Electricity Class 10 Notes: CBSE Science Chapter 11

Introduction

• The Greek words "Electrica" and "Elektron" were used to describe electricity.

• Thales, a Greek philosopher, was the first to notice how certain elements attract other materials when rubbed together.

• These materials were divided into two categories by Gilbert: vitreous and resinous, as Positive charges and Negative charges.

Frictional Electricity

• Fur, flannel, wax, glass, cotton, paper, silk, human skin, wood, metals, rubber, resin, amber, sulphur, and ebonite are just a few of the materials used.

• If any two materials in this series are rubbed together, the element that occurs initially in the series will get positive charge, while the element that occurs later in the series will gain a negative charge.

Fundamental Laws of Electrostatics

• Positive and negative charges are the two types of charges.

• Charges that are similar repel each other, while charges that are dissimilar attract each other.

Coulomb’s Law

• \[\text{F }\alpha \text{ }\dfrac{\left( {{\text{q}}_{\text{1}}}{{\text{q}}_{\text{2}}} \right)}{{{\text{r}}^{\text{2}}}}\]

• The electrostatic attraction or repulsion force between two charges is proportional to the product of their charges and inversely proportional to the square of their distance.

• \[\text{F = }\dfrac{\left( \text{K}{{\text{q}}_{\text{1}}}{{\text{q}}_{\text{2}}} \right)}{{{\text{r}}^{\text{2}}}}\]

• \[\text{K}\] is the constant of proportionality and is equal to \[\text{9 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{9}}}\text{N}{{\text{m}}^{\text{2}}}\text{/}{{\text{C}}^{\text{2}}}\] for free space.

• The force is repulsive for similar charges and attractive for dissimilar ones.

Charge Conservation

• When an ebonite rod is brushed with fur, the ebonite gains a negative charge, while the fur gains a positive charge. 

• This indicates that electrons have transferred from fur to ebonite. 

• The system's net charge stays unchanged. 

• As a result, charges are not generated or destroyed; rather, they are moved from one material to another.

Insulators and Conductors

• Insulators are poor conductors of electricity, yet they can be quickly charged through friction.

• Charges can flow freely through conductors..

Current

• The rate of charge flow is referred to as current.

• If the charge \[\text{q}\] is in coulomb and \[\text{t}\] is the time in seconds then, 

current is

\[\text{I = }\dfrac{\text{q}}{\text{t}}\]

• The \[\text{SI}\] unit of current is ampere \[\left( \text{A} \right)\].

• Current is a scalar quantity.

• Example:

A current of \[1\text{A}\] is drawn by a filament of an electric bulb for \[20\]

minutes. Find the amount of electric charge that flows through the

circuit.

Ans:

The given data is,

\[\text{I = 1A}\] and

\[\text{t = 20 minutes}\]

\[\text{t = 20}\times \text{60}\]

\[\text{t = 1200 seconds}\]

Therefore,

\[\text{I = }\dfrac{\text{q}}{\text{t}}\]

Electric charge is

\[\text{q = It}\]

\[\text{q = 1 }\!\!\times\!\!\text{ 1200}\]

\[\text{q = 1200 C}\]

Symbols Used in Electric Circuit

Electric Component	Function/ Description	Symbol
Connecting Wire	A straight line represents a connecting wire.  It's commonly constructed of copper and has insulation on both ends to link two places electrically.
Resistor	A zip zap line is used to depict the resistor.  Brass terminals are represented by two heavy dots at the ends, to which a wire is attached. Alloys like as nichrome, manganin, constantan, and eureka are commonly used to make resistor wire.
Cell	The positive terminal of a cell is represented by a thin long line, while the negative terminal is represented by a thick and short line.  Electrical current source.
Fuse	In an electric circuit, to limit the current.
Plug key	To build or break an electric circuit for an extended period of time.
Battery	A cell that is made up of two or more cells.  The cells are placed in a row here.  Electrical current source.
Electric bulb	When voltage is placed across the terminals of an electric device, such as an incandescent lamp, glow lamp, or fluorescent lamp, it creates light.
Connecting wires	Wires are linked together.
Connecting wires	Crossing wires that aren't linked.
Voltmeter	It's a device that measures the difference in potential between two locations in an electric circuit.
Ammeter	It's a tool for determining the current in an electric circuit.
Alternating current	Alternating current is a type of current that changes direction fast on its own.

