
01. Chemical Reactions
8
Lecture1.1

Lecture1.2

Lecture1.3

Lecture1.4

Lecture1.5

Lecture1.6

Lecture1.7

Lecture1.8


02. Acids, Bases and Salts
12
Lecture2.1

Lecture2.2

Lecture2.3

Lecture2.4

Lecture2.5

Lecture2.6

Lecture2.7

Lecture2.8

Lecture2.9

Lecture2.10

Lecture2.11

Lecture2.12


03. Metals and Non  metals
11
Lecture3.1

Lecture3.2

Lecture3.3

Lecture3.4

Lecture3.5

Lecture3.6

Lecture3.7

Lecture3.8

Lecture3.9

Lecture3.10

Lecture3.11


04. Periodic Classification of Elements
6
Lecture4.1

Lecture4.2

Lecture4.3

Lecture4.4

Lecture4.5

Lecture4.6


05. Life Processes  1
9
Lecture5.1

Lecture5.2

Lecture5.3

Lecture5.4

Lecture5.5

Lecture5.6

Lecture5.7

Lecture5.8

Lecture5.9


06. Life Processes  2
6
Lecture6.1

Lecture6.2

Lecture6.3

Lecture6.4

Lecture6.5

Lecture6.6


07. Control and Coordination
9
Lecture7.1

Lecture7.2

Lecture7.3

Lecture7.4

Lecture7.5

Lecture7.6

Lecture7.7

Lecture7.8

Lecture7.9


08. How do Organisms Reproduce
9
Lecture8.1

Lecture8.2

Lecture8.3

Lecture8.4

Lecture8.5

Lecture8.6

Lecture8.7

Lecture8.8

Lecture8.9


09. Heredity and Evolution
9
Lecture9.1

Lecture9.2

Lecture9.3

Lecture9.4

Lecture9.5

Lecture9.6

Lecture9.7

Lecture9.8

Lecture9.9


10. Light (Part 1) : Reflection
8
Lecture10.1

Lecture10.2

Lecture10.3

Lecture10.4

Lecture10.5

Lecture10.6

Lecture10.7

Lecture10.8


11. Light (Part 2) : Refraction
6
Lecture11.1

Lecture11.2

Lecture11.3

Lecture11.4

Lecture11.5

Lecture11.6


12. Carbon and Its Compounds
9
Lecture12.1

Lecture12.2

Lecture12.3

Lecture12.4

Lecture12.5

Lecture12.6

Lecture12.7

Lecture12.8

Lecture12.9


13. The Human Eye and The Colorful World
9
Lecture13.1

Lecture13.2

Lecture13.3

Lecture13.4

Lecture13.5

Lecture13.6

Lecture13.7

Lecture13.8

Lecture13.9


14. Electricity
8
Lecture14.1

Lecture14.2

Lecture14.3

Lecture14.4

Lecture14.5

Lecture14.6

Lecture14.7

Lecture14.8


15. Magnetic Effect of Current
10
Lecture15.1

Lecture15.2

Lecture15.3

Lecture15.4

Lecture15.5

Lecture15.6

Lecture15.7

Lecture15.8

Lecture15.9

Lecture15.10


16. Sources of Energy
5
Lecture16.1

Lecture16.2

Lecture16.3

Lecture16.4

Lecture16.5


17. Our Environment
6
Lecture17.1

Lecture17.2

Lecture17.3

Lecture17.4

Lecture17.5

Lecture17.6


18. Management of Natural Resources
4
Lecture18.1

Lecture18.2

Lecture18.3

Lecture18.4

NCERT Solutions – Electricity
Intext Questions
Q.1 What does an electric current mean?
Sol. The flow of electric charge is known as electric current. Electric current is carried by moving electrons through a conductor. Electric current flows in opposite direction to the movement of electrons.
Q.2 Define the unit of current.
Sol. SI unit of electric current is ampere (A).
Ampere is the flow of electric charges through a surface at the rate of one coulomb per second, i.e. if 1 coulomb of electric charge flows through a cross section for 1 second, it would be equal to 1 ampere.
Therefore, 1 ampere = 1C1s
Q.3 Calculate the number of electrons constituting one coulomb of charge.
Sol. We know that charge over 1 electron = 1.6 × 10^{–19} coulomb
Thus, 1.6 × 10^{–19} C of charge = 1 electron
Therefore, 1 C of charge = 11.6×10−19 Electrons
= 10191.6 electrons = 10×10181.6 electrons = 6.25 × 10^{18} electrons
Q.4 Name a device that helps to maintain a potential difference across a conductor.
Sol. Battery or a cell
Q.5 What is meant by saying that the potential difference between two points is 1 V?
Sol. This means 1 joule of work is done to move a charge of 1 coulomb between two points.
