ohm (Ω).Class 10 Science Electricity

Resistivity (ρ)

Definition: Resistivity is an intrinsic property of a material that quantifies its ability to offer opposition to current flow. It’s the “capacity to resist”.
Denotion: Denoted by ρ (rho).
Factors affecting: Primarily depends on the nature of the material itself and temperature. It does not depend on the length or cross-sectional area of the specific piece of material.
SI Unit: Ohm-meter (Ωm).

Think of it this way: if resistance is an “action” (the act of opposing flow), then resistivity is an “adverb” or an “adjective” (describing the material’s inherent ability to perform that action). A material with high resistivity inherently has a high capacity to resist, and therefore, a piece of that material will exhibit high resistance.Class 10 Science Electricity

Resistivity Values and Material Classification

The value of resistivity (ρ) is a key indicator of a material’s electrical behavior.

Good Conductors (e.g., Silver, Copper, Aluminium): Have very low resistivity (e.g., in the range of 10⁻⁸ Ωm). Low resistivity means low inherent ability to resist, so they conduct current very well.
Alloys (e.g., Nichrome, Manganin): Have higher resistivity than pure metals (e.g., in the range of 10⁻⁶ Ωm). Their resistivity is about 100 to 10,000 times higher than that of conductors.
Insulators (e.g., Glass, Rubber, Paper): Have extremely high resistivity (e.g., in the range of 10¹² to 10¹⁷ Ωm). Their very high resistivity means they are very effective at blocking current flow.

These values are for analysis; memorizing them is not necessary.Class 10 Science Electricity

Example: Effect of Stretching a Wire on Resistance and Resistivity

Query: A wire of length L and resistance R is stretched so that its length is doubled. The area of cross-section is halved. How will its resistance change, and will its resistivity change?.Class 10 Science Electricity

Part A: Change in Resistance Initial state:

Length = L
Area = A
Resistivity = ρ (material constant)
Initial Resistance, R = ρL / A (Equation 1).

Stretched state:

New Length, L’ = 2L (length is doubled).
New Area, A’ = A/2 (area is halved).
New Resistivity, ρ’ = ρ (material doesn’t change by stretching).
New Resistance, R’ = ρ’L’ / A’. R’ = ρ(2L) / (A/2). R’ = 4 (ρL / A). Substituting Equation 1 into this: R’ = 4R. So, the new resistance becomes four times the original resistance.

Part B: Change in Resistivity

Resistivity (ρ) depends only on the nature of the material and its temperature, not on its physical dimensions (length or area).
Since the wire is simply stretched and its material remains the same (e.g., copper wire remains copper), its resistivity will not change.
Therefore, the answer is no change in resistivity.Class 10 Science Electricity

10. Series and Parallel Circuits

Resistors can be connected in a circuit in different configurations, primarily series and parallel, which affect the total resistance and the distribution of current and voltage.

10.1 Series Circuit

Definition: When resistors are connected end-to-end in a single wire such that there is only one path for the current to flow, they are said to be in series.
Current and Voltage Distribution:Class 10 Science Electricity
- Current (I) remains the same through all resistors in series, as they are on the same wire.
- Voltage (V) divides across the individual resistors. The series circuit acts as a potential divider. If the total voltage across the series combination is V, and individual voltages are V₁, V₂, V₃ across R₁, R₂, R₃, then V = V₁ + V₂ + V₃.
Derivation of Equivalent Resistance (R_total or R_series): Starting with V = V₁ + V₂ + V₃. Using Ohm’s Law (V = IR), we substitute for each voltage: I R_total = I R₁ + I R₂ + I R₃. (Since current I is the same throughout the series circuit) I R_total = I (R₁ + R₂ + R₃). Dividing by I on both sides: R_total = R₁ + R₂ + R₃ + …. The equivalent resistance in a series circuit is the sum of the individual resistances.Class 10 Science Electricity