A circuit diagram is a diagram that displays how different components in a the circuit has been connected using traditional component symbols.

Electrical Potential

• Electric potential is the work done in carrying a unit positive charge from infinity to a point.

• If \[\text{W}\] is the work done \[\text{q}\] is the charge, then electric potential

\[\text{V = }\dfrac{\text{W}}{\text{q}}\]

• The \[\text{SI}\] unit of electric potential is Volts \[\left( \text{V} \right)\]

Electric Potential Difference

• The work done required to move a unit charge from one location to the other is defined as the electric potential difference between two points in an electric circuit carrying some current.

\[\text{Potential difference }\left( \text{V} \right)\text{ between two points = }\dfrac{\text{Work done }\left( \text{W} \right)}{\text{Charge }\left( \text{Q} \right)}\]

• The electric potential difference between points \[\text{A}\] and \[\text{B}\]is,

\[{{\text{V}}_{\text{AB}}}\text{ = }\dfrac{\text{Work done to carry charge q from A to B}}{\text{Charge }\left( \text{Q} \right)}\]

• The \[\text{SI}\] unit of electric potential difference is Volts \[\left( \text{V} \right)\] .

• Example:

How much work is done in moving a charge of \[\text{4 C}\] across two points

having a potential difference \[\text{24 V}\] ?

Ans:

Given data is,

Charge, \[\text{Q = 4 C}\]

Potential difference, \[\text{V = 24 V}\]

Therefore, the amount of work \[\text{W}\] done in moving the charge can be calculated as,

\[\text{W = V }\!\!\times\!\!\text{ Q}\]

\[\text{W = 24}\times \text{4}\]

\[\text{W = 96 J}\]

Electric Potential Energy

• The work required to transport charges to their proper places against the electric field using a source of energy is known as electric potential energy.

• The potential energy of the charges is used to store the work done.

Ohm’s Law

• Under similar physical conditions, the current flowing through a wire is directly proportional to the difference in potential applied across its ends.

\[\text{V }\alpha \text{ I}\]

\[\dfrac{\text{V}}{\text{I}}\text{ = Constant}\]

\[\dfrac{\text{V}}{\text{I}}\text{ = R}\]

Therefore,

\[\text{V = IR}\]

Where \[\text{R}\] is the resistance offered.

Resistance

• Resistance is the opposition to the flow of current.

• The \[\text{SI}\] unit of resistance is Ohm \[\left( \Omega  \right)\].

• \[\text{1 Ohm}\]is the resistance offered by a wire carrying \[\text{1 A}\]current when \[\text{1 V}\]  is applied across its ends.

• \[\text{        R = }\dfrac{\text{V}}{\text{I}}\]

\[\text{1 Ohm = }\dfrac{\text{1 Volt}}{\text{1 Ampere}}\]

• Variable resistance is a component that regulates current without changing the voltage source. 

• A rheostat is a device that is commonly used to adjust the resistance in an electric circuit.

Factors Affecting Resistance

• A conducting wire's resistance is determined by:

1. Nature of the material of the wire \[\left[ \text{Resistivity ( }\!\!\Omega\!\!\text{ )} \right]\]

2. Length of the wire \[\left( \text{l} \right)\]

3. Cross-sectional area of the wire \[\left( \text{A} \right)\]

• Resistance is directly proportional to its length \[\left( \text{l} \right)\] and inversely proportional to the area of cross-section \[\left( \text{A} \right)\]. 

