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Important Questions for CBSE Class 10 Science Chapter 11 - Electricity 2024-25

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CBSE Class 10 Science Chapter-11 Electricity Important Questions with Answers - Free PDF Download

Class 10 is an essential stage in every student's career. Their performance in Class 10 acts as a base for their future studies. So it's important to score well in Class 10. The most important and challenging subject in Class 10 is Science and many students seem to get confused here and lose marks. If students want a successful career in future, then they can't afford to lose marks in this stage of their life. It's important to understand every chapter in science thoroughly to score good marks. Chapter 11 of Class 10 Science which is about electricity is one of the difficult chapters. A student who is incapable of understanding this chapter must practice Important Questions for Class 10 Science Chapter 11. These important questions of Electricity Class 10 can make the students through on the concepts of this chapter. Vedantu is a platform that provides free CBSE Solutions (NCERT) and other study materials for students. Maths Students who are looking for the better solutions, they can download Class 10 Maths NCERT Solutions to help you to revise complete syllabus and score more marks in your examinations.

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Study Important Questions for Class 10 Science Chapter 11 – Electricity

Very Short Answer Questions (1 Mark)

1. Which two circuit components are connected in parallel in the following circuit diagram? 

(a) R1 and R2 only

(b) R1,R2 only 

(c) R2 and V only 

(d) R1 and Vonly 


circuit components

Ans: (a) The two circuit components that are connected parallel in the circuit diagram is  and R2 only.


2. A metallic conductor has loosely bound electrons called free electrons. The metallic conductor is 

(a) negatively charged 

(b) positively charged 

(c) neutral 

(d) Either positively charged or negatively charged 

Ans: (c) The metallic conductor is neutral.


3. Which of the following expressions does not represent the electric power in the circuit? 

(a) VI.

(b) I2/R

(c) V2/R

(d)  I2R

Ans: (b) The expression which does not represent the electric power in the circuit is I2/R .


4. Resistivity of a metallic wise depends on 

(a) its length 

(b) its shape 

(c) its thickness 

(d) nature of material 

Ans: (d) Resistivity of a metallic wire depends on the nature of the material.


5. If the current I through a resistor is increased by 100% the increased in power dissipation will be (assume temperature remain unchanged) 

(a) 100%

(b) 200%

(c) 300%

(d) 400%

Ans: (c) The increase in power dissipation will be 300% .



6. For the circuit arrangement shown below, a student would observe.


seo images

(a) Some reading in both ammeter and voltmeter. 

(b) No reading in either the ammeter or the voltmeter. 

(c) Some reading in the ammeter but no reading in the voltmeter. 

(d) Some reading in the voltmeter but no reading in the ammeter. 

Ans: (c) A student will observe some reading in the ammeter but no reading in the voltmeter.


7. A wire of resistance R is cut into five equal pieces. These pieces are connected in parallel and the equivalent resistances of the combination are R . Then the ratio RR is 

(a) 15

(b) 5

(c) 125

(d) 25

Ans: (d)The ratio RR is 25 .


8. The resistance of the conductor is R . If its length is doubled, then its new resistance will be 

(a) R

(b) 2R

(c) 4R

(d) 8R

Ans: (c) The new resistance is 4R .


9. A student carries out an experiment and plots the V-I graph of three samples of nichrome wire with resistances R1,R2,R3 respectively as shown in the figure. Which of the following is live? 


seo images

(a) R3>R2>R1

(b) R2>R3>R1

(c) R1>R2>R3

(d) R1=R2=R3

Ans: (a)According to the graph , R3>R2>R1


10. The nature of the graph between potential difference and the electric current flowing through a conductor is 

(a)parabolic 

(b) circle 

(c) straight line 

(d) hyperbolic 

Ans: (c)The nature of the graph between potential difference and the electric current flowing through a conductor is a straight line.


11. An electric heater is salted at 1500 w. How much heat is produced per hour? 

(i) 5400

(ii) 54000

(iii) 5.4×105

(iv) 5.4×106.

Ans: (iv) The electric heater produces 5.4×106 J per hour.


12. A student says that the resistance of two wires of the same length and same area of  cross section is the same. This statement is correct if 

(a) Both wires are of different materials 

(b) Both wires are made of the same material and are at different temperatures.

(c) Both wires are made of the same material and are at the same temperature.

(d) Both wires are made of different materials and are at the same temperature. 

Ans: This statement is correct if (c)The resistance of two wires of the same length and same area of cross section is the same if both wires are made of the same material and are at the same temperature.


13. In an experiment ohm, s law a student obtained a graph as shown in the diagram. The value of resistance of the resistor is 


seo images

(a) 0.1Ω

(b) 1.0Ω

(c) 10Ω

(d) 100Ω

Ans: (d) The value of resistance of the resistor is 100Ω .


14. Work done to move 1 coulomb charge from one point to another point on a charged conductor having potential 10 volt is 

(a) 1 Joule 

(b) 10 Joule 

(c)  zero 

(d) 100 Joule 

Ans: (c) Work done to move 1 coulomb charge from one point to another point on a charged conductor having potential 10 volt is zero. 


15. Three resistors are shown in the figure. The resistance of the combination is 


seo images

(a)3Ω

(b) 6Ω

(c) 9Ω

(d) 7Ω

Ans: (c) The resistance of the combination is 9Ω .


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

Ans: A device that helps to maintain a potential difference between conductors is the battery.


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

Ans: The rate at which energy is delivered by a current is determined by electric power. 


18. A wire of resistance R is cut into five equal pieces. These pieces are connected inparallel and the equivalent resistances of the combination are R . Then the ratio RR is 

(a) 15

(b) 5

(c) 125

(d) 25

Ans: (d)A wire of resistance R is cut into five equal pieces. These pieces are connected inparallel and the equivalent resistances of the combination are R'. In this cases, the ratioRR is 25 .


19. Which of the following terms does not represent electrical power in a circuit?

(a) I2R

(b) IR2

(c)  VI

(d) V2/R

Ans: (b) The term that does not represent electrical power in a circuit is IR2 .


20. An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be: 

(a) 220

(b) 75

(c) 50

(d) 25

Ans: (d)The power consumed will be 25 W.


