Class 10 Science Chapter 12 Electricity

Class 10 Science Chapter 12 Electricity answer to each chapter is provided in the list so that you can easily browse throughout different chapters NCERT Class 10 Science Chapter 12 Electricity and select need one.

Class 10 Science Chapter 12 Electricity

Join Telegram channel

Also, you can read the SCERT book online in these sections Solutions by Expert Teachers as per SCERT (CBSE) Book guidelines. These solutions are part of SCERT All Subject Solutions. Here we have given Assam Board Class 10 Science Chapter 12 Electricity Solutions for All Subjects, You can practice these here.


Chapter – 12

Textual Questions and Answers:

Page – 200 

Q.1. What does an electric circuit mean? 

Ans: The closed-loop or path of electrical components in which electrons are able to flow. This path consists of electrical components source, like a battery, resistance, wire, bulb, etc.

Electric circuits are useful in many ways like all of our electrical appliances run by a particular type of circuit.

Q.2. Define the unit of current? 

Ans: SI unit of electric current is ampere. Current is said to be 1 ampere (1A), if 1 coulomb charge flows per second across a cross section of a conductor.

i.e. 1 amp = 1 coulomb/1 second

Q.3. Calculate the number of electrons constituting one coulomb of charge. 

Ans: Charge on one electron (e) = 1.6×10⁻¹⁹C 

                           Total charge Q  = 1C                                         

      ∴     Number of electrons,  n  = Q/C

                                                     = 1C/1.6×10⁻¹⁹C

                                                     = 6.25×10¹⁸

Page – 202 

Q.1. Name a device that helps to maintain a potential duo difference across a conductor. 

Ans: Battery. 

Q.2. What is meant by saying that the potential difference between two points is 1V? 

Ans: The potential difference between two points is said to be 1 volt if 1 joule of work is done in moving 1 coulomb of electric charge from one point to the other.

Therefore 1 volt = 1 joule/1 coulomb

Q.3. How much energy is given to each coulomb of charge passing through a 6v battery? 

Ans: Required energy = Charge x potential difference 

                                    = 1c x 6v 

                                    = 6 Joule

Page – 209 

Q.1. On what factors does the resistance of a conductor depend? 

Ans: Resistance of a conductor depends upon:

  • Length of the wire
  • Area of cross-section of the wire
  • Nature of material of the wire
  • Physical conditions like temperature.

Q.2. Will current flow more easily through a thick wire or a thin wire the same material. When connected to the same source? Why? 

Ans: According to the ohm’s law of electrical resistance,we came to know that the thickness of the conductive wire is inversely proportional with the amount of resistance. (because,the amount of conducting electrons are lesser in amount in the thin wire,due to the less conductive surface) So,a certain amount of electricity will easily flow through a thick wire rather than the thin wire made with the same material.

Q.3. Let the resistance of an electrical component remains constant while the potential difference across the two Isikn ends of the component decreases to half of its for men value. What change will occur in the current through it? 

Ans: Let the resistance of an electrical component remain constant while the potential difference across the two ends of the component decreases to half of its former value. The current through the component will become half.

According to Ohm’s law, “The current flowing through a component is directly proportional to the potential difference between its terminals, i.e., 

V ∝ I

V ∝ I.” If the potential difference increases, the current will increase and vice-versa. Here, the potential difference across the two ends of the component decreases to half. Hence, the current will also decrease to half of its former value.

Q.4. Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?

Ans: The resistivity of an alloy is higher than the pure metal and it does not corrode easily. Moreover, even at high temperatures, the alloys do not melt readily. Hence, the coils of heating appliances such as electric toasters and electric irons are made of an alloy rather than a pure metal.

Q.5. Use the date in Table 12.2 to answer the following- 

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

Ans: Resistivity of iron = 10.0 x 10⁻⁸Ωm 

Thus iron is a better conductor because it has lower resistivity than mercury.

(b) Which material is the best conductor? 

Ans: As silver has the lowest resistivity, so silver is the best conductor.

Page – 213 

Q.1. Draw a schematic diagram of a circuit consisting of a battery of three cells of 2v each, a 5Ω resistor, and 8Ω resistor, and a 12Ω resistor, and a plug key, all connected in series. 