That is,

\[\text{R  }\!\!\alpha\!\!\text{  l }\]

\[\text{R  }\!\!\alpha\!\!\text{  }\dfrac{1}{\text{A}}\text{ }\]

Therefore,

\[\text{R  }\!\!\alpha\!\!\text{  }\dfrac{\text{l}}{\text{A}}\text{ }\]

\[\text{R = }\rho \dfrac{\text{l}}{\text{A}}\]

Where,

\[\text{ }\!\!\rho\!\!\text{  }\left( \text{rho} \right)\] is a proportionality constant that refers to the electrical resistivity of the conductor's substance.

Resistivity

• The resistance offered by a wire of unit length and unit cross-sectional area is called resistivity.

• Resistivity is also known as specific resistance.

• \[\Omega \text{m}\] ohm-meter is the \[\text{SI}\] unit for resistivity.

• Metals and alloys have very low resistivity which is in the range of \[{{10}^{-8}}\Omega \text{m to }{{10}^{-6}}\Omega \text{m}\] .

• The resistivity of insulators such as rubber and glass is on the order of \[{{10}^{12}}\Omega \text{m to }{{10}^{17}}\Omega \text{m}\].

• Temperature affects both a material's resistance and its resistivity.

• An alloy's resistivity is usually higher than that of its constituent metals.

• At high temperatures, alloys do not easily oxidise (burn) and hence, they're frequently used in electrical heating devices like electric irons and toasters.

• Tungsten is nearly exclusively utilised in electric bulb filaments, while copper and aluminium are commonly used in electrical transmission lines.

• Reciprocal of resistivity is called conductivity.

• Conductivity can be calculated as;

\[\mho \text{ = }\dfrac{\text{l}}{\text{ }\!\!\Omega\!\!\text{ }}\]

• \[\text{SI}\] unit of conductivity is \[\text{oh}{{\text{m}}^{\text{-1}}}{{\text{m}}^{\text{-1}}}\] or  \[\text{mho-}{{\text{m}}^{\text{-1}}}\]

Effect of Temperature

• A conductor's resistance increases linearly as the temperature rises.

• As the temperature rises, an insulator's resistance rises as well.

• As a semiconductor's temperature rises, its resistivity falls.

• The resistivity of an alloy increases as the temperature rises.

Semiconductors and Superconductors

• Semiconductors are materials with resistivity that fall between those of an insulator and a conductor.

• Materials which lose their resistivity at low temperatures are called superconductors.

Examples:

1. The potential difference between the terminals of an electric heater is \[\text{45 V}\] when it draws a current of \[\text{3 A}\] from the source. What current will the heater draw if the potential difference is increased to \[\text{120 V}\] ?

Ans:

Given data is, 

Potential difference \[\text{V = 45 V}\]and current \[\text{I = 3 A}\]

According to Ohm’s law, 

\[\text{R = }\dfrac{\text{V}}{\text{I}}\]

\[\text{R = }\dfrac{45}{3}\]

\[\text{R = 15 }\Omega \]

When the potential difference is increased to \[\text{120 V}\]the current is

given by

\[\text{I = }\dfrac{\text{V}}{\text{R}}\]

\[\text{I = }\dfrac{120}{15}\]

\[\text{I = 8 A}\]

As a result, the heater's current becomes \[\text{8 A}\].

2. A wire of given material having length \[\text{l}\] and area of cross-section \[\text{A}\]has a resistance of \[\text{8 }\Omega \] . What would be the resistance of another wire of the same material having length \[\dfrac{\text{l}}{4}\] and area of cross-section \[\text{2A}\]?

