Class 10 Chapter 12 Science - Electricity NCERT Solutions

Class 10 Chapter 12 Science Electricity NCERT Solutions are important for Class 10 Board Exams. Class 10 Science NCERT solutions of Chapter 12 Electricity are important not only for academic exams but the NCERT solutions of Chapter 12 electricity of the NCERT textbook are also important for competitive exams for collecting general awareness questions based on electricity. After you study chapter 12 electricity you will clear all your doubts on the problems based on electricity. Chapter 12 electricity is part of the physics portion of the class 10 NCERT science textbook.

Download Class 10 Chapter 12 Science - Electricity PDF

Get the detailed solutions for Chapter 3 of Class 10 Science, focusing on Metals and Non-metals. This comprehensive blog is available in a convenient PDF format, allowing you to study offline whenever you need, especially during exams or when facing network issues. Download the Class 10 Science Chapter 3 Solutions PDF today and ensure you have all the answers at your fingertips, anytime, anywhere.

Class 10 Chapter 12 Science - Electricity Introduction

Chapter 12, “Electricity,” is a fundamental part of the Class 10 Science NCERT textbook. It is one of the four physics chapters and is crucial for the CBSE Board exam. Understanding this chapter is essential for students who want to excel in physics as they progress to higher classes.

These NCERT solutions cover important topics like electric circuits, electrical charges, electric current, Ohm’s law, resistivity, conductance, the combination of resistances, the heating effect of electric current, and electric power. Each question is thoroughly explained with the help of circuit diagrams to help you grasp the concepts easily.

NCERT Solutions Class 10 Science from chapter 1 to 16

In Class 10 Science, the NCERT textbook includes four physics chapters:

Chapter 10: Light
Chapter 11: The Human Eye and the Colorful World
Chapter 12: Electricity
Chapter 13: Magnetic Effects of Current

Chapters 12 and 13 are particularly important as they are included in the first term examination according to the CBSE curriculum.

Our expert teacher has crafted these solutions to help you understand each question from Chapter 12 deeply. By studying this content, you’ll be well-prepared to answer questions in your exams.

After reading through these solutions, please feel free to leave a comment. You can also subscribe to our website for free to receive updates directly on your mobile.

Access answers to all in-text and exercise questions for NCERT Class 10 Chapter 12 Science

In-Text Questions 12.1 (Page No 200) – Class 10 Chapter 12 Science

Q1. What does an electric circuit mean?

Ans. An electric circuit continuous closed path that allows the flow of electric current through it, the component of an electric circuit are battery, plug key and conducting wire.

Q2. Define the unit of current.

Ans. The S.I. unit of electric current is ampere (A),1 ampere is the flow of 1-coulomb electric charge per second. The relationship between electric current and time is as follows.

$Q=it\Rightarrow i=\frac{Q}{t}$

Let Q = 1 C, t = 1 s then i= 1 coulomb/second = 1 A, If 1 coulomb charge is flowed for 1 second then the amount of current flowed within the circuit is known as 1 A.

Q3. How many electrons are in 1-coulomb charge?

Ans. The charge in 1e is = $1.6\times 10^{-19}C$

$1.6\times 10^{-19}C$ charge is equivalent to 1e

Then 1 C is equal to = $\frac{1}{1.6\times 10^{-19}C}\times 1C=6.25\times 10^{18}$

Therefore 1-coulomb = $6.25\times 10^{18}e's$

In-Text Questions 12.2 (Page No 202) – Class 10 Chapter 12 Science

Q1. Name a device that helps to maintain a potential difference across a conductor.

Ans. Any source of electricity like a battery or electric generator helps to maintain a potential difference across a conductor.

Q2. What is meant by saying that the potential difference between two points is 1V?

Ans. The relationship between voltage, work done and charge is V = W/Q, if V =1V which means 1 joule of work done by 1 coulom charge from one end to another end of a conductor is known as 1V.

Q3. What will be the energy required to 1C charge in moving from one point to another if the potential difference between the points is 6V?

Ans. The work done is also defined as the transfer of energy, s0 let energy E is required to transfer 1C charge from one point to another

The given potential difference = 6V and charge(Q) = 1C

The relationship between voltage, work done and the charge is

$V =\frac{W}{Q}$

As work done(W) =energy transfer(E)

$V =\frac{E}{Q}$

E = VQ = 6 V × 1C = 6

∴ The energy required to transfer 1 C charge from one point to another is 6 joules.