21. Two conducting wires of the same material and of equal lengths and equal 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 combination would be: 

(a) 1:2

(b) 2:1

(c) 1:4

(d) 4:1

Ans: (c) The ratio of heat produced in series and parallel combination will be 1:4 .


22. A wire of resistance R is bent in the form of a closed circle, what is the resistance across a diameter of the circle? 

Ans: The resistance across a diameter of the circle 

=R/4


23. A charge of 6 C is moved between two points P and Q having, potential 10 V and 5 V respectively. Find the amount of work done. 

Ans: The amount of work done, W=q(V2V1)

= 6(105)

= 30 joule


24. Name the physical quantity whose SI unit is JC 

Ans: The physical quantity whose SI unit is JC is Potential.


25. Why are copper wires used as connecting wires? 

Ans: Copper wires are used as connecting wires because in case of copper the electrical resistivity for it is low.


26. A wire of resistivity p is stretched to double its length. What is its new resistivity? 

Ans: When a wire of resistivity p is stretched to double its length, then the new resistivity remains the same because resistivity depends on the nature of material.


27. What is the resistance of the connecting wire? 

Ans: The resistance of a connecting wire made of a good conductor is extremely low.


28. What is the resistance of an ammeter? 

Ans: An ammeter's resistance is very minimal, and in an ideal ammeter, it is zero.


29. What is the resistance of a Voltmeter? 

Ans: An ideal voltmeter's internal resistance is infinite.


Short Answer Questions (2 Marks)

1. How does use of fuse wire protect electrical appliances? 

Ans: When a large quantity of current passes through the circuit, the temperature of the wire rises and the fuse wire melts. This prevents current from flowing into the house's other circuits, saving electrical appliances.


2. Calculate the resistance of an electric bulb which allows a 10A current when connected to a 220 V power source? 

Ans: It is given from the question that an electric bulb allows a 10A current when connected to a 220 V power source.

Therefore,

I=10 A ,

V=220V

R=VI

=22010

=22 ohm   


3. (i) Identify the VI graphs for ohmic and non-ohmic materials.

Ans: The VI graphs for ohmic and non-ohmic materials respectively can be represented as shown below:


seo images
 
seo images

(ii) Give one example of each.

Ans: (ii) Some examples of ohmic material are Copper, Nichrome and some examples of Non-ohmic material   are Diode, Transistor.


4. What do the following symbols represent in a circuit? Write the name and one function of each? 


(i)

seo images


Ans: (i) It symbolises a battery that maintains a potential difference across the circuit element to allow current to flow.


(ii)

seo images


Ans: It's an ammeter that measures how much current is flowing across a circuit.


5. Define the term “volt”? 

Ans: If 1joule of energy is transferred between two points A and B, the potential difference between them is one volt. In an electric circuit, work is done to move one coulomb of charge from one point to another field.


6. Why does the connecting rod of an electric heater not glow while the heating element does? 

Ans: As its resistance is lower than that of the heating element, the connecting cord of an electric heater does not glow. As a result, the heating element produces more heat than the connecting cord, and it glows


7. A number of n resistors each of resistance R are first connected in series and then in parallel. What is the ratio of the total effective resistance of the circuit is series combination and parallel combination? 

Ans: Total effective resistance of  the circuit when in series combination Rs=nR

And for parallel combination is Rp=Rn  and 

RsRp=nRRn

=n2

The ratio will be n2 .


8. Draw a schematic diagram of a circuit consisting of 3 V battery, 5 ohm, 3Ω  and 1Ω resistor, an ammeter and a plug key, all connected in series. 

Ans: The circuit diagram of a circuit consisting of 3 V battery, 5 ohm, 3Ω  and 1Ω resistor, an ammeter and a plug key, all connected in series can be represented as show below,


seo images

9. A copper wire has diameter 0.5 mm and Resistivity of 1.6×108Ωm What is the length of this wire to make its resistance ? How much does the resistance change if diameter is doubled?

Ans: Diameter of the copper wire, D=0.5×103 m

P=1.6×108

R=10

Then,

R=ρlA

=ρlπr2

=4ρlπD2

l=πRD24ρ

l=3.14×10×(5×104)24×1.62×108

=121.14

Length of the wire,  l=121.14 m

New R=4ρlπ(D)2

=144ρlπD2

=14R

Length of the wire to make its resistance 10Ω is 121.14m and when the diameter is doubled the new resistance will be one fourth that of the old one.


10. Alloys are used in electrical heating devices rather than pure metals. Give a reason. 

Ans: Alloys are utilised in electricity heating devices rather than pure metals because alloys have a higher resistivity and hence produce more heat. Furthermore, alloy is non-combustible (or oxidises easily at higher temperature).


11. On what factor does the resistance of a conductor depend? 

Ans: The factors on which Resistance depends are:-

(a) Length of the conductor

(b) Area of cross - section

(c)  Temperature

(d) Nature of material 


12. Calculate the number of electrons consisting of one coulomb of charge? 

Ans: Let x= no. of electrons 

Charge on 1 electron =1.6×1019 C , that is 

x=11.6×1019

x=6.25×1018

The number of electrons consisting of one coulomb of charge is 6.25×1018 .


13. What does an electric circuit mean? 

Ans: An electric circuit is a current route that is both continuous and closed. Current can flow through an electric circuit if it is complete.


14. Define the unit of current. 

Ans: The ampere is the SI unit for electric current. If 1 coulomb charge flows per second across a conductor cross-section, the current is said to be 1 ampere.


15. Calculate the number of electrons constituting one coulomb of charge. 

Ans: The charge on one electron =1.6×1019 coulomb.

Number of electrons in one coulomb of charge=11.6×1019

=6.25×1018

The number of electrons consisting of one coulomb of charge is 6.25×1018 .


16. What is meant by saying that the potential difference between two points is 1 v? 

Ans: If 1 joule of labour is required to move a charge of 1 coulomb from one location to another, the potential difference between the two points is said to be 1 volt.


17. Ammeter burns out when connected in parallel. Give reasons. 

Ans: When a low-resistance wire is connected in series, a huge quantity of current travels through it, causing it to be burned, or short-circuited.