Ans: The circuit diagram is given below: 

Q.2. Redraw the circuit of questions 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12Ω resistor. What would be the readings in the ammeter and voltmeter? 

Ans: The circuit diagram is given below:

Total voltage,

                           V = 3×2 = 6v 

Total resistance, 

                          R = 5+8+12 = 25 Ω

Reading of ammeter, 

                                  I = V/R

                                    = 6/25

                                    = 0.24A

Reading of voltmeter, 

                                V =  I.R 

                                   =  0.24×12

                                   =  2.88V

Page – 216 

Q.1. Judge the equivalent resistance when the following are connected in parallel – 

(a) 1Ω and 16⁶Ω

(b) 1Ω and 10³Ω and 10⁶ Ω

Ans: (a) Given that, 

                             R₁ = 1Ω

                             R₂ = 10⁶ Ω

                 ∴         1/Rp = 1/ R1 + 1/R2

                                    = 1/1 + 1/106 Ω 

                                    = 166 + 1/166Ω                           

                                    = 1000000+1/1000000Ω

                                    = 1000001/1000000Ω                     

                            ∴ Rp  = 1000000/1000001Ω

                                    = 0.999999 Ω

                            ∴ Rp  = 1Ω (approx)

(b) Given that,

                              R₁ = 1 Ω

                              R₂ = 10³Ω     

                              R₃ = 10⁶Ω       

               ∴            1/Rp 1/R₁ 1/R₂ 1/R₃Ω           

                                  = 1/1 + 1/103 + 1/106Ω

                                  = 10⁶ + 10³ + 1/10⁶Ω

                                  = 1000000 + 1000 + 1/1000000Ω

                                  = 1001001/1000000Ω

               ∴            Rp = 1000000/1001001Ω

                                  = 0.99900001Ω

               ∴            Rp  = 1 Ω (approx)

Q.2. An electric lamp of 100 Ω, toaster of resistance 50 Ω and a water filter of resistance 500 Ω are connected in parallel to a 220V source. What is the resistance of the electric iron connected to the same source that takes as much current as all the three appliances and what is the current through it? 

Ans: Given that,    

                         R₁ = 100Ω

                         R₂ =  50Ω

                         R₃ =  500Ω

          ∴         1/RP = 1/R₁ + 1/R₂ + 1/R₃

                             = 1/100 = 1/50 + 1/50Ω

                             = 5+10+1/500Ω                   

                             = 16/500Ω

          ∴          RP   = 500/16Ω               

                              = 31.25 Ω

          ∴  Resistance of the electric iron = 31.25 Ω

2nd part: 

          Given that,

                        V = 220 Volt 

                        R = 31.25 Ω

                        I  = ?

We have,

                        I = V/R 

                          = 220/31.15 

                          = 7.04 A

Q.3. What are the advantages of connecting electric devices in parallel with the battery instead of connecting them series?

Ans: Advantages of connecting electrical devices in parallel with the battery are as follows:

(a) Each device gets the fall battery voltage. 

(b) The parallel circuit divides the current through the electrical devices. Each device gets proper current depending on its resistance. 

(c) If one device is switched OFF or ON, others are not affected. 

(d) If one electrical device is damaged, it will not affect the other devices connected parallel, hence the other devices continue to work properly.

Q.4. How can three resistors of resistances 2Ω, 3Ω and 6Ω be connected to give a total resistance of 

(a) 4Ω 

(b) 1Ω?

Ans: (a) First 3Ω and 6Ω are connected in 

we have, 

                 1/RP1 = +

                          = 2 + ⅙

                          = 3/6

                          = ½

             ∴  RP1  =  2Ω

Now RP1 is connected with the 3rd resistor of resistance 2Ω in series.

∴  The total resistance, 

                                 = RP1 + 2Ω

                                 = 2Ω + 2Ω

                                 = 4Ω

(b) If we connected all three in parallel then 

                         1/RP = 1/R₁ + 1/R₂ + 1/R₃

                                 = 1/2 + 1/3 + 1/6Ω 

                                 = 3 + 2 + 1/6Ω

                                 = 5/6Ω

                                 = 1Ω

Q.5. What is the (a) highest; (b) lowest total resistance that can be secured by combination of four coils of resistances 4Ω, 8Ω, 12Ω, 24Ω? 