In-Text Questions 12.5 (Page No 209) – Class 10 Chapter 12 Science

Q1. On what factors does the resistance of a conductor depend?

Ans. Resistance of the conductor depends on

(i) Resistivity of the matter the conductor is made up of

(ii) Length of the conductor

(iii) Area of the cross-section of the conductor

(iv) The type of matter

Q2. Will current flow more easily through a thick wire or thin wire of the same material when connected to the same source? Why?
Ans. The relationship between length, the cross-sectional area of the conductor is given by $R=\rho \frac{l}{A}$ which shows that resistance is inversely proportional to the area,ρ is constant for same kind of material since the area of cross-section of the thicker wire is more so its resistance will be lesser than the thinner wire , since the conductance is inverse of the resistance so current flows more easily through the thicker wire as compared to thinner wire.

Q3. Let the resistance of an electrical component remains constant while the potential difference across two ends of the component decreases to half of its former value. What change will occur in the current through it?

Ans. We are given the resistance of an electrical component(R) →constant, Let initial voltage is = V and current= $i_{1}$

Modified voltage = $\frac{V}{2}$ and let the modified current = $i_{2}$

According to ohm’s law, voltage = current × rasistance

$i_{1}=\frac{V}{R}....\left ( i \right )$

$i_{2}=\frac{V/2}{R}=\frac{V}{2R}....\left ( ii \right )$

Solving (i) and (ii) we have

$\frac{i_{1}}{i_{2}}=\frac{V/R}{V/2R}=\frac{2}{1}$

$i_{2}=\frac{i_{1}}{2}$

Therefore the electric current through the electrical component becomes half of the former value of the electric current.

Q4. Why are the coils of electric toasters and electric irons made of an alloy rather than a pure metal?

Ans. The coils of electric toasters and electric irons made of an alloy because an alloy is differed from metal in the following characteristics.

(i) Higher resistivity compared to the metal

(ii) It is not oxidized easily like metals (iii) The alloy doesn’t melt readily at the higher temperature

Q5. Use the data in Table 12.2 of the NCERT book to answer the following questions :

	Material	Resistivity(Ω m)
Conductors	Silver	$1.60\times 10^{-8}$
	Copper	$1.62\times 10^{-8}$
	Aluminum	$2.63\times 10^{-8}$
	Tungsten	$5.20\times 10^{-8}$
	Nickel	$6.84\times 10^{-8}$
	Iron	$10.0\times 10^{-8}$
	Chromium	$12.9\times 10^{-8}$
	Mercury	$94.0\times 10^{-8}$
	Manganese	$1.84\times 10^{-6}$
Alloys	Constantan (Alloy of Cu and Ni)	$49\times 10^{-6}$
	Manganin (Alloy of Cu, Mn, and Ni)	$44\times 10^{-6}$
	Nichrome (Alloy of Ni, Cr, Mn, and Fe)	$100\times 10^{-6}$
Insulators	Glass	$10^{10}-10^{14}$
	Hard rubber	$10^{13}-10^{16}$
	Ebonite	$10^{15}-10^{17}$
	Diamond	$10^{12}-10^{13}$
	Paper (Dry)	$10^{12}$

(a) which among iron and mercury is a better conductor?

(b) Which material is the best conductor?

Ans.

(a) As seen in the table the resistivity of both elements is given below:

Iron: $10.0\times 10^{-8}$ Ωm

Mercury: $94.0\times 10^{-8}$ Ωm

The resistivity of mercury is more than the iron and as we know the conductivity is inverse of resistivity so iron is a better conductor than mercury.

(b) From the above table, silver has the lowest resistivity which implies that it is the best conductor.

In-Text Questions 12.6 (Page No 213) – Class 10 Chapter 12 Science

Q1. Draw a schematic diagram of a circuit consisting of a battery of three cells of 2V each, a 5Ω resistor, a 8 Ω resistor and a 12 Ω resistor and a plug key, all connected in series.

Ans.

