NCERT solutions of class 10 science chapter 12- Electricity - Future Study Point

NCERT solutions of class 10 science chapter 12- Electricity

Electricity class 10

 NCERT solutions of class 10 science chapter 12- Electricity

Class 10 science chapter 12 electricity NCERT solutions are important for class 10 students for their preparation of SA-1 and SA-2 exams. class 10 science NCERT solutions of chapter 12 electricity is important not only for academic exams but the NCERT solutions of chapter 12 electricity of the NCERT textbook is also very 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   class 10 NCERT science textbook.

Electricity class 10

Class 10  science chapter 12 electricity NCERT solutions

NCERT Solutions for Class 10 Science chapter 12 –electricity are the solutions of chapter -12 the electricity of NCERT  textbook which is a basic chapter of physics science NCERT textbook. There are four physics chapters in class 10 science NCERT text book , chapter 12 electricity is the most important for the CBSE Board exam,the concept of chapter 12 electricity is mandatory to clear for upgrading to higher class physics.

These NCERT solutions have questions-related to an electric circuit, electrical charge, electric current, Ohm’s law, resistivity and conductance, the combination of electric resistance, the heating effect of electric current, electric power, all the questions of NCERT solutions are illustrated here with the help of circuit diagram, these well explained NCERT solutions of the textbook unsolved questions will guide students in getting acquainted with the concepts.

NCERT Solutions Class 10 Science from chapter 1 to 16

The class 10 science NCERT book contains 4 chapters of physics chapter 10 light, chapter 11 the human eye and colorful world, chapter 12 electricity and chapter 13 magnetic effect of current, according to CBSE curriculum the chapter 12 and chapter 13 are two lessons part of first term examination. Here 10 class science NCERT solutions of chapter 12 electricity is written by an expert teacher for your help. Each question within the pages and of back exercise is explained readily so that you could understand the depth of the subject scientifically. After you go through the whole post of 10 class science NCERT solution of chapter electricity you will become able to write your answer correctly in your upcoming exams. At the end of this post-NCERT solution to lesson 12 electricity don’t forget to write your comment. You can subscribe to our website at free of cost for updated information on your mobile.

See the Important Numerical Questions –Class 10  Science Chapter 12 Electricity CBSE Board Exam-2023-24

NCERT solutions of class 10 science chapter 12- Electricity

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PDF-Class X science best explained NCERT solutions of the chapter 12

Page no. 200

Q1.    What does an electric circuit mean?

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

2- Define the unit of current.

Ans. 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.

Let Q = 1 C, t = 1 s then i= 1 coloumb/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 =

charge is equivalent to  1e

Then 1 C is equal to =   

Therefore 1-coulomb =

Page no. 202

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 potential difference between two point 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

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

E = VQ = 6 V × 1C = 6

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

   PAGE NO. 209

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

Ans.    Resistance of 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     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=

Modified voltage  =    and let the modified current =

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

Solving (i) and (ii) we have

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 :

 

MaterialResistivity(Ω m)
ConductorsSilver
Copper
Aluminum
Tungsten
Nickel
Iron
Chromium
Mercury
Manganese
AlloysConstantan

(Alloy of  Cu and Ni)

Manganin

(Alloy of Cu, Mn, and Ni)

Nichrome

(Alloy of Ni, Cr, Mn, and Fe)

InsulatorsGlass
  Hard rubber
Ebonite
Diamond
Paper (Dry)

 

(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 are given bellow

Iron-  Ωm

Mercury – 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.

 

Page 213

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.

Page 213 Q1 electricity

 

.

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.

 

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

page 213 Q2 electricity

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

            Page-216

Q1.Judge the equivalent resistance when the following resistors are connected in parallel.

(a) 1Ω and 106

(b) 1Ω, 10³Ω, and 106

(a)   When both 1Ω and 106Ω are connected in parallel then net resistance will be less than the lower resistance 1Ω.

Let the equivalent resistor of both the resistors is = R

Therefore equivalent resistor ≈ 1 Ω

(b) Let the equivalent resistor is R

Therefore equivalent resistance is 0.999 Ω

The equivalent resistance in both cases is slightly less than the minimum individual resistance.

 

Q2. An electric lamp of  100 Ω, a toaster of resistance 50 Ω and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances and what is the current flows through it?