Q.6 How much energy is given to each coulomb of charge passing through a 6 V battery?
Sol. Given, Charge Q = 1C, Potential difference, V = 6V
Therefore, Energy i.e. Work done, W =?
We know that, V = WQ
Therefore, 6V = W1C
⇒ W = 6V × 1C = 6J
Thus, required energy = 6J
Q.7 On what factors does the resistance of a conductor depend?
Sol. Resistance of a conductor depends upon:
(a) Nature of conductor
(b) Length of conductor
(c) Area of cross section of conductor
Q.8 Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?
Sol. Since, resistance is indirectly proportional to the area of cross section, thus current flows easily through a thick wire compared to a thin wire of the same material.
Q.9 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. What change will occur in the current through it?
Sol. Since Resistance (R) = PotentialDifference(V)Electriccurrent(I)
Therefore, if potential difference between two ends of the component will be halved, and resistance remains constant, then electric current would also be halved.
Q.10 Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
Sol. Since, alloys have higher melting point than pure metal so coils of electric toasters and electric irons are made of an alloy rather than a pure metal to retain more heat without melting.
Q.11 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?
Sol. (a) Iron
(b)Silver
Q.12 Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Ω resistor, an 8 Ω resistor, and a 12 Ω resistor, and a plug key, all connected in series.
Sol.
Q.13 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 12 Ω resistor. What would be the readings in the ammeter and the voltmeter?
Sol.
The total resistance in the circuit = Sum of the resistances of all resistors
= 5 Ω + 8 Ω + 12 Ω = 25 Ω
We know;
R=VI or, 25Ω=6VI or, I=6V25Ω=0.24A
Since, resistances are connected in series, thus electric current remains the same through all resistors.
Here we have,
Electric current, I = 0.24A
Resistance, R = 12Ω
Thus, potential difference, V through the resistor of 12Ω = I x R
Or, V = 0.24A x 12Ω = 2.88 V
Thus, reading of ammeter = 0.24A
Reading of voltmeter through resistor of 12Ω = 2.88V
Q.14 Judge the equivalent resistance when the following are connected in parallel –
(a) 1 Ω and 106 Ω, (b) 1 Ω and 103 Ω, and 106 Ω.
Sol.
Since 1R=1R1+1R2+1R3+..+1Rn
when resistors are connected in paralle
(a) 1 Ω and 106 Ω
Thus, 1R=11Ω+1106Ω=106+1106Ω=107106Ω
Thus, R=106107Ω=0.99Ω
Thus, equivalent resistance of 1Ω and 106Ω are connected in parallel = 0.99Ω
(b) 1 Ω and 103 Ω, and 106 Ω
Thus, 1R=11Ω+1103Ω+1106Ω=10918+106+103103×106Ω=1112710918Ω
Thus, R=1091811127Ω=1.02Ω
Thus, equivalent resistance of 1 Ω, 103 Ω and 106 Ω are connected in parallel = 1.02Ω
Q.15 An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω are connected in parallel to a 220 V source.
What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?
Sol.
Given:
R1 = 100Ω, R2 = 50 Ω, R3 = 500 Ω all are connected in parallel
Potential difference = 220V
Thus, 1R=1100Ω+150Ω+1500Ω=5+10+1500Ω=16500Ω
Therefore, R=50016Ω=31.25Ω
Electric current (I) through the circuit = VR
⇒I=220V31.25Ω=7.04A
For electric iron
Since it takes as well current as three appliances, thus electric current through it = 7.04A
The electric current = 7.04 A and potential difference = 220 V
Thus, Resistance of electric iron = Total resistance of three appliances = 31.25 Ω
Thus, electric current through the electric iron = 7.04A
Resistance of electric iron = 31.25 Ω
Q.16 What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
Sol.
Advantages of connecting electrical appliances in parallel instead of connecting in series:
(a) Voltage remains same in all the appliances.
(b) Lower total effective resistance
Q.17 How can three resistors of resistances 2 Ω, 3 Ω, and 6 Ω be connected to give a total resistance of (a) 4 Ω, (b) 1 Ω?
Sol.
(a) When resistors having resistance equal to 3 Ω and 6 Ω are connected in parallel and one having resistance equal to 2 Ω is connected in series
Let total resistance due to resistors having resistance equal to 6 Ω and 3 Ω = R_{1
}Therefore, 1R1=16Ω+13Ω=1+26Ω=36Ω=12Ω
Thus R1=2Ω
Now, total effective resistance in the circuit = R_{1} + 2 Ω = 2 Ω + 2 Ω = 4 Ω
Hence, when resistors having resistance equal to 3 Ω and 6 Ω are connected in parallel and one having resistance equal to 2 Ω is connected in series, then the total effective resistance in the circuit = 4 Ω
(b) When all the three resistance is connected in parallel then
1R=12Ω+13Ω+16Ω=3+2+16Ω=66Ω=1Ω
Thus, R=1Ω
When all the three resistance will be connected in parallel, then the total effective resistance in the circuit = 1 Ω
Q.18 What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?
Sol.
When all the resistors are connected in series
Then the 1R=14Ω+18Ω+112Ω+124Ω=6+3+2+124Ω=1224Ω=12Ω
Thus, R = 2 Ω
(b) When all the resistors are connected in series
Thus, total resistance = 4 Ω + 8 Ω + 12 Ω + 24 Ω = 48 Ω
Thus, highest resistance = 48 Ω
Lowest resistance = 2 Ω
Page Number – 218
Q.19 Why does the cord of an electric heater not glow while the heating element does?
Sol.
The resistance of heating element is very high compared to the cord of an electric heater. Since, heat produced is directly proportional to the resistance, thus element of electric heater glows because of production of more heat and cord does not.
Q.20 Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.
Sol.
Given, Potential difference (V) = 50V,
Charge (Q) = 96000 coulomb
Time, t = 1 hour = 60 X 60 s = 3600 s
Heat produced =?
We know that electric current (I) = Qt=96000C3600s=960003600A
We know that heat produced (H) in the given time (t) = VIt
H=50V×960003600A×3600=50×96000J=4800000J=4.8×106 Joule
Q.21 An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s.
Sol.
Given, Electric current (I) = 5A, Resistance (R) = 20 Ω, Time (t) = 30s
Therefore, Heat produced (H) =?
Since, V=I×R=5A×20Ω=100V
We know that, Heat produced (H) = VIt
⇒H=100V×5A×30s=15000J=1.5×104J Answer
Q.22 What determines the rate at which energy is delivered by a current?
Sol.
Since electric power is the rate of consumption of electric energy in any electrical appliance. Hence, rate at which energy is delivered by a current is the power of electric appliance.
Q.23 An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.
Sol. Given, Electric current (I) = 5A,
Potential difference (V) = 220V,
Time (t) = 2h = 2 x 60 x 60 s = 7200 s
Power (P) =?
Energy consumed =?
We know that, P = VI = 220V × 5A = 1100W
We know that energy consumed by the electric appliance = P × t
⇒ Energy consumed = 1100 W × 7200 s = 7920000 J = 7.92 × 10^{6} J
Exercise
Q.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
Sol. (d) 25
Solution:
The piece of wire having resistance equal to R is cut into five equal parts.
Therefore, resistance of each of the part will be equal to R5
When all parts are connected in parallel, then equivalent resistance is = R’ as given
Therefore, 1R′=5R+5R+5R+5R+5R=5+5+5+5+5R=25R
Or, R′=R25
Thus, RR′=RR25=R×25R=25
Q.2 Which of the following terms does not represent electrical power in a circuit?
(a) I^{2}R (b) IR^{2} (c) VI (d) V^{2}/R
Sol. (b) IR^{2}
Solution:
We know that Power (P) = VI
After substituting the value of V = IR in this we get
