Wednesday, 13 August 2014

Chapter 12. Electricity | CBSE Class X (10th) Science | Solved Exercises

Multiple Choice Questions - MCQs : Chapter 12. Electricity | CBSE Class 10th Science


Question 1. What happens to the resistance of circuit if current through it is doubled ?
(a) Resistance is doubled
(b) Resistance becomes half
(c) Resistance remains same
(d) None

Answer. (c) Resistance remains same

Question 2. The work done in moving a unit positive charge, across two points in an electric circuit is measure of _____ ?
(a) Current
(b) Potential Difference
(c) Resistance
(d) Power



Answer. (b) Potential Difference

Question 3. A electric circuit contains two UN-equal resistance in parallel
(a) Current is same in both
(b) More current flows through larger resistor
(c) Potential difference across each is same
(d) Smaller resistor has less conductance.

Answer. (c) Potential difference across each is same

Question 4. Ohm's law relates potential difference with ______?
(a) Power
(b) Energy
(c) Current
(d) Time

Answer. (c) Current

Question 5. Two bulbs marked 200 watt - 250 volts and 100 watt-250 Volts are joined in series to a 250 Volts supply. Power consumed in the circuit is ____?
(a) 33 Watts
(b) 67 Watts
(c) 100 Watts
(d) 300 Watts

Answer. (b) 67 Watts

Question 6. In what combination, the 3 resistors of 3 Ohms each should be connected to get effective resistance of 1 Ohm ?
(a) Series
(b) Parallel
(c) Any way
(d) None

Answer. (b) Parallel

Question 7. Two light bulbs are marked 230 V- 75 W and 230 V - 150 W, if the first bulb has resistance R, then the resistance of second is __ ?
(a) 4 R
(b) 2 R
(c) 1/4 R
(d) 1/2 R

Answer. (d) 1/2 R

Question 8. A  copper wire has diameter of 0.5 mm and resistivity of 1.6 × 10−8 Ω m. What will be the length of copper wire if the resistance of wire is 10 Ω
(a) 1.227m
(b) 12.27m
(c) 122.7m
(d) 0.1227m

Answer. (c) 122.7m

Question 9. What should be the resistence of a Voltmeter ?
(a) Infinity
(b) Zero
(c) Can not say
(d) None of the above

Answer. (a) Infinity

Question 10. Three equal resistors, when combined in series gives an equivalent resistance of 90 Ω. What will be their equivallent resistance when combined in parallel ?
(a) 270 Ω
(b) 30 Ω
(c) 810 Ω
(d) 10 Ω

Answer. (d) 10 Ω

Question 11. A material which easily allows the flow of electrical charge through it, is called ___
(a) Insulator
(b) Conductor
(c) Semi Conductor
(d) Super Conductor

Answer. (b) Conductor

Question 12. Which property of electricity is responsible for use of fuse wire in household wiring ?
(a) Chemical effect
(b) Magnetic effect
(c) Heating effect
(d) All of above

Answer. (c) Heating effect

Question 13. A piece of wire of resistance R is cut into 4 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) 1/16
(b) 1/4
(c) 4
(d) 16

Answer. (d) 16

Question 14. When electric current is passed through a bulb, the bulb gives light because of
(a) Electric effect of current
(b) Magnetic effect of current
(c) Lighting effect of current
(d) Heating effect of current

Answer. (d) Heating effect of current

Question 15. Resistivity of a wire depends upon __
(a) Length of the wire
(b) Material of the wire
(c) Cross-section area of the wire
(d) All of the above

Answer. (d) All of the above

Question 16. The SI unit of resistivity is _
(a) Ω m
(b) Ω / m
(c) mho
(d) None

Answer. (a) Ω m (Ohm - m)

Question 17. The metals and alloys have very low resistivity, in the range of ___
(a) 10–4 Ω m to 10–2 Ω m
(b) 10–6 Ω m to 10–4 Ω m
(c) 10–8 Ω m to 10–6 Ω m
(d) 10–10 Ω m to 10–8 Ω m