10.2 Parallel Circuit

Definition: When resistors are connected such that their ends are joined at common points (forming junctions or branches), providing multiple paths for the current to flow, they are said to be in parallel. Each resistor has an independent connection to the battery.
Current and Voltage Distribution:
- Voltage (V) remains the same across all resistors connected in parallel because they are connected between the same two common points (which are connected to the battery terminals).
- Current (I) divides among the individual resistors. If the total current is I, and individual currents are I₁, I₂, I₃ through R₁, R₂, R₃, then I = I₁ + I₂ + I₃.
**Derivation of Equivalent Resistance both sides: 1 / R_total = 1 / R₁ + 1 / R₂ + 1 / R₃ + …. The reciprocal of the equivalent resistance in a parallel circuit is the sum of the reciprocals of the individual resistances.

Example: Calculating Total Current in a Series Circuit

Query: Six resistors (1 Ω, 2 Ω, 3 Ω, 4 Ω, 5 Ω, 6 Ω) are connected in series to a 10 V battery. Find the total current in the circuit. Given:

R₁ = 1 Ω, R₂ = 2 Ω, …, R₆ = 6 Ω (all in series).
Total Voltage, V = 10 V. To find: Total Current, I.

Step 1: Calculate Total Resistance (R_total) in series. R_total = R₁ + R₂ + R₃ + R₄ + R₅ + R₆. R_total = 1 + 2 + 3 + 4 + 5 + 6 = 21 Ω.

Step 2: Use Ohm’s Law (V = IR_total) to find total current. 10 V = I × 21 Ω. I = 10 / 21 A. The total current is 10/21 Amperes.

Example: Calculating Total Current in a Parallel Circuit

Query: Two resistors (3 Ω and 6 Ω) are connected in parallel to a 10 V battery. Find the total current in the circuit [ = 6: 1 / R_parallel = 1/6 + 2/6 = 3/6 = 1/2. R_parallel = 2 Ω.

Shortcut for two parallel resistors: For two resistors, R₁ and R₂, in parallel, their equivalent resistance can be found as (R₁ × R₂) / (R₁ + R₂). For 3 Ω and 6 Ω: (3 × 6) / (3 + 6) = 18 / 9 = 2 Ω.

Shortcut for N identical parallel resistors: If ‘N’ identical resistors, each of value ‘X’ Ω, are connected in parallel, the total resistance is X / N. (e.g., three 6 Ω resistors in parallel: 6/3 = 2 Ω).

Step 2: Use Ohm’s Law (V = IR_parallel) to find total current. 10 V = I × 2 Ω. I = 10 / 2 = 5 A. The total current is + 6) = 18 / 9 = 2 Ω.

Now, this 2 Ω equivalent resistor is in series with the original 2 Ω resistor.Class 10 Science Electricity
Total resistance = 2 Ω + 2 Ω = 4 Ω.

Example 2: Three 3 Ω resistors. Two are in series, and this combination is in parallel with the third resistor.

First, solve the series combination of two 3 Ω resistors: 3 Ω + 3 Ω = 6 Ω.
Now, this 6 Ω equivalent resistor is in parallel with the third 3 Ω resistor.Class 10 Science Electricity
Total resistance = (6 × 3) / (6 + 3) = 18 / 9 = 2 Ω.

Example: Arranging Resistors for a Specific Total Resistance

Query: How can three resistors, each of resistance 3 Ω, be connected to give a total resistance of 4.5 Ω and 2 Ω?.

There are typically four ways to connect three identical resistors:

All three in series: R_total = 3 + 3 + 3 = 9 Ω.
All three in parallel: R_total = 3 / 3 = 1 Ω (using X/N shortcut).
Two in parallel, with the combination in series with the third:
- Parallel part: (3 × 3) / (3 + 3) = 9 / 6 = 1.5 Ω.
- Total (series with third): 1.5 Ω + 3 Ω = 4.5 Ω. (This matches one of the required values!)
Two in series, with the combination in parallel with the third:
- Series part: 3 Ω + 3 Ω = 6 Ω.
- Total (parallel with third): (6 × 3) / (6 + 3) = 18 / 9 = 2 Ω. (This matches the other required value!)