Ans. Resistance of electric lamp R = 100 Ω, resistance of toaster R2 = 50 Ω, resistance of water filter R3 = 500Ω

Voltage = 220 V

 

Therefore  net resistance R will be as following

R =500/16

Applying the Ohm’s law V = IR, where I is total current flowing through the circuit

page 216 Q2 electricity

Hence total current flowing through the circuit =7.04 A

Let the resistance of electric iron = , current flowing through it is (I)= 7.04 A, Voltage across it(V) = 220 V

Applying Ohm’s law

Hence the resistance of  electric iron = 31.25Ω and the current  will flow through it = 7.04 A

Q3. What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?

Ans. The advantages of connecting electrical devices in parallel with the battery instead of connecting them in series are the following.

(i) The net resistance of the circuit is reduced when all the appliances are connected in parallel and thus less electrical energy is consumed.

(ii) In a parallel connection, if one appliance is disconnected others are unaffected while in series if one appliance is disconnected others are also disconnected.            


Q4.How can three resistors of resistances 2Ω, 3Ω and 6Ω be connected to give a total resistance of (a) 4Ω (b) 1Ω?


Ans.(a) In order to get the equivalent resistance of 4 Ω, 3Ω and 6Ω resistor should be connected parallel and 2Ω resistor in series

(b) In order to get the equivalent resistance of 2Ω, 3Ω and 6Ω should be connected in parallel

adjustment of resistors

 

Q5. What is  (a) the highest (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8Ω, 12 Ω and 24  Ω respectively.

Ans- (a) The highest resistance is secured by combining all four coils of resistance in series

R = 4Ω + 8Ω +12Ω +24Ω = 48 Ω

(b) The lowest resistance is secured by combining all four coils of resistance in parallel.

1/RP = ¼ + 1/8 + 1/12 + 1/24

R = 24/12

RP = 2Ω

Therefore highest and lowest resistance can be secured by the combination of the given coils of resistance are 48 Ω and 2 Ω respectively.

page no 218

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

 

Page no. 220

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

 

                          Textbook questions

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

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

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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 ⇒   

           (Resistivity of the wire) and resistance, R = 10Ω

Using the following relationship

= 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

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

From (i) and (ii) we have

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)0.51.02.03.04.0
V (volts)1.63.46.710.213.2

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

graph electricity class 10

 

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

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, RT = 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 

1/RT =  1/R + 1/R + 1/R ……1/Rn

1/RT = n × 1/R

RT  = R /n

Applying the Ohm’s law, Current, I  = V/ RT = 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.

Q11 electricity back ex

 

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

1/RP = (1/6 + 1/6) ⇒ RP = 3Ω

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

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

RS = 6 + 6  = 12Ω

1/RT = 1/12 + 1/6⇒ RT = 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

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, RT = 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, PS = 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, PP = 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 .

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.

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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

18-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 aluminum 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 resistanceof 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.

Summery of the chapter 12-Electricity where from these NCERT questions are taken

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 6×1018 electrons constitutes 1 coulomb of charge

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

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.

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

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.

Where 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.

series connection of resistors

 

Let the net resistance of the circuit is

So,

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.

parallel connection resistors

 

Let the net resistance of three resistors is

 

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.

Replacing W by H(heat dissipated)

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.

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.

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NCERT Solutions of Science and Maths for Class 9,10,11 and 12

NCERT Solutions for Class 10 Science

Chapter 1- Chemical reactions and equationsChapter 9- Heredity and Evolution
Chapter 2- Acid, Base and SaltChapter 10- Light reflection and refraction
Chapter 3- Metals and Non-MetalsChapter 11- Human eye and colorful world
Chapter 4- Carbon and its CompoundsChapter 12- Electricity
Chapter 5-Periodic classification of elementsChapter 13-Magnetic effect of electric current
Chapter 6- Life ProcessChapter 14-Sources of Energy
Chapter 7-Control and CoordinationChapter 15-Environment
Chapter 8- How do organisms reproduce?Chapter 16-Management of Natural Resources

NCERT Solutions for class 9 maths

Chapter 1- Number SystemChapter 9-Areas of parallelogram and triangles
Chapter 2-PolynomialChapter 10-Circles
Chapter 3- Coordinate GeometryChapter 11-Construction
Chapter 4- Linear equations in two variablesChapter 12-Heron’s Formula
Chapter 5- Introduction to Euclid’s GeometryChapter 13-Surface Areas and Volumes
Chapter 6-Lines and AnglesChapter 14-Statistics
Chapter 7-TrianglesChapter 15-Probability
Chapter 8- Quadrilateral