18. Judge the equivalent resistance when the following are connected in parallel: 

(a) Equivalent resistance of 1Ω and 106Ω

Ans: When the resistances are connected in a parallel arrangement, the resultant resistance is given by:

1R=1R1+1R2+1R3+

1R=11+1106

=1+106

R=1Ω

The equivalent resistance is 1Ω .


(b) Equivalent resistance of  1Ω , 103Ω and 106Ω

Ans: 1R=11+1103+1106

=1+103+106

R=1Ω

The equivalent resistance is 1Ω .


19. An electric iron of resistance 20 takes a current of 5 A. Calculate the heat developed in 30 s.

Ans: Resistance of electric iron, R=20Ω , current, I=5 A and time =30 s.

Heat generated H=I2Rt

=52×20×30

=15000 j

The heat developed in 30 second is 15000 j .


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

Ans: The given information is as shown below,

Charge transferred, Q=96000

Potential Difference, V=50  V.

Heat generated, H=VQ

=50×96000

=4800000

=4.8×106 j

The heat generated while transferring 96000  coulomb of charge in one hour

through a potential difference of 50 V is 4.8×106 j


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

Ans: Given that current drawn by electric motor I=5 A.

The line voltage V=220

Time, t=2 h

 Power of motor , P=VI

=220×5

=1100 W and 

the energy consumed E=Pt

=2×1100

=2.2 KWh

The power of the motor and energy consumed in 2 h are 1100 W and 2.2 kWh respectively.


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

Ans: A voltmeter is connected in parallel to the resistance across the place where the potential difference is to be determined.


23. 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: Given that Voltage of battery, V=12  V, 

Current, I=2.5 mA 

=2.5×103 A

Resistance, R=V/I

=122.5×103

=4800Ω

The value of the resistance of the resistor is 4800Ω .


24. 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 5 A? 

Ans: The given information is as shown below

Each bulb is rated as 10 W, 220 V, 

It draws current, I=P/V

=10220 V

=1/22 A.

The maximum allowable current is 5 A and all lamps are connected in parallel. Therefore the maximum number of bulbs joined in parallel with each other =5×22whichis110.


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

Ans: Current drawn by 1stlamp rated 100 W at 220V=P/V

=100/220

=5/11 A. 

Current drawn by 2ndlamp rated 60 W at 220 , V=60/220=3/11A.

In parallel arrangement the total current = 3/11+ 5/11

= 8/11

= 0.73 A.

Current drawn from the line if the supply voltage is 220 V is 0.73 A .


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

Ans: Energy used by a TV set of power 250 W in 1 hour =Pt

=250Wh

Energy used by toaster of power 1200 W in 10 minute (10/60h)= 200 Wh.

A 250 W TV set in 1 hour uses more energy.


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

Ans: Resistance of electric heater, R = 8Ω

current, I = 15 A.

Rate at which heat developed in the heater =I2Rtt

=1800 W.

The rate at which heat is developed in the heater is 1800 W.


28. In the given figure what is the ratio of current in  A


seo images

Ans: Observe that it I clearly known to us that V=IR

I1I2=R2R

=12

The ratio of current is1/2.


29. Two wires of equal cross sectional area, one of copper and other of managing have the same resistance. Which one will be longer? 

Ans: Using the equation  ρ=RAl , where ρ is the resistivity, R is the resistance and A the area

We can see that copper wire has a lower resistance than manganin, hence copper will last longer.


30. A Rectangular block of iron has dimensions L×L×b . What is the resistance of the block measured between the two square ends? Given p resistivity. 

Ans: R=pbL2is the resistance of the block measured between the two square ends


31. Three equal resistances are connected in series then in parallel. What will be the ratio of their Resistances? 

Ans: Rseries=3R

Rparallel=R/3

The ratio of Resistances is 9 .


32. Justify for any pair of resistance the equivalent resistance equivalent resistance in parallel.

Ans: As R=V/I


seo images

From the graph for any pair of resistance the equivalent resistance in series is greater than equivalent resistance in parallel.

A= Series, B= Parallel 


33. How many bulbs of 81 should be joined in parallel to draw a current of 2 A from a battery of 4 V? 

Ans: R=V/I

=4/2

=2Ω

let n be the number of bulbs.

1/R=1/R1+1/R2++1/Rn=n8

12=n8

n=4

Number of bulbs are 4 .


34. Two cubes A and B are of the same material. The side of B is thrice as that of A . Find the ratio RA/RB

Ans: The value of  RA=pLA and

RB=p3L9A

RA:RB=3:1


35. If there are 3×1011 electrons are flowing through the filament of the bulb for two minutes. Find the current flowing through the circuit. Charge on one electron 1.6×1019

Ans: Observe as shwn below,

Using the equation

q=ne

=3×1011×1.6×1019C

=4.8×108C

I=q/t

=4.8×1082×60

=4×107 A

The current flowing through the circuit is 4×107 A.


36. A nichrome wire of resistivity 100 W m and copper wire of resistivity 1.62 M ohm-m of the same length and same area of cross section are connected in series , current is passed through them, why does the nichrome wire get heated first? 

Ans:  Looking at the equation

Q=I2RT

Q=I2(pL/A)t

Henceforth, because nichrome wire has a higher resistance than copper wire, it must be heated first.


37. What is represented by joule/coulomb? 

Ans: The potential difference is represented by the joule/coulomb.


38. A charge of 2 C moves between two plates, maintained at a p.d of IV . What is the energy acquired by the charge? 

Ans: The energy acquired by the charge, W=QV

=2J

The energy acquired is 2 J


39. Which has more resistance: 100 W bulb or 60 W bulb? 

Ans: As, it is clearly known thatR1P, thus the resistance of 60 W bulb is more.


40. What happens to the current in a circuit if its resistance is doubled? 

Ans: As current and resistance are inversely proportional, the current is reduced to half of its previous value.


41. What happens to the resistance of a circuit if the current through it is doubled? 

Ans: Resistance is unaffected since the circuit's resistance is independent of the current flowing through it.


42. How does the resistance of a wire depend upon its radius? 

Ans: As R1A

RV

Resistance of a wire is directly proportional to its radius.