This systematic approach helps in finding the correct configuration.

11. Rheostat (Variable Resistor)

A rheostat is a variable resistor, meaning its resistance can be changed.

Description and Function

A rheostat typically consists of a coil of wire wound around a cylindrical insulator.
It has a sliding contact (or clip) that can be moved along the wire coil.
By moving this sliding contact, the effective length of the wire through which the current flows can be increased or decreased.
Since resistance is directly proportional to length (R ∝ L), changing the length of the wire in the circuit directly changes the resistance of the circuit.
If the slider is moved to include a longer section of the coil, resistance increases; if moved to include a shorter section, resistance decreases.
This allows for manual control over the current in a circuit, as changing resistance affects current (I = V/R).Class 10 Science Electricity

12. Heating Effect of Electric Current (Joule’s Law of Heating)

When current flows through a resistor, some electrical energy is converted into heat. This phenomenon is known as the heating effect of electric current, described by Joule’s Law.

Statement of Joule’s Law of Heating

Joule’s Law of Heating states that the heat (H) produced in a conductor is directly proportional to:

The square of the current (I²) flowing through it.
The resistance (R) of the conductor.
The time (T) for which the current flows.

Combining these proportionalities, we get: H ∝ I²RT. Introducing a constant of proportionality (which is experimentally found to be 1), the formula becomes: H = I²RT. The heat produced is measured in Joules (J).

Derivation of H = I²RT

The formula for heat (energy) can be derived from basic electrical definitions:Class 10 Science Electricity

Potential Difference (V) = Work Done (W) / Charge (Q). Rearranging, W = VQ.
Current (I) = Charge (Q) / Time (T). Rearranging, Q = IT.
Ohm’s Law: V = IR.

Now, substitute Q = IT into the work done equation: W = V (IT). Next, substitute V = IR into this equation: W = (IR) (IT). W = I²RT. Since work done is a form of energy, and in this context, it manifests as heat energy: H = I²RT.Class 10 Science Electricity

Example: Calculating Heat Developed

Query: An electric iron has a resistance of 20 Ω and takes a current of 5 A. Calculate the heat developed in 30 seconds. Given:

Resistance, R = 20 Ω.
Current, I = 5 A.
Time, T = 30 s. To find: Heat developed, H.

Using Joule’s Law, H = I²RT: H = (5 A)² × 20 Ω × 30 s. H = 25 × 20 × 30 J. H = 500 × 30 J. H = 15,000 J. The heat developed is 15,000 Joules, or 15 kJ.

Practical Applications of Heating Effect

While heat is often seen as wasted energy, it has several practical and essential applications:

Electric Bulb:
- Principle: The filament inside a bulb heats up due to the flow of current (I²RT) to such a high temperature that it starts to glow and emit light.
- Filament Material: The filament is typically made of Tungsten.
- Reasons for Tungsten:
  - High Resistance: Ensures significant heat generation (H ∝ R).
  - Very High Melting Point: Tungsten has the highest melting point of all naturally occurring substances (around 3380 °C). This prevents the filament from melting even at very high temperatures, allowing it to glow for a long time.
- Inert Gas: The bulb is filled with an unreactive gas like Argon or Nitrogen to prevent the filament from oxidizing and prolong its life.
Electric Fuse:
- Principle: A fuse protects electrical circuits and appliances from damage due to excessive current (overloading or short-circuiting). It works on the principle of the heating effect.
- Fuse Wire Material: The fuse wire is usually made of a Tin-Lead alloy (SnPb alloy).
- Reasons for Tin-Lead Alloy:
  - Low Melting Point: When the current exceeds a safe limit, the fuse wire quickly heats up (H ∝ I²) and melts due to its low melting point.
  - High Resistance: Ensures sufficient heat generation to melt the wire when current is excessive.
- Function: When the fuse wire melts, it breaks the circuit, stopping the flow of current and protecting the connected appliances from damage due to excessive heat.Class 10 Science Electricity
Heating Elements (e.g., Electric Heater, Iron, Geyser):
- Principle: These devices are designed to produce a large amount of heat to perform tasks like cooking, ironing, or heating water.
- Heating Element Material: The heating element is commonly made of Nichrome alloy.
- Reasons for Nichrome Alloy:
  - High Resistivity (ρ): Alloys generally have much higher resistivity than pure metals. This high resistivity translates to high resistance (R) for a given length and area.
  - High Resistance (R): High resistance leads to significant heat production (H ∝ R) when current flows.Class 10 Science Electricity
  - Does Not Oxidize Easily at High Temperatures: Nichrome can withstand high temperatures without readily burning or corroding, ensuring durability.