Ans: (a) Resistance is maximum when connected is series.

∴  Total resistance, 

                             = 4Ω + 8Ω + 12Ω + 24Ω

                             = 48Ω

(b) Resistance is minimum when connected in parallel. 

∴   1/RP = 1/R₁ + 1/R₂ + 1/R₃ + 1/Ry 

             = 1/4 + 1/8 + 1/12 + 1/24

             = 6 + 3 + 2+ 1/24Ω

             = 12/24Ω           

             = 1/2Ω            

∴    RP   = 2Ω

∴    Total resistance = 2

Page – 218 

Q.1. Why does the cord of an electric heater not glow while the heating element does? 

Ans: The heating element of an electric heater is made of an alloy which has a high resistance. When the current flows through the heating element, the heating element becomes too hot and glows red. The cord is usually made of copper or aluminum which has low resistance. Hence the cord doesn’t glow.

Q.2. Compute the heat generated while transferring 96,000 coulomb of charge in one hour through a potential difference of 50v. 

Ans: Given that,

                          Q = 96,000c 

                           t  = 1 hour 

                              = 36 sec 

                           V = 50V

Heat generated is,    

                            H = VQ

                                =  50×96,000J

                                =  48,00000J 

Q.3. An electric iron of resistance 20 Ω tkes a current of 5A. Calculate the heat developed in 30 Sec. 

Ans: Given that, 

                         R = 20 Ω

                          I = 5A 

                          t = 30 sec 

∴  Heat developed,  

                              = I²Rt 

                              = 5²x20x30J 

                              = 25x20x30J 

                              = 15000J 

                              = 15 KJ

Page – 220 

Q.1. What determines the rate at which the energy is delivered by a current? 

Ans: The rate at which the energy is delivered by a current is called power. 

The power P is given by,   

                                      P =  VI 

                                 or  P = I²R 

                                          = v²/R 

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

Ans: Given that,

                                     I = 5A 

                                    V = 220v

                                    P = ? 

We have,    

                                  P = VI 

                                      = 220 x 5 

                                      = 1100 w 

Energy consumed,  

                                      =  P x t 

                                      = 1100w x 2h 

                                      = 1100w x 2 x 60 x 60 sec 

                                      = 1100w x 7200 sec 

                                      = 7920000J 


Q.1. A piece of wire of resistance R is cut into five equal parts. These the equivalent resistance of this combination is Rᴵ, then the ratio R/Rᴵ is 

(a) 1/25 

(b) 1/5     

(c) 5

(d) 25

Ans: Given that,

            1/Rᴵ = 1/R/5 + 1/R/5 + 1/R/5 + 1/R/5 + 1/R/5 

                   = 5/R + 5/R + 5/R + 5/R + 5/R   

                   = 25/R

        ∴    Rᴵ = R/25

Now R/Rᴵ 

                  = R/R/25

                  = 25R/R

                  = 25 

Hence answer is (d) 

Q.2. Which of the following terms does not represent electrical power in the circuit? 

(a) I²R?    

(b) IR²    

(c) VI    

(d) V²/R?

Ans: (b) IR² is not power.

Q.3. An electrical bulb is rated 220v and 100w. When it is operated on 110v, the power consumed will be

(a) 100w 

(b) 75w 

(c) 50w 

(d) 25w

Ans: (d) 25w.


                   R = V2/P R  

                       = (220)2/100 = 484Ω

When operated on 110v, the power consumed will be 

Pᴵ = vᴵ²/R = (110)2/484

               = 25w

Q.4. Two conducting wires of the same material and of equal lengths and equal diameters are first connected is vlivi series and then parallel in circuit across the same potential difference. The ratio of heat produced in series ai and parallel combinations would be 

(a) 1:2     

(b) 2:1     

(c) 1:4    

(d) 4:1

Ans: Since the wires are of the same material and are equal in lengths and diameters therefore if R is resistance of each, resistance in series and parallel would be R₅ and RP respectively given by 

   RS = R + R + 2R      

1/RP = 1/R + 1/R = 2/R   

∴ RP = R/2           

When connected to source v, then heat produced in series H₅ and when in parallel in HP and is given by

HS = v²t/R2 and HP = v²t/RP 

  ∴ = HS/HP = v²t/RS x RP/v²t

     = RP/RS 

     = R/2/2R

     = R/4R = ¼

⇒  = HS : HP = 1:4                       

Hence (c) is correct.