43. Two wires are of the same length, same radius, but one of them is of copper and the other is of iron. Which will have more resistance. 

Ans: Since  R = p1/A ,

but A and I are the same. It is solely determined by resistivity, hence iron has a higher resistance.


44. Two wires of same material and same length have radii R and r Compare their resistances. 

Ans: Suppose R and r are resistances, then R=r as p and I are the same.


Short Answer Questions (3 Marks)

1. Two metallic wires A and B are connected, wire A has length I and radius r , while  wire B has  length  2l  and  radius 2r .  Find  the  ratio  of  total  resistance  of  series combination and the resistance of wire A , if both the wires are of the same material? 

Ans: Observe as shown below,

Resistance of metallic wire A , R1=ρlA

=ρlπr2

Resistance of metallic wire B , R2=ρ2l4πr2

Total resistance in series is R=R1+R2

=ρlπr2+2ρl4πr2

=3ρl2πr2

The ratio of the total resistance is series to the resistance of A is

RR1=ρlπr23ρl2πr2

=2/3

The ratio of the total resistance is series to the resistance of A is 2/3 .


2. Should  the heating  element  of  an  electric  iron be made  of  iron,  silver  or nichrome wire? Justify giving three reasons? 

Ans: The following reasons can be found, why the heating element of an electric iron is composed of nichrome wire.

(1) Due to the high resistance, the passage of current generates additional heat.

(2) High melting point.

(3)  At high temperatures, it does not easily oxidised (or burn).


3. (a) Define electric resistance of a conductor? 

Ans: A conductor's electric resistance is defined as the resistance it provides to the flow of current.

That is R=V/I and its S.I. unit is ohm , Ω .

(b) A wire of  length L and resistance R  is stretched so that  its  length  is doubled and the area of the cross section is halved. How will its 

(i) resistance change 

Ans: It is clearly known that resistance,  R=ρlA

New length L=2L and

A=A2

Therefore R1=ρLA

=4R

Therefore, the resistance of a wire becomes 4 times its original resistance.

(ii) resistivity change? 

Ans: The size of a wire has no bearing on its resistance. As a result, resistance does not vary.


4. Two  resistors  of  resistance  R  and  2R  are  connected  in parallel  in  an  electric  circuit. Calculate the ratio of the electric power consumed by R and 2R

Ans: Power consumed by R , ρ1=V2R

Power consumed by2R , ρ2=V22R

Ratio ρ1ρ2=V2RV22R

.=2:1.

The ratio of the electric power consumed by R and 2R is 2:1 .


5. The length of different metallic wires but of same area of cross section and made of the same material are given below 

Wire

Length

A

1m

B

1.5m

C

2.0m

(i) Out of these two wires which wire has higher resistance. 

Ans: As Rl (length of the conductor) and since length of wire C is more than A and B ,  wire C has higher resistance.


(ii) Which wire has higher electrical resistance? Justify your answer. 

Ans: The electrical resistivity of a wire is determined by the nature of the material, not by its dimensions. As a result, the resistivity of all wires is the same as the substance of all wires.


6. Two resistors of resistances R and 2R are connected in series with an electrical circuit? Calculate the ratio of the electric power consumed by R and 2R

Ans: It is clearly known thatElectric power consumed by R , P1=I2R

Also, electric power consumed by 2R , P2=I22R

P1P2=1/2

The ratio of the electric power consumed by R and 2R is1:2.


7. Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then in parallel in an electric circuit. 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: (c)Let resistance of each wire is R

In series, resistance is  =2R

Heat produced, H1=V22Rt

In parallel total resistance =R/2

Heat produced, H2=(V2R2)t

=2V2Rt

H2=4H1

H1H2=1/4

The ratio of heat produced in series and parallel combinations is 1:4 .


8. Calculate the following of a circuit shown in the figure.


seo images

(i) effective resistance 

Ans: Effective resistance, R=R1+R2

=5+10

=15

The effective resistance is 15Ω .

(ii) current 

Ans: Current, I=V/R

=2/15

=0.133

The current is 0.133 A.

(iii) Potential difference across 10Ω resistor

Ans: Potential difference across 10Ω

V=IR

=215×10

=1.33

The potential difference across 10Ω is 1.33 volt.


9. A Piece of wire of resistance 20Ω is drawn out so that its length is increased to twice its original length to calculate the resistance of the wire is the new situation? 

Ans: Resistance of wire=20Ω

As R=ρlA

And as length of a wire is increased, its area of cross- section decreases, and volume of the wire remains constant.

l=2l

A=A/2

R=ρlA

=4ρlA

RR=41

R=80

The resistance of the wire is the new situation is 80Ω


10. A  battery  made  of  5  cells  each  of  2 V  and  have  internal  resistance 0.1Ω,0.2Ω,0.3Ω,0.4Ω and 0.5Ω  is  connected  across 10Ω resistance.  Draw a circuit diagram and calculate the current flowing through10Ωresistance?


seo images

Ans: The internal resistance , 0.1+0.2+0.3+0.4+0.5=1.5Ω

Total resistance =1.5+10

=11.5

I=V/R

=10/11.5

I= 0.869 A

Current flowing through 10Ω resistance is 0.869 A.


11. In the circuit diagram given here Calculate- 


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(a) The total effective resistance 

Ans: As resistances are in parallel

1R=1R1+1R2+1R3

=12+15+110

=810

R=108

Total effective resistance is 108Ω

(b) The total current 

Ans: Total current, I=V/R

=610/8

=4.8

The total current is 4.8 A.

(c) The current through each resistor. 

Ans: From the circuit diagram, I1,I2 and I3 be the current through 2Ω,5Ω and 10Ω  respectively

Therefore, I1=VR1

=3

I2=VR2

=1.2

I3=VR3

=0.6

Current through each2Ω,5Ω and 10Ω  is 3 A ,1.2 A, and 0.6 A respectively.


12. You have two circuits Compare the power used in 2Ω resistor in each case. 

(i) a 6V battery is series with 1Ω and 2Ω resistor

Ans: Potential difference, V=6V

R1=1Ω

R2=2Ω

Total Resistor =R1+R2

=1+2

=3

I=VR

=63

=2

P1=V2R

=8

The power used in 2Ωresistor  is 8 W.