Answer. (c) 10–8 Ω m to 10–6 Ω m

Question 18. Insulators like rubber and glass have very high resistivity, in the range of __
(a) 102 Ω m to 107 Ω m
(b) 104 Ω m to 1011 Ω m
(c) 1022 Ω m to 1027 Ω m
(d) 1012 Ω m to 1017 Ω m

Answer. (d) 1012 Ω m to 1017 Ω m

Question 19. Two electric bulbs have resistance in the ratio of 1:2, if they are joined in series, the ratio of their energy consumption will be __
(a) 1:2
(b) 2:1
(c) 4:1
(d) 1:1

Answer. (b) 2:1

Question 20. A device used for measuring potential difference is known as ______
(a) Potentiometer
(b) Ammeter
(c) Voltmeter
(d) Galvanometer

Answer. (c) Voltmeter

True & False : Chapter 12. Electricity | CBSE Class 10th Science


Mark the below given statements as True or False

Question 1. Electric current can flow through metals

Answer. True

Question 2. Conductance is the property of a conductor to resist the flow of charges through it

Answer. False

Question 3. SI unit of electrical Potential Difference is Ampere (A)

Answer. False

Question 4. A voltmeter is used for measuring the electric current in the circuit.

Answer. False

Question 5. Resistance of conductor does not depend upon its length

Answer. False

Question 6. One volt is a potential difference between two points in current carrying conductor, when 1 joule of work is required to be done, in moving 1 coulomb of charge from one point to another.

Answer. True

Question 7. Electric cell is source of electrical energy
Answer. True

Question 8. The passage of electric current through a solution causes chemical reaction

Answer. True

Question 9. When current is passed through copper sulphate solution, the free copper gets accumulated on the electrode connected to positive terminal of the battery

Answer. False

Question 10. In an electrical circuit, potential difference (V) across the ends of given metallic wire, is inversely proportional to current flowing through it, provided its temperature remains the same. This is called Ohm's law.

Answer. False

Question 11. Materials which do not allow electric current to pass through them are called insulators

Answer. True

Question 12. In an electric circuit, the direction of the flow of current is from negative to positive terminal of the electrical cell.

Answer. False

Question 13. The filament of an electrical lamp is heated to such a high temperature that it starts glowing.

Answer. True

Question 14. The ammeter reading is reduced to approximately one half when length of the wire in the circuit is doubled.

Answer. True

Question 15. The process of depositing layer of any desired metal on another material by means of electricity is called electroplating.

Answer. True

Question 16. S.I. unit of electrical charge is Coulomb (C), which is equivalent to charge contained in nearly 6 ×1018 electrons

Answer. True

Question 17. The main disadvantage of a series circuit is that, when any one of the components breaks down, all other components stop working

Answer. True

Question 18. For the flow of charge in conducting metallic wire, the gravity has no role to play, the electrons move only if there is difference of electric pressure - called potential difference, along the conductor

Answer. True

Question 19. When electric current is passed through copper sulphate solution, the copper gets transferred from one electrode to other.

Answer. True

Question 20. Electrical current can pass through glass sheet.

Answer. False

Question 21. Pure water is good conductor of electricity

Answer. False

Question 21. Electrical Fuses are used in the circuit or appliances for lighting and heating purpose

Answer. False

Question 22. Conventionally, in an electrical circuit, the direction of electrical current is taken in the same direction as the flow of the electrons, which are negatively charged.

Answer. False

Question 23. Instead of a metallic wire a cotton thread can be used to make a circuit

Answer. False

Question 24. The metals and alloys have low electrical resistivity

Answer. True

Question 25. In order to maintain the electric current in the given circuit, the electric cell makes use of mechanical energy stored in it.

Answer. False

Question 26. ρ (rho) is a constant of proportionality and is called the electrical resistivity of the material of the conductor.

Answer. True

Question 27. The resistivity of an alloy is generally higher than that of its constituent metals

Answer. True

Question 28. Alloys do not oxidise (burn) readily at high temperatures. For this reason, they are commonly used in electrical heating devices, like electric iron, toasters etc

Answer. True

Question 29. The voltmeter is always connected in series across the points between which the potential difference is to be measured.