Ans:

For the first wire, resistance is;

\[{{\text{R}}_{1}}\text{  =   }\!\!\rho\!\!\text{  }\dfrac{\text{l}}{\text{A}}\text{  = 8 }\!\!\Omega\!\!\text{ }\]

Now, for the second wire, resistance is

\[{{\text{R}}_{2}}\text{  =   }\!\!\rho\!\!\text{  }\dfrac{\dfrac{\text{l}}{4}}{\text{2A}}\]

\[{{\text{R}}_{\text{2}}}=\rho \dfrac{\dfrac{1}{4}}{2\text{A}}\]

Since, \[\text{l = }\dfrac{\text{l}}{4}\]

Therefore,

\[{{\text{R}}_{2}}\text{ =  }\!\!\rho\!\!\text{ }\dfrac{\text{l}}{\text{8A}}\]

\[{{\text{R}}_{\text{2}}}\text{ = }\dfrac{\text{l}}{\text{8}}{{\text{R}}_{\text{1}}}\]

Since, \[{{\text{R}}_{1}}=8\Omega \]

Therefore, \[{{\text{R}}_{\text{2}}}\] becomes

\[{{\text{R}}_{\text{2}}}\text{  =  }\dfrac{\text{l}}{\text{8}}\times \text{8}\]

\[{{\text{R}}_{2}}=1\Omega \]

As a result, the new wire's resistance is \[\text{1}\Omega \]

Resistances in Series

The sum of the individual resistances of multiple resistors in series equals their equivalent resistance.

If resistances \[{{\text{R}}_{\text{1}}}\text{, }{{\text{R}}_{\text{2}}}\] and \[{{\text{R}}_{3}}\] are connected in series, then the equivalent resistance can be calculated as,

\[{{\text{R}}_{\text{s}}}\text{ = }{{\text{R}}_{1}}+{{\text{R}}_{2}}+{{\text{R}}_{3}}\]

When numerous resistors are connected in series, the combined resistance \[{{\text{R}}_{\text{s}}}\]equals the total of their individual resistances \[{{\text{R}}_{\text{1}}}\text{, }{{\text{R}}_{\text{2}}},{{\text{R}}_{3}}\] and hence, \[{{\text{R}}_{\text{s}}}\]is greater than any individual resistance.

In a series circuit, the current remains the same and potential differences vary.

• The current in a series circuit is constant throughout the electric circuit. As a result, connecting an electric lamp and an electric heater in series is clearly impractical, as they require currents of vastly different values to function effectively.

• The main disadvantage of a series circuit is that if one component fails, the circuit is broken and none of the other components works.

• Example:

An electric lamp, whose resistance is \[\text{25 }\Omega \] , and a conductor of \[\text{5 }\Omega \]

resistance are connected to a \[\text{9 V}\] battery as shown in below figure. Calculate 

1. The total resistance of the circuit,

2. The current through the circuit.

Ans:

The resistance of electric lamps, 

\[{{\text{R}}_{\text{1}}}\text{ = 25 }\Omega \]

The resistance is connected in series, 

\[{{\text{R}}_{2}}\text{ = 5 }\Omega \]

Then the total resistance in the circuit can be calculated as,

\[{{\text{R}}_{\text{s}}}\text{ = }{{\text{R}}_{1}}+{{\text{R}}_{2}}\]

\[{{\text{R}}_{\text{s}}}\text{ = 25}+5\]

\[{{\text{R}}_{\text{s}}}\text{ = 30 }\Omega \]

The overall difference in potential across the battery is

\[\text{V = 9 V}\]

According to Ohm’s law, the current through the circuit can be calculated as;

\[\text{I = }\dfrac{\text{V}}{{{\text{R}}_{\text{s}}}}\]

\[\text{I = }\dfrac{\text{9}}{\text{30}}\]

\[\text{I = 0}\text{.3 A}\]

Resistances in Parallel

• If resistances \[{{\text{R}}_{\text{1}}}\text{, }{{\text{R}}_{\text{2}}}\] and \[{{\text{R}}_{3}}\] are connected in parallel, then the equivalent resistance \[{{\text{R}}_{\text{p}}}\] is given by

\[\dfrac{\text{1}}{{{\text{R}}_{\text{p}}}}\text{ = }\dfrac{\text{1}}{{{\text{R}}_{\text{1}}}}\text{+}\dfrac{\text{1}}{{{\text{R}}_{2}}}\text{+}\dfrac{\text{1}}{{{\text{R}}_{3}}}\]

• The sum of the reciprocals of the individual resistances is equal to the reciprocal of the equivalent resistance of a group of resistances linked in parallel.