(ii) a 4V battery in parallel with  12Ω and 2Ω resistor 

Ans: Potential difference, V=4V

R1=12Ω

R2=2Ω

P2=V2R

=8

The power used in 2Ωresistor  is 8 W.

The ratio of both power is 1:1 .


13. How  much energy  is  given  to  each coulomb  of charge  passing  through  a  6  volt battery? 

Ans: Potential difference, V=6V

 Charge, Q=1C

Energy = total work done  

= Q x V

= 1x6

= 6 joule.

Energy  given to each coulomb of charge passing through a 6 volt battery is 6 joules.


14. On what factor does the resistance of a conductor depend? 

Ans: A conductor's resistance is determined by the following factors:

(i) length of conductor

(ii) Area of cross-section

(iii) Temperature

(iv) Conductors are made from a variety of materials.


15. Will  current  flow  more  easily  through  a  thick  wire  or  a  thin  wire  of  the  same material, when connected to the same source? Why? 

Ans: When linked to the same source, current flows more freely through a thick wire than via a small wire of the same material. It's because resistance rises as thickness decreases.


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

Ans: The electrical component's resistance R remains unchanged, but the potential difference across its ends falls to half of its original value. As a result of Ohm's law, new current is reduced to half of its initial value.


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

Ans: The following reasons are why the coils of electric toasters and electric irons are built of an alloy rather than a pure metal:

(i) An alloy's resistivity is higher than that of pure metal.

(ii) An alloy does not rust quickly at high temperatures.


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

Ans: The diagram of circuit is as follows


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19. An electric lamp of 100Ω , a toaster of resistance 50Ω and a water filter of resistance 500Ω are connected in parallel to a 220V 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 through it? 

Ans: It can be found from the question, voltage, V=220V

R1=100Ω

R2=50Ω

R3=500Ω

1/R=1/100+1/50+1/500

R = 500/16

= 31.25Ω

The resistance of electric iron, which draws as much current as all three appliances take together is 31.25Ω

Current passing through electric iron,  I=V/R

=220/31.25

=7.04 A.

That is the current passing is 7.04 A.


20. What  is (a) the  highest total  resistance  that  can  be  secured  by a combination of four resistance of 4Ω,8Ω,12Ω and 24Ω ?

Ans: When all four resistances must be connected in series, the highest resistance is achieved. In that instance, the outcome

R=R1+R2+R3+R4

= 4+8+12+24

= 48Ω

The highest resistance is 48Ω .

(b) The  lowest  total  resistance  that  can  be  secured  by a combination of four resistance of 4Ω,8Ω,12Ω and 24Ω ?

Ans: All four resistances must be connected in parallel to produce the lowest resistance.

1/R=1/R1+1/R2+1/R3+1/R4

= 1/4 +1/8 +1/12 + 1/24

= 12/24

The lowest resistance is 2Ω .


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

Ans: When connected to the voltage source, the cord of a heater and the cord of an electric heater are connected in series and carry the same current.

Because the resistance of the cord is so low in comparison to the resistance of the heater element.

As a result, the amount of heat created in the cord is extremely low, but significantly higher in the heater element. As a result, the heating element glows, but the cord does not.


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

Ans: The diameter of wire, d = 0.5 mm, 

Resistivity, ρ=1.6×108

resistance  R= 10 Ω

R=ρL/A

L=πD2R4ρ

=22×(5×104)27×4×1.6×108

= 122.5 m

As resistance is inversely proportional to the cross-section area of wire, when the diameter is doubled for a given length of material, the resistance reduces.


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

Ans: Potential difference V=9V

Total resistance   = 0.2 +0.3 + 0.5 + 0.5 + 12 

= 13.4 Ω

Current in the circuit I = V/R 

= 9 V / 13. 4  A. 

= 0.67

In a series circuit the same current  flows  through all  the resistance, hence current of 0.67 A will flow through 12 Ω resistor. 


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

Ans: Let the resistors of 176Ω be joined in parallel. 

Their combined resistance,

1/R=1/176+1/176 times

=n/176 or

R=176/nΩ

Given that V=220V and 

I=5

R=V/I

=176/n

=220/5

=44Ω

n = 176/44 

= 4

A number of 4 resistors should be joined in parallel.


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

(i) 9Ω

Ans: From the question, R1=R2

=6

Join three resistors as below to get net resistance of 9Ω :


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(ii) 4Ω

Ans: Join three resistors as below to get 4Ω net resistance :


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26. A hot plate of an electric oven connected to a 220V line has two resistance coils A and B . Each of 24Ω resistances, which may be used separately, in series or in parallel. What are the currents in the three cases? 

Ans: From the question, potential differenceV= 220 V .

Resistance of coil A= Resistance of coil B

=24 Ω

When coil is used separately, the circuit I=V/R

=220V/24Ω

= 9.2 A

The current is9.2 A.

When coils are used in series total resistance R=R1+R2

=24+24

=48Ω

The current flowing, I=V/R

=220V/48Ω

=4.6 A

The current is 4.6 A.

Two coils are joined in parallel. 

Total resistance R= 1/24 + 1/24

= 2/24

R=12 Ω .

Current I=V/R

=220V/12Ω

=18.3 A.

The current is 18.3 A.


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

(i) a 6 volt battery in series with 1Ω and 2Ω resistors and, 

Ans: Suppose a 2Ω resistor is joined to a 6 V battery in series with 1Ω and 2Ω resistors.

Total resistance R=2+1+2

=5Ω

Current I=6V/5Ω

=1.2 A

Power used in 2A resistor =I2R

=2.88 W

Power used is 2.88 W.


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

Ans: Suppose 2Ω resistor is joined to a 4V battery in parallel with 12Ω resistor and 2Ω resistors, 

the current flowing in 2Ω=4V/2Ω

=2A

Power used in 2Ω resistor =I2R

=8

Ratio =2.88/8

=0.36:1

The ratio of power used is 0.36:1 .