Q2. Redraw the circuit of the above question, putting in an ammeter to measure the current through the resistors and a voltameter to measure the voltage across the 12Ω resistor. What would be the reading in the ammeter and the voltameter?

Ans. Total voltage = 2×3 = 6V

Total resistance of the circuit = 5 + 8 + 12 = 25Ω

Using the Ohm’s law V = IR⇒ I = V/R = Voltage/Total resistance

Total current = V/R = 6/25 = 0.24A

Hence voltage across the12Ω resistor = I XR= 0.24 X 12 =2.88V

So, the reading of the ammeter =0.24 A and the reading of the voltameter = 2.88Ω.

In-Text Questions 12.7 (Page No 218) – Class 10 Chapter 12 Science

Q1.Why does the cord of an electric heater not glow while the heating element does ?

Ans. The cord of an electric heater is made up of metallic wire such as copper or aluminum which has low resistance while the heating element is made up of an alloy (nichrome) which has more resistance than its constituent metals.As we know heat produced(H) is H = I²Rt, where R is the resistance of the conductor ,so heating element glows while the cord of heater does not.

Q2. Compute the heat generated while transferring 96000C of charge in one hour through a potential difference of 50 V.

Ans. q =96000 C, V = 50V, t = 1h

H = I²Rt = VIt = vq

= 50 x 96000

=48 x 10⁵J

Q3.An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s.
Ans. R = 20Ω, I = 5A, t = 30 s

H = I²Rt

= 5² x 20 x 30

= 15000J

= 1.5 X 10⁴J

In-Text Questions 12.8 (Page No 220) – Class 10 Chapter 12 Science

Q1.What determines the rate at which energy is delivered by a current?

Ans. Electric power determines the rate at which energy is released by a current.

Q2- An electric motor takes 5A from a 220 V line. Determine the power of the motor and the energy consumed in 2h.

Ans. I = 5A. V = 220 V, t = 2h

P = VI = 220 X 5 = 1100 W

Energy consumed = Pt = 1100 x 2 = 2200wh

Exercise Questions – Page 221 (Class 10 Chapter 12 Science)

Q1. 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

Ans. After cutting the wire in five equal parts each of piece will have the R/5 resistance, When connected in parallel their equivalent resistance R’ will be as following.

1/R’ = 1/(R/5) + 1/(R/5) + 1/(R/5) + 1/(R/5) + 1/(R/5)

1/R’ = (5/R)x 5 = 25/R

R/R’ = 25

Class 10 Chapter 12 Science: Solutions for Questions 1 and 2 (Video)

Q3. 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

Ans. (d) It is the resistance of bulb which will remain constant so for calculating it, we have

Power = Voltage × Current

P = V × I = V × V/R = V²/R ( Ohm’s law I = V/R)

R = V²/P = 220²/100

So power will be consumed at 110 V

P = V²/R = 110²/(220²/100)

= (110²/220²) ×100

= 25 W

Q4. Two conducting wires of the same material and of equal lengths and 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

Ans. When both the wire connected in series their equivalent resistance = R + R = 2R

Let the heat produced when wires are connected in series and in parallel is H1 and H2 respectively.

Heat = Power × time = Voltage × Current × time = Voltage × (Voltage/Resistance)×time

So, heat consumed, H₁ = V²t/R = V²t/2R

When both are connected in parallel, the equivalent resistance will be as following.

1/R’ = 1/R + 1/R

1/R’ = 2/R

R’ = R/2

In case of parallel connection heat produced , H₂ = V²t/(R/2)= 2V²t/R

Therefore

H₁/H₂ = (V²t/2R) X(R/2V²t)

= 1/4

H₁ : H₂ = 1 : 4

Q5. How is the voltameter connected in the circuit to measure the potential difference between two points ?

Ans. A voltmeter is connected in parallel across any two points in a circuit to measure the potential difference between them with its positive terminal to the point of higher potential and negative terminal to the point showing lower potential of the source.

Q6. A copper wire has a diameter 0.5 mm and a resistivity of 1.6 x 10^-8 Ωm. What will be the length of this wire to make its resistance 10Ω? How much does the resistance change if the diameter is doubled?