Answer. False

Question 30. The ammeter reading is increased when a thicker wire of the same material and of the same length is used in the circuit.

Answer. True

Question 31. When more than one cell is connected to each other, the combination is called a battery.

Answer. True

Question 32. when several resistors are joined in series, the resistance of the combination Rs equals the sum of their individual resistances

Answer. True

Question 33 Chromium plating is done on many object, because it has shiny appearance, it does not corrode, it resists scratches.

Answer. True

Question 34. Total potential difference across a combination of resistors in series is equal to the sum of potential difference across the individual resistors.
That is, V = V1 + V2 + V3


Answer. True

Question 35. A wire gets hot, when electric current is passed through it. This is due to magnetic effect of current

Answer. False

Solved Exercises : Chapter 12. Electricity | CBSE Class 10th Science


Question 1. 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

Answer. (d) 25

Question 2. Which of the following terms does not represent electrical power in a circuit?
(a) I2R
(b) IR2
(c) VI
(d) V2/R

Answer. (b) IR2

Question 3. 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

Answer. (d) 25 W

Explanation :
V2
Electric Power P=
R
V2
R=
P
220 ×220
=
100
R=484 Ω
Now, When this bulb with Resistance 484 Ω is operated at 110 V
V2
Electric Power P=
R
110 × 110

=
484
Power consumed by bulb=25 W
Question 4. 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 combinations would be –
(a) 1:2
(b) 2:1
(c) 1:4
(d) 4:1

Answer. (c) 1:4

Explanation :
Suppose the conducting wire has resistance R Ω
When ther wires are connected in series :
Equivalent resistance R(series)=R + R= 2R
When ther wires are connected in parallel :
R ×R R2R
Equivalent resistance R(parallel)=
=
=
R + R  2R2
When same potential difference (V) is applied across series and parallel combination of resistors, let the the Power produced in the form of heat be P(series) and P(parallel)
V2

P(series)R(series)V2R(parallel)R / 21

=
=
×
=
=
P(parallel)V2R(series)V22R4

R(parallel)


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

Answer. Voltmeter is connected in parallel in the circuit to measure the potential difference between two points.

Question 6. A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10–8 Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?

Answer. As we know :
l
R=ρ
A
Where;
R=Resistance of conductor
ρ=Electrical resistivity of copper
l=Length of conductor
A=Area of cross-section of conductor
R=10 Ω
Diameter of wire=0.5 mm = 5/(10×1000) m
Radius of wire r=(5 × 10−4)/2
=2.5 × 10−4 m
=25 × 10−5 m
A=Πr2
=(22/7)(25 × 10−5 m )2
=3.14 ×625 × 10−10 m2
=1962.50 × 10−10 m2
=19.625 × 10−8 m2
RA
l=
ρ
10 × 19.625 × 10−8
l=
m
1.6 × 10− 8
196.25
l=
m
1.6
l=122.65625 m
Hence; the required length of wire l is 122.65625 m
When the diameter 'D' is doubled, the radius 'r' will become 2r and its area A' will increase by 4 times and will become 4A.
A'=Π(2r)2
∴ A'=4× Πr2 = 4 × A
Since, as we know :
l
R
A
∴ Resistance of wire with double the diameter
l
R
4A
Hence the resistance of the wire will be reduced to 1/4 th of the earlier, if diameter of the wire is doubled

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

I (amperes)0.51.02.03.04.0
V (volts)1.63.46.710.213.2

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


Answer. Let us plot given values of Potential difference 'V' along Y-axis and that of current 'I' along X-axis.

V (volts)I (amperes)R(resistance) =V/I
1.60.53.2
3.41.03.4
6.72.03.35
10.23.03.4
13.24.03.3
While plotting a graph with given corresponding values of V and I, we have found that approximately the same value for V/I = 3.33 (approx.) is obtained in each case. Thus the V–I graph is a straight line that passes through the origin of the graph, as shown in Fig., V/I is a constant ratio.
A straight line plot shows that as the current through a wire increases, the potential difference across the wire increases linearly – this is Ohm’s law.

Question 8. 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.