28. In  the given  figure what  is  ratio of ammeter  reading when J is connected  to A and then to B

Ans: Connect  J  to A  ,     then


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I=V/R

=3/5

=0.6A

When J is connected to BV=1+2+3+4

=10V

I=10/5

=2A

Ratio of ammeter reading when J is connected to A and then to B is 3:10 .


29. Given  a  resistor  each  of  resistors  R .  How  will  you  combine  them  to  get  the 

(i) maximum effective resistance?  

Ans: For maximum resistance R=nr  , this is the same as combining a series of numbers.

(ii) minimum effective resistance? What is the ratio of the maximum to minimum resistance? 

Ans: For minimum resistance R=r/n  , this is the same as combining a series of numbers.The ratio of the maximum to minimum resistance is R/R=n2.


30. A wire of  length L and resistance R  is stretched so that  its  length  is doubled. How will its

(a) Resistance change 

Ans: The resistance of a wire is determined by its length, cross-sectional area, and resistivity asR = ρlA

Hence, if the length is doubled and area is halved, then we have

R2R1=ρl2A2ρl1A1

=l2A1l1A2

=4

Therefore, R2=4R1

Hence, resistance of the wire becomes four times the original value.


(b) Resistively change?

Ans: The substance from which wire is formed has a property called wire resistivity. As a result, changing the wire's size has no effect on its resistivity.


31. Two students perform the experiments on series and parallel combinations of two given resistors R1  and R2 and plot the following VI graphs. 

Ans: Both students are  correct  because  

AV/A1= resistance R  and  

A1/AV=l/R

The term "series" refers to high resistance, while "parallel" refers to low resistance.


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32. A household uses the following electric appliances. Calculate the electricity bill of the household for the month of June if the cost per unit of electric energy is Rs. 3.00

(i) Refrigerator of rating 4 for ten hours each day.

Ans: Month of June has 30 days. 

Refrigerator of 400 W is running 2 hours each day.

Total hours it is run in 30 days =2×30

=60 h

Energy consumed in kWh is =400 × 60/1000 

= 24 kWh


(ii) Two electric fans of rating 80each for twelve hours each day. 

Ans: Two electric fans of 80 W are run 12 hours each day.

Total hours they are run in 30 days =12×30

=360

Energy consumed in kWh is =2×80×360/1000

=57.6 kWh


(iii) Six electric tubes of rating 18 W each for 6 hours each day. 

Ans: Six electric tubes each of 18 W are run 6 hours daily.

Total hours it is run in 30 days =6×30

=180 h

Energy consumed in kWh is =6×18×180/1000

=19.44  kWh

Net energy consumed in the month of June is =24+57.6+19.44

=101.04 kWh 

Thus, the electric bill is =3×101.04

=Rs303.12


Long Answer Questions (5 Marks)

1. Two wires  A  and  B  are  of  equal  length,  different  cross  sectional  areas  and made  of the same metal. 

(a) (i) Name the property which is same for both the wires, 

Ans: Resistivity - As resistivity is a property of a substance, it is constant for both wires.

(ii) Name the property which is different for both the wires. 

Ans: Resistances - As the cross sectional areas of each wires are different, they are treated as separate objects.


(b) If the resistance of wire A is four times the resistance of wire B , calculate 

(i) the ratio of the cross sectional areas of the wires and 

Ans: Since R=ρlA

For wire A , R1=ρlA1

For wire B , R2=ρlA2

R2R1=A1A2

Since R1=4R2

A1A2=1:4

A1A2=πr12πr22

=(r1r2)2

Ratio is (r1r2)2 .

(ii) The ratio of the radii of the wire. 

Ans: (r1r2)2=1/4

Ratio is 1:2 .


2. (a) State ohm’s law? 

Ans: If the temperature and other physical conditions of the conductor stay constant, the electric current flowing through the conductor is precisely proportional to the potential difference across the conductor's end.

(b) The value of (I) current flowing through a conductor for the corresponding values of (V) potential difference are given below 

I (Amperes)

0.5

1.0

1.5

2.5

3

V (Volts)

1

2

3

4.5

5

Plot a graph between V and I and also calculate resistance. 

Ans: Along x -axis IV=1 cm 


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R=5/3

=1.67

The resistance  is 1.67Ω .


3. (a) Define electrical energy with S.I. unit? 

Ans: The effort done by a source of electricity to sustain current in a circuit is known as electrical energy. The joule is its SI unit.

(b) A household uses the following electric appliance; Calculate the electricity bill of the household for the month of June if the cost per unit of electric energy is Rs. 3.00

(i) Refrigerator of rating 400 w for ten hour each day. 

Ans: Electricity consumed by refrigerator in one day = power × time

=400×10

= 4000 Wh

=4 kwh

Therefore the electricity consumed is 4  KWh. 

(ii) Two electric fans of rating 80 w each for twelve hours each day. 

Ans: Electricity consumed by 2 electric fans in 1 day, power × time

=2×80×12

= 1.92 kwh

Therefore the electricity consumed is 1.92 KWh. 

(iii) Six electric tubes of rating 18 w each for 6hours each day. 

Ans: Electricity consumed by 6 electric tubes in 1 day =6×18×6

= 0.648 kwh

Therefore the electricity consumed is 0.648 KWh. 

Total energy consumed in one day =4+1.92+0.648

=6.548 kwh

Total energy consumed in one month =6.568×30

=197.04 kwh

Cost of 1 unit (kwh) = Rs 3.00

Cost of 197.04 kwh =197.04×3

Electricity bill =Rs591.12

The electricity bill of the household for the month of June is Rs. 591.12.


4. Redraw  the  circuit  of  question 1 ,  putting  in  an  ammeter  to  measure  the  current through the resistors and a voltmeter to measure the potential difference across the 12Ω resistors. What would be the reading in the ammeter and voltmeter? 

Ans:


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Ammeter A has been joined in series with circuit and voltmeter V is joined in parallel to 12 ohms resistor.

Total voltage of battery V=3x2

=6 V.

Total resistance R=R1+R2+R3

=5Ω+8Ω+12Ω

=25Ω

Ammeter reading =I

=V/R

=6/25

=0.24 A.