P = (IR) I = I x R x I = I^{2}R, Thus P = I^{2}R
Since I=VR, therefore, after substituting this value in above equation we get P=(VR)2 R=V2R2=V2R
Thus, P cannot be expressed as IR^{2}
Q.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
Sol. (d) 25 W
Solution: Given,
Potential difference, V = 220V, Power, P = 100W
Therefore, power consumption at 100V =?
To solve this problem, first of all resistance of the bulb is to be calculated.
We know that P=V2R
Therefore, 100W=(220V)2R ⇒ R=220×220100Ω=484Ω
Now, when bulb is operated at 110V
The power consumption P=V2R=(110V)2484Ω=12100V2484Ω=25W
Thus, bulb will consume power of 25W at 110V
Q.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
Sol. (c) 1 : 4
Solution:
Let the potential difference = V,
Resistance of the wire = R
Resistance when the given wires connected in series = Rs
Resistance when the given wires connected in parallel = Rp
Heat produced when the given wires connected in series = Hs
Heat produced when the given wires connected in parallel = Hp
Thus, resistance Rs when the given two wires connected in series = R + R = 2R
Resistance RP when the given two wires connected in parallel
=1R+1R=2R or, RP=R2
We know that, H=I2Rt=(VR)2Rt,=VRt,(since,I=VR)
Thus, ratio of heat produced in given two conditions
HS:HP=VtRS:VtRP=VtRS×RPVt=RPRS
After subsituting the value of RP and RS we get
HSHP=R22R=R2×2R=14
Thus, HS:HP=1:4
Q.5 How is a voltmeter connected in the circuit to measure the potential difference between two points?
Sol. Voltmeter is connected into parallel to measure the potential difference between two points in a circuit.
Q.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?
Sol.
Given, Diameter of wire = 0.5 mm
Therefore, radius = 0.5mm2=0.25mm=0.25m1000=0.00025m
Resistivity, ρ=1.6×10−8Ωm
Resistance (R) = 10Ω, therefore length (I) = ?
Resistance (R_{1}) when diameter is doubled = ?
We know that, R = ρ=lA
Therefore, l=RAρ=Rπr2ρ
⇒ l=10Ω×3.14×(0.00025m)21.6×10−8Ωm
= 10Ω×3.14×0.00025m×0.00025m1.6×10−8Ωm
= 10×3.14×0.0000000625×1081.6m
= 10×108×0.0000001962501.6m
= 196.251.6m=122.656m=122.7m
When diameter of the wire is doubted, i.e. diameter = 0.5mm × 2 = 1 mm
Therefore, radius = 1mm2=0.5mm=0.51000m=0.0005m
Therefore, R1=ρlA=ρ1πr2
= 1.6×10−8Ωm×122.7m3.14×0.0005m×0.0005m
1.6×10−8×122.73.14×0.000000025Ω=1.6×122.73.14×0.000000025×108Ω
196.3278.5Ω=2.5Ω
Thus, length of the wire =122.7 m
Resistance = 2.5Ω (When diameter becomes double)
Q.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 –Plot a graph between V and I and calculate the resistance of that resistor.
Sol.
Since slope of the graph will give the value of resistance, thus
Let consider two points A and B on the slope.
Draw two lines from B along Xaxis and from A along Yaxis, which meets at point C
Now, BC = 10.2 V – 3.4 V = 6.8 V
AC = 3 – 1 = 2 ampere
Slope = 1R=ACBC=26.8=13.4
Thus, resistance, R = 3.4Ω
Q.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.
Sol.
Given, Potential difference, V = 12V
Current ( I ) across the resistor = 2.5mA = 2.5 x 10 3 = 0.0025 A
Resistance, R =?
We know that R=VI=12V0.0025A=4800Ω
Q.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?
Sol.
Given, potential difference, V = 9V
Resistance of resistors which are connected in series = 0.2 Ω, 0.3 Ω, 0.4 Ω , 0.5 Ω and 12 Ω respectively
Current through resistor having resistance equal to 12Ω =?
Total effective resistance, R = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω = 13.4 Ω
We know that current (I)=VR=9V13.4Ω=0.671A