Q.5. How is voltmeter connected in the circuit to measure the potential difference between two points? 

Ans: Voltmeter is basically an electronic device which is used for measuring the potential difference between two separate points in a circuit.

We always connect a voltmeter in the parallel formation, to an electrical circuit.

To, avoid the high amount of resistance and to get the most accurate measurement of the potential difference.

Q.6. A copper wire has diameter 0.5 mm and resistivity 1.6×10⁸Ω m. What will be the length of this wire to make its resistance 10Ω? How much does the resistance charge if the diameter is doubled?

Ans: Here d = 0.5 mm

                    = 0.5 x 10⁻³m 

                    = 5×10⁻⁴Ωm 

              p    = 1.6 x 10⁻⁸Ωm 

              I     = ? 

               R = 100Ω

We have,

          R = pᴵ/A

      ⇒ R = pᴵ/πd²/4

      ⇒ R = 4pᴵ/πd²

      ⇒ R = Rπd²/4p

      ⇒  I  = 10×22/7x(5×10⁻⁴)2/4 × 1.6×10⁻⁸

              = 10×22×25×10⁻⁸/4×7×1.6×10⁻⁸

              = 122.77 m

If the diameter is doubled, resistance will become 1/4 of the original. 

∴  Resistance will be,

                                  = 100Ω/4

                                  = 2.5Ω

Q.7. The values of the current I flowing in a given resistor for corresponding values of potential differences V across the resistor are given below 


Plot a graph between V and I and calculate the resistance of the resistor. 

Ans: The graph between V and I for the given data is

Resistance of the resistor, 

R = v² –v₁/I₂ –I₁

    = 13.2 – 1.6/4.0 – 0.5

    = 11.6/3.5

    = 3.214 Ω

Q.8. When a 12v 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: Given that,

                          V = 12 volt 

                           I = 2.5 mA 

                             =  2.5×10⁻³A 

                          R = ?

We have,  

R = V/I                

   = 12/2.5×10⁻³Ω

   = 4800Ω

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

Ans: Given that,  

                        R = 0.2+0.3+0.4+0.5+12

                            = 13.4 Ω

                         v  =  9 volt 

                         I   = ?

We have,

              I = V/R

                = 9/13.4A

                = 0.67A

∴ Current will low through 12Ω resistor is 0.67A. 

Q.10. How many 176Ω resistors (in parallels) are required to carry 5A on a 220v line? 

Ans: Suppose n resistors are required. 

                     = N/176

       ∴   R       = 176/NΩ

Now, we have, 

                          R = V/I

               ⇒ 176/N = 220/5

               ⇒ N = 176/44 = 44

                       = 4

Q.11. Show how you would connect three resistors, each of resistance 6Ω, so that the combination has a resistance of 

(i) 9Ω    

(ii) 4Ω

Ans: Given that, 

                          R₁  =  R₂  =  R₃ = 6Ω

(a) Combine R₂ and R₃ in parallel and connected R₁ in series.


∴ Equivalent resistance = R₁ + 1/R₂ + 1/R₃

                                      = R₁ + R₂R₃/R₂ +  R₃ 

                                      = 6 + 6-6/6 + 6Ω

                                      = 6 + 36/12Ω

                                      = 6  + 3Ω

                                       = 9Ω

(b) First R₁ and R₂ are connected in series and then connected R₃ in parallel. 


∴ Equivalent resistance = 1/12 + 1/6


                                      = 1 + 2/12

                                      = 12/3

                                      = 4Ω

Q.12. Several electric bulbs designed to be used on a 220v electric supply line, are rated 10w. How many lamps can be connected in parallel with each other across the two wires of 220v line if the maximum allowable current io t is 5A? 

Ans: Ut current through each bulb be I 

                             P = vI 

                      ⇒ 10 = 220 I 

                      ⇒    I = 1/22A

Let n such bulbs be connected in series combined current through n bulbs = 5A

                                                 ⇒  n×I = 5

                                                 ⇒ n x 1/22 = 5

                                                 ⇒ n = 5×22

                                                        = 110

∴ 110 such bulbs can be lighted within allowable limit of 5A. 