Example: Identifying Materials for Different Components

Query: Which element is used to make: a) Filament, b) Connecting wire, c) Heating element, d Heater Cord (Connecting Wire): * Made of pure metal, typically copper. * Copper has very low resistivity (ρ). * This means it has very low resistance (R). * Consequently, the heat produced (H = I²RT) in the cord is minimal, and it does not glow.

Heater Coil (Heating Element):
- Made of an alloy, specifically Nichrome [5 at which work is done or electrical energy is consumed/transferred per unit time**. Mathematically: P = Work Done (W) / Time (T). Or, P = Electrical Energy (E) / Time (T).

SI Unit of Electric Power

The SI unit of power is Watt (W). From the formula, 1 Watt = 1 Joule per Second (J/s).

Relationship Between Energy and Power

From P = E/T, we can derive the formula for * Since I = Q / T (current is charge per unit time), substitute Q/T with I: P = VI.

P = I²R (Current² × Resistance):
- Start with P = VI.
- From Ohm’s Law, V = IR.
- Substitute V = IR into P = VI: P = (IR)I. P = I²R.
**P = V²/R (Voltage² Energy Since E = PT, we can combine this with the power formulas to get different expressions for energy (which are also expressions of Joule’s Law of Heating):
E = VIT (substituting P = VI).
E = I²RT (substituting P = I²R). This is the same as Joule’s Law of Heating we discussed earlier.
E = (V²/R)T (substituting P = V²/R).Class 10 Science Electricity

All three energy formulas are equivalent and represent the heat generated or energy consumed.

Example: Calculating Total Current for Bulbs in Parallel

Query: A 60 W bulb and a 40 W bulb are connected in parallel across a 220 V battery. Find the amount of current flowing in the circuit. Given:

Bulb 1: P₁ = 60 W, connected to V = 220 V.
Bulb 2: P₂ = 40 W, connected to V = 220 V (since in parallel). To find: Total current (I_total).

Step 1: Calculate current through each bulb separately. Since the bulbs are in parallel, the voltage across each is the same as the battery voltage, 220 V. Using P = VI:

For Bulb 1: P₁ = V × I₁ 60 W = 220 V × I₁. I₁ = 60 / 220 A.
For Bulb 2: P₂ = V × I₂ 40 W = 220 V × I₂. I₂ = 40 / 220 A.

Step 2: Calculate total current (I_total). In a parallel circuit, total current is the sum of individual currents: I_total = I₁ + I₂. I_total = (60 / 220) + (40 / 220) A. I_total = 100 / 220 A. I_total = 10 / 22 A. I_total = 5 / 11 Amperes.Class 10 Science Electricity

Example: Calculating Power of a Bulb at a Different Voltage

Query: A bulb has a power rating of 60 W when connected to 220 V. What power will it consume if it is connected to 110 V?. Given:

Rated Power, P_rated = 60 W at V_rated = 220 V.
New Voltage, V_new = 110 V. To find: New Power, P_new.