In a parallel circuit, the potential difference remains the same and the current will vary.

The current flowing through the electrical devices is divided in a parallel circuit.

This is especially useful when each device has a different resistance and requires a varied amount of current to function properly.

Example:

In the circuit diagram given in below figure, suppose the resistors \[{{\text{R}}_{\text{1}}}\text{, }{{\text{R}}_{\text{2}}}\] and \[{{\text{R}}_{3}}\] have the values \[2\Omega ,4\Omega ,6\Omega \] respectively, which have been connected to a battery of \[\text{9 V}\]. 

Calculate 

1. The current through each resistor, 

2. The total current in the circuit, and 

3. The total circuit resistance.

Ans:

Given that

\[{{\text{R}}_{1}}\text{ = 2 }\Omega \]

\[{{\text{R}}_{2}}\text{ = 4 }\Omega \]

\[{{\text{R}}_{3}}\text{ = 6 }\Omega \]

The total potential difference across the battery is

\[\text{V = 9 V}\]

According to Ohm’s law,

The current \[{{\text{I}}_{1}}\], through \[{{\text{R}}_{\text{1}}}\] \[\text{= }\dfrac{\text{V}}{{{\text{R}}_{\text{1}}}}\]

\[{{\text{I}}_{1}}\text{ = }\dfrac{\text{V}}{{{\text{R}}_{\text{1}}}}\]

\[{{\text{I}}_{1}}\text{ = }\dfrac{9}{2}\text{ A}\]

\[{{\text{I}}_{1}}\text{ = 4}\text{.5 A}\]

The current \[{{\text{I}}_{2}}\], through \[{{\text{R}}_{2}}\] \[\text{= }\dfrac{\text{V}}{{{\text{R}}_{2}}}\]

\[{{\text{I}}_{2}}\text{ = }\dfrac{\text{V}}{{{\text{R}}_{2}}}\]

\[{{\text{I}}_{2}}\text{ = }\dfrac{9}{4}\text{ A}\]

\[{{\text{I}}_{2}}\text{ = }2.25\text{ A}\]

The current \[{{\text{I}}_{3}}\], through \[{{\text{R}}_{3}}\] \[\text{= }\dfrac{\text{V}}{{{\text{R}}_{3}}}\]

\[{{\text{I}}_{3}}\text{ = }\dfrac{\text{V}}{{{\text{R}}_{3}}}\]

\[{{\text{I}}_{3}}\text{ = }\dfrac{9}{6}\text{ A}\]

\[{{\text{I}}_{3}}\text{ = 1}\text{.5 A}\]

Therefore, the total current in the circuit can be calculated as;

\[\text{I = }{{\text{I}}_{1}}+{{\text{I}}_{2}}+{{\text{I}}_{3}}\]

\[\text{I = }4.5+2.25+1.5\]

\[\text{I = 8}\text{.25 A}\]

The total resistance \[{{\text{R}}_{\text{p}}}\] can be calculated as;

\[\dfrac{\text{1}}{{{\text{R}}_{\text{p}}}}\text{ = }\dfrac{\text{1}}{{{\text{R}}_{\text{1}}}}\text{+}\dfrac{\text{1}}{{{\text{R}}_{2}}}\text{+}\dfrac{\text{1}}{{{\text{R}}_{3}}}\]

\[\dfrac{\text{1}}{{{\text{R}}_{\text{p}}}}\text{ = }\dfrac{\text{1}}{\text{2}}\text{+}\dfrac{\text{1}}{4}\text{+}\dfrac{\text{1}}{6}\]

\[\dfrac{\text{1}}{{{\text{R}}_{\text{p}}}}\text{ = }\dfrac{\text{11}}{\text{12}}\]

Therefore,

\[{{\text{R}}_{\text{p}}}\text{  =  }\dfrac{\text{12}}{\text{11}}\text{  }\!\!\Omega\!\!\text{ }\]

Heating Effect of Electric Current

The source energy is constantly dissipated fully in the form of heat if the electric circuit is purely resistive, that is, a configuration of resistors exclusively connected to a battery, this effect is called as the heating effect of electric current. 