Since, there is no division of electric current, in the circuit if resistors are connected in series, thus, resistance through the resistor having resistance equal to 12 Ω = 0.671 A
Q.10 How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?
Sol.
Given, resistance of each of the resistor = 176Ω
Electric current ( I ) = 5A
Potential difference (V) = 220V
Number of resistors connected in parallel =?
Let total x resistance are connected in parallel, and total effective resistance = R
Therefore, 1Rt=x×1176Ω=x176Ω
We know that, R = VI Therefore, 176Ωx=220V5A
⇒ x×220V=176Ω×5A
⇒ ⇒x=176Ω×5A220V=4
Thus, there are 4 resistors are to be connected.
Q.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.
Sol.
Here we have four options to connect the three resistors in different ways.
(a) All the three resistors can be connected in series
(b) All the three can be connected in parallel
(c) Two of the three resistors can be connected in series and one in parallel and
(d) Two of the three resistors can be connected in parallel and one in series
Thus, effective resistance in the case
(a) When all the three resistors can be connected in series
Effective total resistance = 6 Ω + 6 Ω + 6 Ω = 18 Ω. This is not required
(b) All the three can be connected in parallel
Then, 1R=16Ω+16Ω+16Ω=1+1+16Ω=36Ω=12Ω
Thus, effective total resistance R = 2 Ω. This is also not required.
(c) Two of the three resistors can be connected in parallel and one in series
When two resistors are connected in parallel
Then, 1R=16Ω+16Ω=26Ω=13Ω
Therefore, R = 3Ω
And third one is connected in series, then total resistance = 3 Ω + 6 Ω = 9 Ω. This is required
(d) Two of the three resistors can be connected in parallel and one in series
When two resistors are connected in series, then total resistance = 6 Ω + 6 Ω = 12 Ω
And one resistor is connected in series with two in parallel
Then, 1R=112Ω+16Ω=1+212Ω=14Ω
Thus, R = 4 Ω. This is required.
Thus, when two resistors are connected in series and one in parallel then total effective resistance = 9 Ω
When two resistors are connected in parallel with one in series then total effective resistance = 4 Ω
Q.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?
Sol.
Given, potential difference (V) = 220 V
Power input (P) = 10W
Allowable electric current (I) = 5A
Number of lamps connected in parallel =?
To calculate this, first of all resistance of each of the lamp is to be calculated.
We know that, P=V2R
Therefore, 10W=(220V)2R
⇒ R=4840010W=4840Ω
Let x bulb are to be connected in parallel to have the electric current (I) equal to 5A
Therefore, 1R=x×14840Ω=x4840Ω
Therefore, effective resistance R=4840xΩ
Now, we know that, R=VI
⇒4840xΩ=220V5A
⇒ x=4840×5220=110
Thus, total 110 bulbs are to be connected in parallel.
Q.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?
Sol.
Given, Potential difference (V) = 220 V
Resistance of each coil = 24 Ω
The current in given three case, i.e. when used separately, when used in parallel, when used in series =?
Case – 1 –
When used separately, then resistance, R = 24 Ω and V = 220V
We know that electric current, I=VR=220V24Ω=9.16A
Case – 2 – When the two resistors are connected in series.
Total effective resistance = 24 Ω + 24Ω = 48 Ω
Therefore, electric current, I=VR=220V48Ω=4.58A
Case – 3 – When the two resistors are connected in parallel,
Then 1R=124Ω+124Ω=1+124Ω=224Ω=112Ω
Therefore, total effective resistance, R=12Ω
Thus, electric current, I=VR=220V12Ω=18.33A
Thus, electric through the circuit
(a) When coil is used separately = 9.16A
(b) When coils are used in series = 4.58 A
(c) When coils are used in parallel = 18.33 A
Q.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.