Answer. As we know, according to Ohm's law
V
R=
I
V=12 Volts
I=2.5 mA
=2.5×10−3 A

12
R=
Ω
2.5×10−3

12×10−3
R=
Ω
2.5
R=4.8 × 103Ω


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

Answer. When resistors are connected in series, the strength of current 'I' passing through each resistor is same.
REq. = R1+R2+R3+R4+R5

REq. = 0.2+0.3+0.4+0.5+12 Ω = 13.4 Ω

V9V90
∴ I =
=
=
REq.13.4 Ω134
I = 0.67 Ampere
∴ value of current passing through 12 Ω resistor as well others = 0.67 Ampere.


Question 10. How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?

Answer.
Let 'n' be the number of resistances, connected in parallel to 220 V to draw a maximum of 5 A current
111

=
+
+... n times
REq.176176
176
REq.=
Ω
n

Now, according to Ohm's law :
V
R=
 
I

V
Or   I=
 
R

220 × n
5 A=
 
176

5 ×176880
n=
=
=4
220220
Hence, Number of Resistors, which can be connected in parallel across 220 V line to carry 5 A of current is 4


Question 11. Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω, (ii) 4 Ω.

Answer. (i) To get a combination resistance of 9 Ω, we can connect the given three resistance of 6 Ω each as given below in the diagram :

As shown, a parallel combination of two resistance is connected in series with the third resistance.
R1×R2
REq=
+R3
R1 + R2

6×6
REq=
+6
6 + 6

36
REq=
+6
12

REq=3+6=9 Ω
(ii) To get a combination resistance of 4 Ω, we can connect the given three resistance of 6 Ω each as given below in the diagram :

As shown, a series combination of two resistance is connected in parallel with the third resistance.
111

=
+
REqR1 + R2R3

(R1+ R2) × R3
REq=
(R1 + R2)+R3

(6+ 6) × 612×6
REq=
=
=4 Ω
(6 + 6 ) + 618



Question 12. 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?

Answer.
Voltage applied across the bulbs = 220 V
Maximum allowable current = 5 A
Power rating of each bulb W = 10 W
Let R be resistance of each bulb
V
R=
I
P
I=
V
V2
∴ R=
P
220 × 220
∴ R=
10
∴ R=4840 Ω

Let 'n' be the number of bulbs, connected in parallel to 220 V to draw a maximum of 5 A current
111

=
+
+... n times
REq.48404840

4840
REq.=
Ω
n

Now, according to Ohm's law :
V
R=
 
I

V
Or   I=
 
R

220 × n
5 A=
 
4840

5 ×484024200
n=
=
=110
220220
Hence, Number of bulbs which can be connected in parallel with each other across the two wires of 220 V line when the maximum allowable current is 5 A = 110


Alternate Solution: !!
Vlotage applied across the bulbs = 220 V
Maximum allowable current = 5 A
Power rating of each bulb W = 10 W
Max. Power capacity of the line , P = VI
P= 220 × 5 = 1100 W
let 'n' be the Numbers of bulbs of 10 Watt each which can be connected in parallel
Therefor n= Total Power Capacity of line÷Power Rating of each bulb
n= 1100÷10 = 110
Hence 110 bulbs can be connected in parallel across the line



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

Answer.
Case 1. When electric oven is connected with just one resistance coil
V220
I1=

=9.2 A
R24

Case 2. When electric oven is connected with two resistance coils in series
∴ R = 24 + 24 = 48 Ω

V220
I2=
=
=4.6 A approx.
R48

Case 3. When electric oven is connected with two resistance coils in parallel
24 × 2424 × 24
R=
=
=12 Ω approx.
24 + 2448
V220
I3=
=
=18.3 A approx.
R12


Question 14. Compare the power used in the 2 Ω resistor in each of the following circuits: (i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and (ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.

Answer.
Case 1. : When resistor is connected in series :

 
V=6 Volt
R=1 + 2 = 3 Ω
V6
I=
=
R3
I=2 A
Let P1 be power used in 2 Ω resistor
P1=VI=6×2 = 12 W
Case 2. : When resistor is connected in parallel :

 
V=4 Volt
R1×R2
R=
R1+R2
 
12×2
R=
12+2
R=1.71 Ω
V4
I=
=
R1.7
I=2.34 A
Let P2 be power used in 2 Ω resistor
P2=VI=4×2.34= 9.36 W
Comparison of power used : Ratio
P1:P1::12 : 9.36
P1:P1::1 : 0.78


Question 15. 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?