Voltmeter reading = IR

=0.24 x 12

=2.88 V.

The reading in the ammeter and voltmeter is 0.24 A and 2.88 V respectively.


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

Ans: The following are the benefits of connecting electrical equipment in parallel with the battery rather than in series:

(i) Each connecting electrical device will have the same voltage, and the device will take current according to its resistance.

(ii) It is possible to use separate on/off switches.

(iii) As the total resistance in the parallel circuit falls, a large current can be pulled from the cell.

(iv) Even if one electrical gadget is broken, other devices continue to function normally.


6. How  can  three  resistors  of  resistance  2Ω,3Ωand  6Ω  be  connected  to  give  a  total resistance of

(a) 4Ω

Ans: If we connect resistance of 3Ω and 6Ω in parallel and resistance of 2Ω is connected in series with the combination, then total resistance of combination is 4Ω .


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(b) 9Ω

Ans: If all the three resistance are joined in parallel the resulting resistance will be 3Ω .


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7. The  value  of  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: From the given data the IV graph is a straight line as shown below: 


seo images

Resistance of resistor R = VAVB/1A1B

= 12 V  6 V/ 3.6 A  1.8 A

= 6V/ 1.8 A 

= 3.3 Ω


8. Explain the following: 

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

Ans: For the filament of electric lamps, we need a robust metal with a high melting point. Because of its high melting point, tungsten is utilised only for electric lamp filament.

(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? 

Ans: Electric heating device conductors are composed of an alloy because it has a higher resistance than pure metal and a higher melting point, which prevents it from oxidising at high temperatures.

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

Ans: As the current to all appliances remains constant despite varying resistance, each appliance cannot be turned on or off independently.

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

Ans: As resistance of a wire is inversely proportional to its cross-section area, the resistance will decrease when the area of cross section increases.

(e) Why are copper and aluminum wires usually employed for electric transmission? 

Ans: As copper and aluminium wires are good conductors with low resistance, they are commonly utilised for electrical transmission. They can also be drawn into thin wires since they are ductil

Important Questions of Chapter 11 Class 10 Science - Free PDF Download

Students who are weak in Science and do not have a strong core knowledge might find the chapter electricity quite confusing. This chapter is full of theories and diagrams which confuse students and act as a barrier in their way of achieving good marks. Students must plan to avoid this situation so that they can score the highest possible marks in the final exams. The best way to overcome this problem is a continuous practice. Students can solve some of Class 10 Science Chapter 11 Important Questions regularly. They must induct these practice hours in their preparation schedule. So that every day some time is given to revise and practice. This will make the students efficient and thorough.

Students who are unable to solve the questions must take the help of Important Questions of Electricity Class 10 with Solutions which is available on Vedantu for free. Students can download these questions in PDF format. The downloaded CBSE Class 10 Science Chapter 11 Important Questions act as a guide in preparation for the final exams.

Class 10 Science Ch 11 Important Questions

Students will learn about the concepts and theories of electricity when they will study Chapter 11 of Class 10. They will gain a detailed knowledge of this chapter after practising Important Questions for Class 10 Science Chapter 11. Some of the knowledge that the students will learn are as follows:

Electricity

Electricity is considered as a set of physical phenomena which are associated with the presence and motion of electric charge. It is also believed that electricity is somehow related to magnetism which is why both are part of a phenomenon which is called the phenomenon of electromagnetism. This was described in Maxwell's equations. There are many other phenomena which are related to electricity such as lightning, static electricity, electric heating, electric discharges and many others.

An electric field gets produced, when there is some kind of presence of positive or negative charge. The positive and negative charges are considered as electric charges. When the electric charge moves it is known as Electric current, and it produces a magnetic field nearby.

Attributes of Electricity

There are two primary attributes of electricity: voltage and current. They both have different properties and are quite different from each other, but only in electronic circuits, they are interrelated. Absence of any of these attributes can harm the circuit operation and ruin it. Let's discuss both these attributes:

Voltage:

Voltage is considered as a  force which is responsible for making the current flow in a circuit. Voltage gets measured in volts. In order to get a better understanding of voltage, try to think of a water faucet, in which voltage can get compared to the pressure at which the water is coming out of the faucet. A slow-flowing stream from a faucet resembles a low voltage circuit. In contrast, a high flowing stream resembles a high voltage circuit. A voltage is a mandatory requirement for a current to flow.

Current:

The movement of electrical charges is termed as current. When electrons flow through an electronic circuit, it generates a current. Current is measured in terms of amperes (amp). Here also we can take the example of a water faucet to understand the concept of current. It is assumed that the current is at a high level when more water flows in an hour through the faucet, whereas when the flowing water level is low, the current is also low. After the rain, you must have seen that river flows faster than the usual speed because the current remains high at that point of time as more amount of water passes during rains.

Conductors and Insulators

Another thing that students are going to learn while practising the Electricity Class 10 Important Questions is about the significant classification of elements which is done on the basis of their conductivity of electric charge, i.e. conductors and insulators.

Conductors:

In simple terms, conductors is the name given to those materials which allow electricity to flow through them easily without any difficulties. The conductors' property that makes them capable of allowing electricity to flow through them is termed as conductivity. When electrons start flowing in the conductors, they start to produce an electric current.  The force which is required for the current to move through the conductor is known as the voltage. Some examples of conductors are copper, gold and iron. Let's discuss some properties of a Conductor.

  • Materials which are considered as conductors have a minimal resistance because electricity flows through them.

  • The inductance of the electric conductor occurs when there is a high voltage drop in the conductor.

  • Inside a perfect conductor, the electric field is considered to be zero. It is zero because it helps to keep the electrons calm so that they don’t accelerate.

  • There is no electric charge inside the material which is considered as a conductor.

Insulators

Insulators are considered to be materials within which free flow of electrons from one particle of the element to the other particle is interrupted. If a certain amount of charge is transferred to such an element at a given point of time, then the charge does not get distributed in the surface and remains at the same position. The most common process to charge these elements is by rubbing or charging it through induction. Some examples of insulators are wood, plastic and glass. Let's discuss some properties of insulators:

  • There are no free electrons in such material because all the electrons are tightly held with each other.