Sol.
Case – 1 –
Potential difference = 6 V
Resistance of resistors = 1 Ω and 2 Ω
Power used through resistors of 2 Ω =?
Since, resistors are connected in series, thus, total effective resitance, R = 1 Ω + 2 Ω = 3 Ω
We know that, I=VR=6V3Ω=2A
Since, current remains same when resistors are connected in series.
Therefore, current through the resistors or 2Ω=2A
Thus, Power (P)=I2×R=(2A)2×2Ω=8W
Case – 2 –
Potential difference, V = 4V
Resistance of resistors connected in parallel = 12 Ω and 2 Ω
Power used by resistors having resistance = 2 Ω
Since, voltage across the circuit remains same if resistors are connected in parallel.
Thus, Power (P) used by resistance of 2Ω=V2R=(4V)22Ω=162W=8W
Thus, power used by resistance of 2 Ω in both the case = 8 W
Q.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?
Sol.
Since, both the lamps are connected in parallel, thus, potential difference will be equal
Thus, potential difference = 220 V
Power of one lamp, P1 = 100W
Power of second lamp, P2 = 60W
We know that, Power (P) = VI, or I = PV
Thus, total current through the circuit, I = P1V+P2V
⇒ I = 100W220V+60W220V=100+60220A=160220A=0.727A answer
Q.16 Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?
Sol.
Given, power of TV (P) = 250W,
time (t) = 1 hr = 60 x 60 s = 3600 s
Thus, energy used by it =?
We know that energy used by appliance = Power x time
Thus, energy used by TV = 250 W x 3600 s = 900000 J = 9 x 10^{5} J
Power of toaster = 1200W
Time (t) = 10 minute = 60 x 10 = 600 s
Thus, energy used by toaster = P x t = 1200W x 600 s = 720000 J = 7.2 x 10^{5} J
Thus, given TV set will use more energy than toaster.
Q.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.
Sol.
Given, Resistance, R = 8 Ω
Electric current ( I ) = 15A
Time (t) = 2 h = 2 x 60 x 60 s = 7200 s
Rate at which heat is developed in heater =?
We know that rate of heat produced = I^{2}R = (15A)^{2} x 8 Ω = 225 x 8 J/s = 1800 J/s Answer
Q.18 Explain the following.
(a) Why is tungsten used almost exclusively for filament of electric lamps?
(b) Why are the conductors of electric heating devices, such as breadtoasters 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 crosssection?
(e) Why are copper and aluminium wires, usually employed for electricity transmission?
Sol.
(a) The melting point of tungsten is very high, i.e. 3380^{0}C, which enables it not to melt at high temperature and to retain most of the heat. The heating of tungsten makes it glow. This is the cause that tungsten is used almost exclusively for filament of electric lamps.
(b) To produce more heat, the high melting point of conductors is necessary. The alloys of metal have higher melting points than pure metals. Thus, to retain more heat alloy is used rather than pure metal in electrical heating devices, such as breadtoaster, electric iron, etc.
(c) There is loss of voltage in the series arrangement in the circuits because of add on effect of resistances. So, series arrangement is not used for domestic circuits.
(d) Resistance of a wire is indirectly proportional to the area of cross section. Resistance increases with decrease in area of cross section and vice versa.
(e) The resistivity of copper and aluminium wires are lower than that of iron but more than that of silver. These wires are cheaper than silver, that’s why copper and aluminium wires usually employed for electricity transmission.
Exemplar
Multiple Choice Questions :
Q.1 A cell, a resistor, a key and ammeter are arranged as shown in the circuit diagrams of figure. The current recorded in the ammeter will be
(a) Maximum in (i)
(b) Maximum in (ii)