Answer. As given,
Power rating of first lamp=100 W
Power rating of second lamp=60 W
Voltage applied across lamps=220 V
Let, Resistance of first lamp=R1
Resistance of second lamp=R2
V2
As we know, R=
P
220 ×220
R1=
= 480 Ω
100
220 ×220
R2=
= 806.66 Ω
60
When R1 and R2 are connected in parallel, Let be REq equivalent resistance and I be current drawn through R1 and R2 connected in parallel.
111

=
+
REqR1R2
R1×R2
REq=
R1+R2 
484×806.66
REq=
Ω
484+806.66 
390423.44
REq=
= 302.5 Ω
1290.66 
V220
I=
=
R302.5
I=0.727 A
 


Question 16. Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?

Answer.We know that:-
Energy = Power × Time
For TV set :
E1=250 W × 1 × 3600 sec
=900000 J = 9 × 105 J
For Toaster :
E2=1200 W × 10 × 60 sec
=720000 J = 7.2 × 105 J
=>E1>E2
∴ TV Set uses more power.


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

Answer.

Heat energy developed in the heater= I2Rt
Rate of Heat Energy= Power
P= I2Rt / t
P= I2R
P = 152 × 8
P = 1800 Watt
∴ Heat is developed at the rate of 1800 W or 1.8 KW or 1800 joule per sec.

Question 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?


Answer.
(a) Tungsten offers high resistance against electric current and it has high melting point. Mainly for these two characteristics, Tungsten is used almost exclusively for filament of electric lamps. When the electric current flows through Tungsten filament, due to high resistance of Tungsten, it gets heated up to a very high temperature and starts glowing with light. High melting point of Tungsten prevents it from melting away at higher temperature.

(b) The conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal because an alloy has certain favorable characteristics, which are as follows :
  1. It offers high resistance against an electric current
  2. It can withstand high temperature due to high melting poin.
  3. It has high density
As per Joule’s law of heating,
H = I2Rt
The that heat produced in a resistor is (i) directly proportional to the square of current for a given resistance
(ii) directly proportional to resistance for a given current, and (iii) directly proportional to the time for which the current flows through the resistor.
Higher resistance, high melting point and higher density of an alloy in conductors of electric heating devices, all to gather, help in producing a large amount of heat at higher temperature, at the same time offsetting any deformation arising out of temperature variations.

(c) When appliances are connected in series, the value of equivalent resistance will be very large, as shown below :-
Req = R1 + R2 + R3 ......
Due to resulting higher value of Equivalent Resistance, there will be a considerable drop in line voltage and at the same time a large amount of heat will be produced in the domestic circuit As per Joule’s law of heating,
H = I2Rt
This may result in either loss of electric energy due to heating or melting away of electric wiring in domestic circuit due to over heating. Any fault in a series arrangement results in complete shutdown of other devices in circuit. When appliances are connected in parallel, the value of equivalent resistance will be very small hence damage or energy losses due to heating will be minimum. Also break down of a certain device does not affect others in the parallel circuit
This is the main reason, why the series arrangement is not used for domestic circuits

(d) As we know :
l
R=ρ
A
Where;
R=Resistance of conductor
ρ=Electrical resistivity of copper
l=Length of conductor
A=Area of cross-section of conductor
l
R
A
l
=>    R
Πr2
l
=>    R
r2
From the above, it is clear that as the radius of cross sectional area of a conductor increases, the resistance of conductor decreases and vice-versa

(e) Certain elements like Silver, Copper and Aluminum are the best conductors of electricity due to theirs low electric resistivity. Hence, voltage drop and heating loss due to resistance against electric current is minimum. Silver being too costly, Copper and Aluminum wires are usually employed for electricity transmission

Intext Questions|Page 200 |Chapter 12. Electricity | CBSE Class 10th Science

Question 1.