Ans. We are given the diameter of copper wire , d = 0,5 mm ⇒ $r=\frac{0.5}{2}\times 10^{-3}$

$\rho =1.6\times 10^{-8}$ (Resistivity of the wire) and resistance, R = 10Ω

Using the following relationship

$R =\rho \frac{L}{A}$

$\because A=\pi r^{2}$

$\therefore L=R\times \frac{\pi r^{2}}{\rho }$

$L=10\times \frac{3.14\times \left [ \left ( 0.5/2 \right )\times 10^{-3} \right ]^{2}}{1.6\times 10^{-8} }$

= 122.6 m ≈ 123 m

Let the resistance of copper wire become R’ when its diameter is changed

Diameter is doubled, the radius of copper wire is also doubled, applying the following equation

$R =\rho \frac{L}{A}$

$R=\rho \times \frac{L}{\pi r^{2}}.....\left ( i \right )$

Modified radius = 2r and let modified resistance, R’

$R'=\rho \times \frac{L}{\pi \left ( 2r \right )^{2}}=\rho \times \frac{L}{4\pi r^{2}}.....\left ( ii \right )$

From (i) and (ii) we have

$\frac{R}{R'}=\frac{4}{1}$

$R'=\frac{R}{4}$

Hence resistance will become one-fourth of the original resistance when the diameter of wire is doubled.

Q7. The values of the 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.

Ans. Plotting the voltage on Y-axis and current on X-axis

According to Ohm’s law, Resistance (R) =Δ Voltage/ΔCurrent = AB/BC

$R=\frac{AB}{BC}=\frac{13.2-6.7}{4-2}=\frac{6.5}{2}=3.25$

Therefore the resistance of the wire = 3.25 Ω

Q8. 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.

Ans. Applying Ohm’s law, voltage = current × resistance

V = IR

R = V/I, 1 A = 1000 mA, so 2.5 mA = 0.0025 A

R = 12 /0.0025 = 4800

Therefore the resistance of the resistor = 4800 Ω

Q9. A battery of 9V is connected in series with resistors of 0.2 Ω , 0.3Ω , 0.4Ω , 0.5Ω and 12 Ω respectively. How much current would flow through the 12Ω resistor?

Ans. Net resistance, R_T = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4Ω, Total voltage in the circuit,V = 9V

Applying the Ohm’s law, Voltage = current × resistance

I = V/R = 9/13.4 = 0.67A

Since all the resistors are in series, the same current, 0.67 A flows through the 12 Ω resistor

Q10. How many 176Ω resistors ( in parallel) are required to carry 5A on a 220 line ?

Ans. Let the n resistors are connected in parallel, so the net resistance $R_{T}$

1/R_T = 1/R + 1/R + 1/R ……1/R_n

1/R_T = n × 1/R

R_T = R /n

Applying the Ohm’s law, Current, I = V/ R_T = V/R/n = nV/R

n = R I /V = 176 x 5 / 220 = 4

Therefore the required number of resistors are =4

Q11. Show how you would connect three resistors, each of resistance 6Ω so that the combination has a resistance of

(i) 9Ω

(ii) 4Ω

Ans.

(i) When two 6 Ω resistances are in parallel and the third is connected in series, then the equivalent resistance will be 9Ω

1/R_P = (1/6 + 1/6) ⇒ R_P = 3Ω

6Ω in series , then net resistance,R_T = 3Ω + 6Ω = 9Ω

(ii) When two 6Ω resistors are in series and third is in parallel to them ,then

R_S = 6 + 6 = 12Ω

1/R_T = 1/12 + 1/6⇒ R_T = 4Ω

Q12. 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 5A?
Ans. Let n bulbs each of power 10 w and of resistance R are connected in parallel

Power of each bulb, P = V × I = V × V/R = V²/R= 220²/R (where R is net resistance of n bulbs)

P = 220²/R …….. (i)

Net resistance of the circuit, R’ = V/I = 220/5 = 44 Ω

$\frac{1}{R'}=\frac{1}{R}\times n$

R = nR’ = 44n……(ii)

Putting the value of R in eq.(i)

p = 220²/44n

We are given the value of P= 10 W

10 = 220²/44n ⇒ n = (220 × 220)/440 = 110

Hence the 110 bulbs can be connected in parallel

Q14. Compare the power used in the 2Ω resistor in each of the following circuits.