(c) Maximum in (iii)
(d) The same in all the cases
Sol. (d)
Q.2 In the following circuits Figure ,heat produced in the resistor or combination of resistors connected to a 12 V battery will be
(a) same in all the cases
(b) minimum in case (i)
(c) maximum in case (ii)
(d) maximum in case (iii)
Sol. (d)
Q.3 Electrical restivity of a given metallic wire depends upon
(a) Its length
(b) Its thickness
(c) Its shape
(d) Nature of the material
Sol. (d)
Q.4 A current of 1 A is drawn by a filament of an electric bulb. Number of electrons passing through a cross section of the filament in 16 seconds would be roughly
(a) 10^{20}
(b) 10^{16}
(c) 10^{18}
(d) 10^{23}
Sol. (a)
Q.5 Identify the circuit figure in which the electrical components have been properly connected.
(a) (i)
(b) (ii)
(c) (iii)
(d) (iv)
Sol. (b)
Q.6 What is the maximum resistance which can be made using five resistors each of 1/5 Ω?
(a) 1/5 Ω
(b) 10 Ω
(c) 5 Ω
(d) 1 Ω
Sol. (d)
Q.7 What is the minimum resistance which can be made using five resistors each of 1/5 Ω?
(a) 1/5 Ω
(b) 1/25 Ω
(c) 1/10 Ω
(d) 25 Ω
Sol. (b)
Q.8 The proper representation of series combination of cells figure obtaining maximum potential is
(a) (i)
(b) (ii)
(c) (iii)
(d) (iv)
Sol.(a)
Q.9 Which of the following represents voltage?
(a)
(b) Work done × Charge
(c)
(d) Work done × Charge × Time
Sol. (a)
Q.10 A cylindrical conductor of length l and uniform area of cross – section A has resistance R. Another conductor of length 2l and resistance R of the same material has area of cross section
(a) A/2
(b) 3A/2
(c) 2A
(d) 3A
Sol. (c)
Q.11 A student carries out an experiment and plots the VI graph of three samples of nichrome wire with resistances R_{1}, R_{2} and R_{3} respectively figure. Which of the following is true?
(a) R_{1} = R_{2} = R_{3}
(b) R_{1} > R_{2} > R_{3}
(c) R_{3} > R_{2} > R_{1}
(d) R_{2} > R_{3} > R_{1}
Sol. (c)
Q.12 If the current I through a resistor is increased by 100% (assume that temperature remains unchanged), the increase in power dissipated will be
(a) 100 %
(b) 200 %
(c) 300 %
(d) 400 %
Sol. (c)
Q.13 The resistivity does not change if
(a) The material is changed
(b) The temperature is changed
(c) The shape of the resistor is changed
(d) Both material and temperature are changed
Sol. (c)
Q.14 In an electrical circuit three incandescent bulbs A, B and C of rating 40 W, 60 W and 100 W respectively are connected in parallel to an electric source. Which of the following is likely to happen regarding their brightness?
(a) Brightness of all the bulbs will be the same
(b) Brightness of bulb A will be the maximum
(c) Brightness of bulb B will be more than that of A
(d) Brightness of bulb C will be less than that of B
Sol. (c)
Q.15 In an electrical circuit two resistors of 2 Ω and 4 Ω respectively are connected in series to a 6 V battery. The heat dissipated by the 4 Ω resistor in 5 s will be
(a) 5 J
(b) 10 J
(c) 20 J
(d) 30 J
Sol. (c)
Q.16 An electric kettle consumes 1 kW of electric power when operated at 220 V. A fuse wire of what rating must be used for it?
(a) 1 A
(b) 2 A
(c) 4 A
(d) 5 A
Sol. (d)
Q.17 Two resistors of resistance 2 Ω and 4 Ω when connected to a battery will have
(a) Same current flowing through them when connected in parallel
(b) Same current flowing through them when connected in series
(c) Same potential difference across them when connected in series
(d) Different potential difference across them when connected in parallel
Sol. (b)
Q.18 Unit of electric power may also be expressed as
(a) Volt ampere
(b) Kilowatt hour
(c) Watt second
(d) Joule second
Sol. (a)