  • The ability of these materials to stop the electric current from passing through them is known as resistance.

  • Dielectric length of insulators is vast. Dielectric strength is considered as the maximum electric field that an insulator can handle without suffering an electrical breakdown.

  • High air permeability is a feature of good insulators as they allow air to pass through their pores.

Important Questions for Electricity Class 10

To give the students an overview of the Important Questions of Chapter 11 Class 10 Science, we are listing here some of the questions which are most likely to come in the exams:


What is Electricity?

  • What are the attributes of electricity?

  • Write the SI unit of resistivity.

  • What do you mean by electric current? Mention and define the SI unit of electric current.

  • What do you understand by electrical resistivity of a material? Describe an experiment which will show the factors on which the resistance of a conducting wire depends.

  • What do you understand by a conductor? State its properties.

  • What do you understand by an insulator? State its properties.

  • Differentiate between indicator and conductor.

  • State the difference between electric energy and power.

  • Mention the commercial unit of electric energy.

  • Convert the unit of electric energy into joules.


Benefits of Important Questions for Class 10 Science Chapter 11

  • Class 10 Science Chapter 11 Important Questions offer significant advantages for board exam preparation.

  • Vedantu's team conducts thorough research to compile a list with a high probability of exam inclusion.

  • Expert reviews from seasoned professionals with extensive experience ensure accuracy.

  • Questions adhere to the CBSE board format, providing students insight into exam patterns.

  • Detailed and explanatory solutions accompany these important questions for comprehensive understanding.


Conclusion

These important questions serve as a valuable reference, elucidating key concepts and aiding comprehensive exam preparation. Whether unravelling Ohm's Law or comprehending circuit configurations, these important questions offer understanding. For students grappling with scientific complexities, relying on "Electricity Class 10 Important Questions" proves beneficial, promising improved comprehension and heightened confidence. This resource is not just a tool for academic success but a key to unlocking one's potential in understanding and excelling in the captivating subject of electricity.


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Class 10 Science Chapter 11 Electricity Solutions

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FAQs on Important Questions for CBSE Class 10 Science Chapter 11 - Electricity 2024-25

1. How will Vedantu’s important questions for Class 10 Science Chapter 11 will help you to score well?

Ans: Vedantu’s important questions for Class 10 Science Chapter 11 helps the students by providing an excellent exam strategy to prepare for the board examinations. Science is an essential subject for students who are aspiring for various competitive exams like JEE, NEET, etc in future. Every student aspiring to excel in the board examinations requires deep knowledge and thorough understanding of this subject. Therefore, the important questions for CBSE Class 10 Science help the students in preparing for their exams. All the questions present in these materials are framed under the latest CBSE curriculum and guidelines.

2. Can I download the important questions for Class 10 Science Chapter 11 from Vedantu website for free?

Ans:  Yes, the download option is available on Vedantu website and mobile application for Class 10 Science Chapter 11. Any student can download these study materials from Vedantu website in PDF format at absolutely free of cost. These are high-quality study materials created by our in-house subject matter experts as per the latest CBSE curriculum and guidelines.

3. Define electricity.

Ans: Electricity is one of the most important aspects of our society. Electricity is a set of physical phenomena that has been shaping up our civilization ever since the wake of the industrial revolution all over the entire industries and businesses. Today, life without electricity would be unimaginable and would lead to total chaos if we somehow lose this important source of energy.

4. What are the topics and subtopics covered under this chapter of Class 10 Science?

Ans: Class 10 Science Chapter 11 – Electricity carries at least 8 marks according to the question paper patterns observed in the previous years. The main topics and sub topics covered under this chapter are given below. Take a look.

  • Ohm’s law

  • Resistivity and Resistance

  • Factors that affect the Resistance of a Conductor

  • Parallel and Series Combination of Resistors and their applications

  • Heating Effect of Electric Current and its Applications

  • Electric Power

  • The interrelation between P, V, I and R

5. Define the following terms- 

  • One volt 

  • Potential difference 

Ans: One volt- One volt is defined as the potential difference between two points in a conductor that carries current. Here, one joule of work is done to locomote a charge from one place to the other. It is the SI unit of potential difference. 

Potential Difference- Potential difference is defined as the work done to move a charge from one point to another in a current-carrying conductor. The formula of potential difference is 1 volt- 1 joule/ 1coulomb. A voltmeter is used to measure the potential difference. A voltmeter is connected parallelly in a circuit. 


6. Define resistance. 


Ans: Resistance is defined as the internal property of a substance that offers obstruction to the current flowing through it. The electric resistance of a substance is inversely proportional to its area of cross-section (A) and directly proportional to its length (l). This means when the area will increase the resistance will decrease and vice-versa. And if the length increases, the resistance will also increase. 

The formula is: R=⍴*L/A 

Here R is the resistance, L is the length, A is the Area and ⍴ is a constant. The SI unit of resistance is the ohm.


7. Define resistivity. 


Ans: The resistivity can be defined as the resistance offered by a current-carrying conductor that has an area of cross-section equal to one centimetre square with a length equal to one centimetre. This means it is the resistance of a one-centimetre cube of the current-carrying conductor. The formula for resistivity is ⍴=R*L/A 

Here R is the resistance, L is the length and A is the area. The SI unit of resistivity is ohm*metre. To study more about resistivity, students can download the Important Questions free of cost from the Vedantu website or mobile app.

8. Define ohm’s law. 


Ans: Ohm’s law states that the potential difference (V) across a current-carrying conductor is directly proportional to the current (I) flowing through it. Mathematically, V ∝ I or V= IR. 

Here, V is the potential difference, I is the current and R is a constant called resistance of the conductor. Ohm’s law is followed by both electrolytic conductors as well as metallic objects. 


9. Define Joule’s law of heating. 


Ans: Joule’s law of heating states that the heat generated in a current-carrying resistor is directly proportional to the square of the current, resisting capability of the resistor and time till which the current flows through the resistor. The formula of Joule’s Law of Heating is given as 

H= I square RT. Here, H is the heat produced in the resistor, I is current, R is the resistance and T is the time during which the current flows in the resistor. The SI unit of heat is Joule.