Important Concept: The resistance of the bulb filament remains constant regardless of the applied voltage, as it depends on the physical characteristics of the bulb. We can use the rated values to find this constant resistance.

Step 1: Calculate the resistance (R) of the bulb using its rated values. Use the formula P = V²/R (since P and V are given, and R is constant). 60 W = (220 V)² / R. R = (220 × 220) / 60 Ω. (It’s often useful to leave R in this form without calculating a decimal, to avoid rounding errors).

Step 2: Calculate the new power (P_new) using the new voltage and the constant resistance. Use the same formula: P_new = (V_new)² / R. P_new = (110 V)² / [(220 × 220) / 60 Ω]. P_new = (110 × 110 × 60) / (220 × 220) W. P_new = (110/220) × (110/220) × 60 W. P_new = (1/2) × (1/2) × 60 W. P_new = (1/4) × 60 W. P_new = 15 Watts.Class 10 Science Electricity

Notice that halving the voltage (from 220 V to 110 V) did not halve the power (from 60 W to 30 W); instead, it quartered it (to 15 W) because power is proportional to the square of the voltage (P ∝ V²) when resistance is constant.

14. Commercial Unit of Energy

Electrical energy is bought and sold, and for commercial purposes, a larger unit than Joules is used.

The Kilowatt-hour (kWh)

The commercial unit of electrical energy is the kilowatt-hour (kWh).
This unit is commonly referred to as “1 Unit” of electricity.
Relationship to SI Unit (Joules): 1 kWh = 3.6 × 10⁶ Joules. (This conversion value should be memorized).

Derivation of Kilowatt-hour

We know that Energy (E) = Power (P) × Time (T).
In SI units, Power is in Watts and Time is in Seconds, so Energy is in Watt-seconds, which is Joules.
For commercial use, larger units are more practical:
- Watts are converted to kilowatts (kW) (1 kW = 1000 W).
- Seconds are converted to hours (h) (1 hour = 3600 seconds).
Thus, Energy = (kilowatt) × (hour), giving kilowatt-hour (kWh).Class 10 Science Electricity

Example: Calculating Electricity Cost for a Refrigerator

Query: An electric refrigerator rated 400 W operates for 8 hours per day. What is the cost of energy to operate it for 30 days if the cost is ₹3 per kWh?. Given:

Power of refrigerator, P = 400 W.
Daily operation time, T_daily = 8 hours/day.
Total days = 30 days.
Cost per unit = ₹3 / kWh. To find: Total cost for 30 days.

Step 1: Calculate total operating time in hours for 30 days. T_total = 8 hours/day × 30 days = 240 hours.

Step 2: Calculate total energy consumed in kWh. First, convert power from Watts to kilowatts: P = 400 W = 400 / 1000 kW = 0.4 kW. Now, use E = P × T_total: E = 0.4 kW × 240 hours. E = 96 kWh.

Step 3: Calculate the total cost. Cost = Energy consumed (kWh) × Cost per kWh. Cost = 96 kWh × ₹3/kWh. Cost = ₹288. The cost of operating the refrigerator for 30 days will be ₹288.

Conclusion

This comprehensive guide has covered all essential concepts of the “Electricity” chapter for Class 10 Science, from the fundamental nature of electric charge to its various manifestations as current, voltage, resistance, and power. We’ve explored how these concepts are interconnected through Ohm’s Law, how materials behave differently based on their resistivity, and the practical applications of the heating effect of current in everyday appliances. We also tackled numerous problems, including those involving circuit combinations and commercial energy calculations, to strengthen your understanding.

To truly master this chapter, it’s highly recommended to:

Write down all the notes by hand, rather than simply printing them. This active learning approach enhances retention.Class 10 Science Electricity
Re-solve all the example problems discussed. Practice is key to understanding and applying the formulas and concepts.Class 10 Science Electricity

By diligently working through these steps, you will gain confidence and competence in electricity, preparing you well for your exams and future studies.