Electric heaters, electric irons, and other gadgets are operating on this effect.

For a steady current \[\text{I}\] , the amount of heat \[\text{H}\] produced in time \[\text{t}\] can be calculated as;

\[\text{H = VIt}\]

Joule’s Law of Heating:

When a current \[\text{I}\]flows through a resistor \[\text{R}\]heat is produced and this phenomenon is called as Joule’s law of heating.

\[\text{H = }{{\text{I}}^{\text{2}}}\text{Rt}\]

Example:

\[\text{120 J}\] of heat is produced each second in a \[\text{6 }\Omega \] resistance. Find the

potential difference across the resistor.

Solution:

Given data is

\[\text{H = 120 J}\]

\[\text{R = 6 }\Omega \]

\[\text{t = 1 sec}\]

According to Joule’s Law of Heating effect, 

\[\text{H = }{{\text{I}}^{\text{2}}}\text{Rt}\]

Therefore,

\[\text{I = }\sqrt{\dfrac{\text{H}}{\text{Rt}}}\]

\[\text{I = }\sqrt{\dfrac{120}{6\times 1}}\]

\[\text{I = }\sqrt{20}\]

\[\text{I = 4}\text{.47 A}\]

Thus the potential difference across the resistor \[\text{V}\]can be calculated as;

\[\text{V = IR}\]

\[\text{V = 4}\text{.47}\times \text{6}\]

\[\text{V = 26}\text{.82 V}\]

Applications of Heating Effect of Electric Current

Heat creation in a conductor is an unavoidable result of electric current.

Heat is undesirable because it converts useful electrical energy into heat.

The electric laundry iron, electric toaster, electric oven, electric kettle, and electric heater are all examples which are working on Joule’s heating effect.

As in an electric bulb, electric heating is also employed to produce light.

Bulb filaments are made of tungsten, a strong metal with a high melting point.

The fuse used in electric circuits is another prominent application of Joule's heating.

Fuse Wire:

• A fuse wire is a wire that melts, breaks the circuit, and protects various appliances in household connections from damage.

• It protects circuits and appliances by interrupting any too-high electric current flow.

• The fuse is connected to the device in series.

• It consists of a piece of wire made of a metal or an alloy with the proper melting point, as well as a piece of wire made of a metal or an alloy with the appropriate melting point.

• Aluminium, copper, iron, and lead alloys are used to make fuse wires.

• The maximum safe current that can pass through the fuse wire increases as the thickness of the wire increases.

Electrical Energy

• Because of the existence of resistance to the flow of current work has to be done in order to maintain the flow of current.

• Since the potential difference \[\text{V}\] is the work done to carry a unit positive charge from infinity to a point, the work done to carry a charge \[\text{q}\] is given by:

\[\text{W = qV}\]

But \[\text{I = }\dfrac{\text{q}}{\text{t}}\]

Therefore,

\[\text{W = ItV}\]

Since \[\text{V=IR}\]

Therefore,

\[\text{W = }{{\text{I}}^{\text{2}}}\text{Rt}\]

\[\text{W = }\dfrac{{{\text{V}}^{\text{2}}}\text{t}}{\text{R}}\]

• This work done is stored as energy.

• \[\text{SI}\] unit of electrical energy is Joule.

Electric Power

• The rate at which electric energy is consumed is called electric power.

• \[\text{Power }\!\!~\!\!\text{  = }\dfrac{\text{Work done}}{\text{Time}}\]

\[\text{P = }\dfrac{\text{W}}{\text{t}}\]

\[\text{P = V  }\!\!\times\!\!\text{  }{{\text{I}}^{\text{2}}}\text{R}\]

\[\text{P = }{{\text{V}}^{\text{2}}}\text{R}\]

\[\text{P = VI}\]

• \[\text{SI}\] unit of electric power is Watt.