Q.13. A hot plate of an electric oven connected to a 220v line has two resistance coils A and B, each of 24Ω resistance, which may be used separately, in series, or in parallel. What are the current in the three cases? 

Ans: (i) When the two coils A and B are used separately, 

         Given that,

                        R = 24 Ω

                        v  =  220v

                        I   = ?

We have,

                         I = V/R 

                           = 220/24A  

                           = 9.167 A 

(ii) When the two coils A and B are connected in series. 

Given that,

                     R = 24 + 24 = 48 Ω  

                     V = 220v 

                      I = ?

We have, 

                      I =  V/R

                        = 220/48A

                        = 4.58 A 

(iii) When the two coils A and B are connected in parallel 

Given that,

                       1/R = 1/24 + 1/24

                              = 1 + 1/24

                              = 2/24


                ∴      R  = 12 Ω

                        v   = 220v

                         I   = ?

We  have,

                         I = V/R

                           = 220/12

                           = 18.33 A 

Q.14. Compare the power used in the 2 Ω resistor in each of the following circuits: 

(i) a 6v battery in series with 1Ω and 2 Ω resistors.

(ii) a 4v battery in parallel with 12Ω and 2Ω  resistors.

Ans: (i)  1Ω and 2Ω are connected in series.

                  ∴      R =  1 + 2 

                              = 3Ω

                           v = 6 volt 

                  ∴       I = V/R

                             = 6/3

                             = 2 A

∴ The current passing through 2Ω resistor is 2A 

     Power used in 2 Ω resistor  

                                                = I²R 

                                                = 2².2 

                                                = 8w 

(ii) 12Ω and 2Ω are connected in parallel 

     power used in 2 Ω resistor = V2/R

                                                = 42/2

                                                = 8w

Q.15. Two lamps, one rated 100w at 220v, and the other 60w at 220v, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220v? 

Ans: Total power consumed in the circuit 

                                              = 100+60 

                                              = 160w

                                           V =  220 volt 

          We have, power  

                                              =  VI

                                        ⇒ I = Power/V

                                              = 160/220A

                                              =  0.727 A

Q.16. Which appliances use more energy, a 250w TV set in or a 1200w toaster in 10 minutes? 

Ans: For TV set 

                     Energy used  =  250w × 1h

                                           =  250wh

 For toaster 

                    Energy used  = 1200w × 10 minutes 

                                           = 1200w × 10 minutes

                                           = 1200w × 10/60h

                                           = 1200w × 1/6h                                                                                    

                                           = 200wh

Thus, TV set uses more energy than the toaster. 

Q.17. An electric heater of resistance 8Ω draws 15A from service mains for 2 hours. Calculate the rate at which heat is developed in the heater.

Ans: Given that,

                       R = 8Ω

                       I  = 15A 

        Power (P) = ?

        We have, 

                         P  = I²R

                             = 15².8

                             = 225×8

                             =  1800w

Q.18. Explain the following: 

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

Ans: The metal tungsten is almost exclusively used for the filament of the electric lamp because it has a very high melting point due to which it does not melt even when it is heated to high temperatures due to the passage of electric current.

(b) Why are the conductors of electric heating devices, such as bread toasters and electric irons, made of an alloy rather than pure metal? 

Ans: The conductors of electric heating devices such as bread toasters and electric irons are made of alloy because the resistivity of an alloy is more than that of metals which produces a large amount of heat.

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

Ans: In series circuits voltage is divided. Each component of a series circuit receives a small voltage so the amount of current decreases and the device becomes hot and does not work properly. Hence, series arrangement is not used in domestic circuits.

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

Ans: The resistance of a wire is inversely proportional to its area of cross-section i.e. Thus, when the area of cross-section of wire increases (or thickness of wire increases), then its resistance decreases. Similarly, when the area of cross-section of wire decreases (or thickness of wire decreases), then its resistance increases.

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

Ans: Copper and aluminium wires usually employed for electricity transmission because they have low resistivity and they are good conductors of electricity.

Leave a Comment

Your email address will not be published. Required fields are marked *

Scroll to Top