Answer :

Question 1.What does an electric circuit mean?

Answer : An electrical circuit essentially consists of load devices, connected together by conducting material such as wire and electric power source. The naure of load devices, depending upon the application, may be either Resistive or Inductive or Capacitive .

Question 2. Define the unit of current.

Answer : The rate of flow of electric charges, flowing through a particular area of a conductor is called Electric current. S.I. unit of Electric current is ampere (A). One ampere of current represents a flow of one coulomb of charge, passing through a particular area of a conductor, per second. Coulomb is the SI unit of electric charge and one coulomb (C) of charge is equivalent to the charge contained in nearly 6 × 1018 electrons

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

Answer :
We know that an electron possesses a negative charge of 1.6 × 10–19 C
Let 'n' be the number of electrons present in one coulomb of charge

Number of electrons in Coulomb × Charge per Electron = One coulomb of charge

∴ n ×1.6 × 10–19 C = 1 C


11019100×1018
n=
=
=
1.6 × 10–191.616

n = 6.25 × 1018 electrons


Intext Questions|Page 202 |Chapter 12. Electricity | CBSE Class 10th Science

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

Answer : Electric cell is the device that helps to maintain a potential difference across a conductor. The chemical action within a cell generates the potential difference across the terminals of the cell, even when no current is drawn from it. When the cell is connected to a conducting circuit element, the potential difference sets the charges in motion in the conductor and produces an electric current

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

Answer : One volt is the potential difference between two points in a current carrying conductor when 1 joule of work is done to move a charge of 1 coulomb from one point to the other. Volt (V) is the SI unit of electric potential difference

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

Answer : We define the electric potential difference (V) between two points in an electric circuit carrying some current as the work (W) done or Energy in joule to move a unit charge(Q) from one point to the other –

Potential difference (V) between two points = Work done (W)/Charge (Q)
V=
W
Q

W = V×Q

W = 6×1

W = 6 joules



Intext Questions|Page 209 |Chapter 12. Electricity | CBSE Class 10th Science

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

Answer : Resistance of the conductor depends on following factors :

(i) Length of conductor (l)
: Resistance of a uniform metallic conductor is directly proportional to its length

R ∝ l

(ii) The Cross-section area of the conductor (A) :
Resistance of a uniform metallic conductor is inversely proportional to the area of cross-section (A).

R
1
A
(iii) Nature of conducting material or its electrical resistivity (ρ):
Different materials exhibit different amount of electrical resistivity. It is denoted by ρ (rho) and also know as constant of proportionality.
R =ρ
l
A


Question 2. Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?

Answer : As we know :
Current (I)
1
R

Also
R
1
A

∴ Current (I)A (Area of cross-section)
The current is inversely proportional to Resistance and further Resistance of a uniform metallic conductor is inversely proportional to the area of cross-section (A). Which means more the area of cross section, lesser will be the resistance and therefor more will be the current flow. Hence, current Will flow more easily through a thick wire than a thin wire of the same material, when connected to the same source

Question 3. Let the resistance of an electrical component remains 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?

Answer : As we know :
V (Potential difference) = Current (I) × Resistance (R)

As the Resistance (R) is constant.

When V (Potential difference) decreases to half of its former value

ILater=
VLater
R
ILater=
VFormer
2R
=
IFormer
2
Clearly, Resistance being constant, as potential difference decreases to half of its former value, the value of current (I) will also be reduced to half.


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

Answer : The resistivity of an alloy is generally higher than that of its constituent pure metals. Higher resistance results in increased electrical heating. Alloys do not oxidise (burn) easily at high temperatures. For this reason, they are commonly used in electrical heating devices, like electric iron, toasters etc

Question 5. Use the data in Table 12.2 to answer the following –
(a) Which among iron and mercury is a better conductor?
(b) Which material is the best conductor?


Answer : (a) Iron is a better conductor than mercury as its Electrical resistivity is less (10.0 × 10–8Ω m), whereas that of mercury is very high.( 94.0 × 10–8 Ω m)

(b) Silver is the best conductor of electric current because of its Electrical resistivity is 1.60 × 10–8 Ω m, which is least of all other metals.

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