• Example:

An electric bulb is connected to a \[\text{230 V}\] generator. The current is \[\text{0}\text{.75 A}\]. What is the power of the bulb?

Solution:

We have

\[\text{V = 230 V}\]

\[\text{I = 0}\text{.75 A}\]

Thus, Power can be calculated as;

\[\text{P = VI}\]

\[\text{P = 230}\times 0.75\]

Therefore,

\[\text{P = 172}\text{.5 W}\]

Calculation of Power for HouseElectricity

• The commercial unit for electrical energy is the kilowatt hour\[\left( \text{kWh} \right)\] 

\[\text{1kWh = 3,600,000 J}\]

\[\text{1kWh}=\text{3}\text{.6}\times \text{1}{{\text{0}}^{\text{6}}}\text{ J}\]

• No. of units of electricity consumed in a household can be calculated as;\[\text{No}\text{. of units of electricity consumed in a household = }\!\!~\!\!\text{ No}\text{. of kWh}\]

• \[\text{Total cost of electricity }\!\!~\!\!\text{  = Total units  }\!\!\times\!\!\text{  Cost per unit of electricity}\]

5 Important Topics of Science Class 10 Chapter 11 You Shouldn’t Miss!

Important formula in Class 10 Science Chapter 11 Electricity

1. Ohm's Law:

 $V = I \times R$
This formula relates voltage (V), current (I), and resistance (R) in an electrical circuit. It means that voltage is equal to the current multiplied by the resistance.

2. Resistance:

$R = \frac{V}{I}$
This formula helps you calculate resistance when you know the voltage and current. Resistance (R) is equal to voltage (V) divided by current (I).

3. Current:

$I = \frac{V}{R}$
This formula calculates the electric current flowing through a circuit, which is equal to voltage (V) divided by resistance (R).

4. Power:

$P = V \times I$
Power (P) in an electrical circuit is the product of voltage (V) and current (I). It measures how much energy is used per unit of time.

5. Electrical Energy:

$E = P \times t$
Electrical energy (E) is calculated by multiplying power (P) by time (t). This formula helps in finding the total energy consumed or used over a period.

6. Series Circuit Resistance:

$R_{total} = R_1 + R_2 + R_3 + \ldots$

In a series circuit, the total resistance ($R_total$) is the sum of all individual resistances ($R_1, R_2, R_3$, etc.).

7. Parallel Circuit Resistance:

$\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \ldots$

In a parallel circuit, the reciprocal of the total resistance ($R_total$) is the sum of the reciprocals of each individual resistance ($R_1, R_2, R_3$, etc.).

Importance of Class 10 Science Chapter 11 Electricity Notes

• Class 10 Physics Electricity Notes simplify complex electrical concepts, making them easier to understand and remember.

• Class 10 Science Chapter 11 Notes help you prepare effectively for exams by focusing on key topics and important formulas.

• Practical examples included in the notes show how electricity is used in real life, making the subject more relatable.

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Tips for Learning the Class 10 Chapter 11 Science Electricity

• Start by understanding the basic concepts of electric current, voltage, and resistance to build a strong foundation.

• Use and practice important formulas like Ohm's Law regularly to become comfortable with solving problems.

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• Teach the concepts to a friend or family member to test your understanding and clarify any doubts.

Conclusion

The Class 10 Science Chapter 11 Notes are important for understanding the topic. They make complex ideas like electric current, voltage, resistance, and Ohm’s Law easier to understand by breaking them down into simpler parts. Electricity Notes Class 10 offers clear explanations and practical examples that connect the theory to real-world applications, making the subject more engaging. They are designed to aid in exam preparation by highlighting key formulas and concepts, allowing for quick and effective revision. Regularly reviewing these notes will help reinforce your learning and boost your confidence. By covering both theoretical concepts and practical applications, Electricity Chapter Class 10 PDF Notes provide a solid foundation in electricity, making your study sessions more productive and less stressful.

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