NCERT Solutions for class 9 science 

Chapter 1-Matter in our surroundingsChapter 9- Force and laws of motion
Chapter 2-Is matter around us pure?Chapter 10- Gravitation
Chapter3- Atoms and MoleculesChapter 11- Work and Energy
Chapter 4-Structure of the AtomChapter 12- Sound
Chapter 5-Fundamental unit of lifeChapter 13-Why do we fall ill ?
Chapter 6- TissuesChapter 14- Natural Resources
Chapter 7- Diversity in living organismChapter 15-Improvement in food resources
Chapter 8- MotionLast years question papers & sample papers

NCERT Solutions for class 10 maths

Chapter 1-Real numberChapter 9-Some application of Trigonometry
Chapter 2-PolynomialChapter 10-Circles
Chapter 3-Linear equationsChapter 11- Construction
Chapter 4- Quadratic equationsChapter 12-Area related to circle
Chapter 5-Arithmetic ProgressionChapter 13-Surface areas and Volume
Chapter 6-TriangleChapter 14-Statistics
Chapter 7- Co-ordinate geometryChapter 15-Probability
Chapter 8-Trigonometry

CBSE Class 10-Question paper of maths 2021 with solutions

CBSE Class 10-Half yearly question paper of maths 2020 with solutions

CBSE Class 10 -Question paper of maths 2020 with solutions

CBSE Class 10-Question paper of maths 2019 with solutions

NCERT Solutions for Class 10 Science

Chapter 1- Chemical reactions and equationsChapter 9- Heredity and Evolution
Chapter 2- Acid, Base and SaltChapter 10- Light reflection and refraction
Chapter 3- Metals and Non-MetalsChapter 11- Human eye and colorful world
Chapter 4- Carbon and its CompoundsChapter 12- Electricity
Chapter 5-Periodic classification of elementsChapter 13-Magnetic effect of electric current
Chapter 6- Life ProcessChapter 14-Sources of Energy
Chapter 7-Control and CoordinationChapter 15-Environment
Chapter 8- How do organisms reproduce?Chapter 16-Management of Natural Resources

NCERT Solutions for class 11 maths

Chapter 1-SetsChapter 9-Sequences and Series
Chapter 2- Relations and functionsChapter 10- Straight Lines
Chapter 3- TrigonometryChapter 11-Conic Sections
Chapter 4-Principle of mathematical inductionChapter 12-Introduction to three Dimensional Geometry
Chapter 5-Complex numbersChapter 13- Limits and Derivatives
Chapter 6- Linear InequalitiesChapter 14-Mathematical Reasoning
Chapter 7- Permutations and CombinationsChapter 15- Statistics
Chapter 8- Binomial Theorem Chapter 16- Probability

CBSE Class 11-Question paper of maths 2015

CBSE Class 11 – Second unit test of maths 2021 with solutions

NCERT Solutions for Class 11 Physics

Chapter 1- Physical World

chapter 3-Motion in a Straight Line

NCERT Solutions for Class 11 Chemistry

Chapter 1-Some basic concepts of chemistry

Chapter 2- Structure of Atom

NCERT Solutions for Class 11 Biology

Chapter 1 -Living World

NCERT solutions for class 12 maths

Chapter 1-Relations and FunctionsChapter 9-Differential Equations
Chapter 2-Inverse Trigonometric FunctionsChapter 10-Vector Algebra
Chapter 3-MatricesChapter 11 – Three Dimensional Geometry
Chapter 4-DeterminantsChapter 12-Linear Programming
Chapter 5- Continuity and DifferentiabilityChapter 13-Probability
Chapter 6- Application of DerivationCBSE Class 12- Question paper of maths 2021 with solutions
Chapter 7- Integrals
Chapter 8-Application of Integrals

Class 12 Solutions of Maths Latest Sample Paper Published by CBSE for 2021-22 Term 2

Class 12 Maths Important Questions-Application of Integrals

Class 12 Maths Important questions on Chapter 7 Integral with Solutions for term 2 CBSE Board 2021-22

Solutions of Class 12 Maths Question Paper of Preboard -2 Exam Term-2 CBSE Board 2021-22

Solutions of class 12  maths question paper 2021 preboard exam CBSE Solution

 

 

 

 

 

 

 

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