(i) A 6 V battery in series with 1Ω and 2Ω resistors, and

(ii) A 4 V battery in parallel with 12Ω and 2Ω resistors

Ans.

(i) We are given

Voltage, V = 6V, Net resistance of the circuit, R_T = 1 + 2 = 3Ω

Applying Ohm’s law

I = V/R = 6/3 = 2A

current flow through the 2Ω resistor = 2A

So, the power used in 2Ω, P_S = i²R = 2² X 2 = 8 W

(ii)12 Ω and 2Ω are connected in parallel to 4V battery, so the voltage across each one will be= 4V

R = 2 Ω, Voltage across 2 Ω resistor, V = 4V

Power used across 2Ω, P_P = V²/R = 4 ×4/2 = 8 W

Power consumed through 2Ω resistor in each case will be same 8W

Q15. Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to the electric mains supply. What current is drawn from the line if the supply voltage is 220 V?

Ans. We are given, the voltage, V = 220 V, Power of both lamps are 100 W and 60 W, Let the resistance across 100W lamp is and across 60 W lamp is $R_{60}$ .

P = V²/R₁₀₀

100= 220²/R₁₀₀

R₁₀₀ = 22 × 22 = 484 Ω

Similarly R₆₀ = 220²/60 = 4840/6 Ω

Current drawn by 100 W bulb = 220/484 = 0.45 A( Ohm’s law, i= V/R)

Current drawn by 60 W bulb = (220/4840) x 6 = 0.27A

Q16.Which uses more energy,a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?

Ans. Energy consumed by 250 W TV set in 1 hr=Power × time = 250 x 1 = 250 Wh.

Energy consumed by 1200 W toaster in 10 min =Power × time= 1200 × 10/60 =200 Wh(10 min=10/60 hr

So , energy consumed by TV set is more than the energy consumed by toaster in the given timings.

Q17. An electric heater of resistance 8Ω draws 15 A current from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.

Ans. We are given the resistance of heater, R = 8Ω, current through the resistance, i = 15 A, Time, t= 2 hours

Rate of heat developed = H/t = I²Rt/t

= 15² x 8

= 225 x 8

=1800 J/s

Q18. Explain the following:

(a) Why is the tungsten used almost exclusively for filament of electric lamps?

(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?

(c) Why is the series arrangement not used for domestic circuits?

(d) How does the resistance of a wire vary with its area of cross-section?

(e) Why are copper and aluminium wires usually employed for electricity transmission?

Ans.

(a) Tungsten has a high melting point and has the virtue of emitting light at high temperatures.

(b) An alloy has more resistivity, high melting point and it does not oxidize like the metals.

(c) In series connection, all the appliances have a different potential difference as per their resistances.

(d) The resistance of wire is inversely proportional to its area of the cross-cross section.

(e) Aluminum and copper has a lower resistivity, which makes them the best conductors.

Summary of Chapter 12 – Electricity (Class 10 Science)

Electric charge: Electric charge is the property of the matter,it is of two kinds positive and negative. The negative charge is constituted by the electrons revolving around the nucleus of the atom and the positive charge constituted by the protons at the nucleus of the atom. In metal outermost electrons i.e valance electrons are loosely packed, these electrons are free to move from one atom to another atom throughout the body of metal, when a potential difference is produced across the metallic conductor in a closed circuit, these free electrons move from one end of the conductor to another end. The electric charge flows from one end of the conductor to another end intermittently and it has been observed by the scientist that the amount of flowing charge is the integral value of the charge in an electron. The charge is represented by the following equation.

Q = ne

Where Q is the charge, n= number of electrons and e= charge in one electron

If Q = 1 coulomb, e = 1.6 × 10^-19C

So, n ≈ 6×10¹⁸

We can say that the flow of a group of 6×10¹⁸electrons constitutes 1 coulomb of charge

Electric Current: The rate of flow of current is known as current,it is measured in ampere.

$\boldsymbol{i=\frac{Q}{t}}$

Where Q is the total charge flowed in the circuit in certain time, t is time taken by the flowing charge from one end of the circuit to another end and i is amount of current.

Ohm’s Law: A German scientist Jeorge Simon Ohm observed that flow of electric current is proportional to potential difference between across a conductor.

$\boldsymbol{V\propto i}$

V = iR

Where R is the constant of proportionality known as Resistance of the circuit, resistance is the property of the conductor opposing the electric current, it is measured in Ohm(Ω).

Voltage: Voltage is a force known as the electromotive force that compels the electrons to flow from the negative terminal to the positive terminal. The voltage in a battery is produced by the chemical reaction, the electrons produced in chemical reactions attracted by the positive terminal of the battery and repelled by the negative terminal of the battery, this resulted in an electromotive force developed across the battery, when the battery is connected in an electric circuit this e.m.f causes the flow of electrons from one end of the circuit to another end. The amount of voltage is measured in Volt.

Voltage across a conductor is defined as the work done by a unit charge in flowing from one end to another end.It is given by

$\boldsymbol{V=\frac{W}{Q}}$

Where W is work done or electrical energy transferred from one end of the conductor to another another end and Q is total amount of the charge flowed from one end to another end, V is voltage or potential difference

The factors affecting the resistance of a conductor: The resistance of the conductor is affected by its length and cross-sectional area. The resistance of a conductor is proportional to its length and inversely proportional to its cross-sectional area. The resistance of a thicker wire of a certain length is less than the thinner wire of the same length and the resistance of a long wire of a certain diameter is larger than the shorter wire of the same diameter.

Specific resistance (Resistivity): Resistivity is the property of a substance that opposes the flow of current in per unit length of the substance. The resistivity of a substance is constant in a fixed temperature. The resistivity of all substances are different.

The resistivity is given by the following formula.

$\boldsymbol{\rho =\frac{RA}{l}}$

Where $\boldsymbol{\rho }$ is the resistivity of a substance, A is the crossectional area of the conductor, l is the length of the conductor.

The combination of electric resistance: In an electric circuit a group of resistors can be combined in two ways because the resistance is affected by the length of the conductor and by the crossectional area of the conductor.

Series Connection: The resistance is proportional to the length of the conductor,so in series connection, the positive terminal of one resistor is connected to the negative terminal of another resistance and the net resistance of the circuit is the sum of individual resistances in a straight forward way. Let there are 3 resistors connected in series as follows.

Let the net resistance of the circuit is $\boldsymbol{R_{T}}$

So,

$\boldsymbol{R_{T}=R_{1}+R_{2}+R_{3}}$

Parallel Connection: Since the resistance of the circuit is inversely proportional to the cross sectional area of the conductor, so In this type of connection positive terminals of all the resistors are connected to the positive terminal of the battery and the negative terminals of all the resistors are connected to the negative terminal of the battery, the inverse of net resistance is termed as the sum of the inverse of individual resistance. Let 3 resistors are connected in parallel connection as follows.

Let the net resistance of three resistors is $\boldsymbol{R_{T}}$

$\boldsymbol{\frac{1}{R_{T}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}}$

Conductance and Conductivity: Conductance is the opposite to resistance,it is the property of a conductor that allows the flow of electric current through it. Conductivity is the property of the substance that allows the flow of electric current in per unit length of the substance.

Heating effect of electric current: The work done by unit charge from one end of the conductor to another end is dissipated in the form of heat.

$\boldsymbol{V=\frac{W}{Q}}$

Replacing W by H(heat dissipated)

$\boldsymbol{V=\frac{energy\: dissipated}{Q}}$

$\boldsymbol{V=\frac{H}{Q}}$

H = VQ

We know Q = it

H = Vit

From Ohm’s law, V = iR

H = iR.it = i²Rt

H = i²Rt

The above equation is known as Joul’s equation of electric heat

Electric Power: The rate at which the electrical energy is dissipated is known as electric power,it is measured in Watt.

$\boldsymbol{P=\frac{W}{t}=\frac{H}{t}}$

$\boldsymbol{P=\frac{i^{2}Rt}{t}}$

P= i²R

P= iR.i = V×i

Where P is electric power, Voltage across the conductor and i is electric current.

The commercial unit of power is a kilowatt

1KWh = 1000 W

The commercial unit of electrical energy is Kilowatt-hour that is equivalent in joules as follows.

$\boldsymbol{1KWh=3.6\times 10^{6}J}$

$\boldsymbol{1Wh=3600J}$