NCERT Solutions for Class 10 Science Chapter 12 Electricity

Here I have provided you NCERT Solutions for Class 10 Science Chapter 12 Electricity.By going through Class 10 Science Chapter 12 Electricity Question Answers you will acquire a better command on this chapter. I hope that this will certainly help you in your studies and examinations!

NCERT Solutions for Class 10 Science Chapter 12 Electricity

    Page Number: 200

    Question 1

    What does an electric circuit mean ?

    Answer:

    A continuous closed path made of electric components through which an electric current flows is known as an electric circuit. A simple circuit consists of the following components:

    (a) Conductors

    (b) Cell

    (c) Switch

    (d) Load

    Question 2

    Define the unit of current.

    Answer:

    Unit of current is ampere. If one coulomb of charge flows through any section of a conductor in one second then the current through it is said to be one ampere.

    1 A = I C s-1

    Question 3

    Calculate the number of electrons constituting one coulomb of charge.

    Answer:

    Charge on one electron, e = 1.6 x 10-19 C

    Total charge, Q = 1 C

    Number of electrons, n = 

    Qe = 1C1.6x1019 = 6.25 x 1018

    Page Number: 202

    Question 1

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

    Answer:

    A battery.

    Question 2

    What is meant by saying that the potential difference between two points is IV?

    Answer:

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

    Question 3

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

    Answer:

    Energy given by battery = charge x potential difference

    or W = QV = 1C X 6V = 6J.

    Page Number: 209

    Question 1

    On what factors does the resistance of a conductor depend ?

    Answer:

    The resistance of a conductor depends 

    (i) on its length

    (ii) on its area of cross-section and 

    (iii) on the nature of its material

    (iv) on temperature of the conductor

    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:

    The current will flow more easily through a thick wire than a thin wire of the same material. Larger the area of cross-section of a conductor, more is the ease with which the electrons can move through the conductor. 

    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:

    When potential difference is halved, the current through the component also decreases to half of its initial value. This is according to ohm’s law i.e., V ∝ I.

    Question 4

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

    Answer:

    The coils of electric toasters, electric irons and other heating devices are made of an alloy rather than a pure metal because 

    (i) the resistivity of an alloy is much higher than that of a pure metal, and 

    (ii) an alloy does not undergo oxidation (or burn) easily even at high temperature, when it is red hot.

    Question 5

    Use the data in Table 12.2 (in NCERT Book on Page No. 207) to answer the following :

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

    (ii) Which material is the best conductor ?

    Answer:

    (i) Resistivity of iron = 10.0 x 10-8 Ω m

    Resistivity of mercury = 94.0 x 10-8 Ω m

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

    (ii) Because silver has the lowest resistivity (=1.60 x 10-8 Ω m), therefore silver is the best conductor.

    Page Number: 213

    Question 1

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

    Answer:

    The required circuit diagram is shown below :

    NCERT Solutions for Class 10 Science Chapter 12 Electricity

    Question 2

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

    Answer:

    An ammeter should be connected in the circuit in series with the resistors. To measure the potential difference across the resistor it should be connected in parallel, as shown in the following figure.

    NCERT Solutions for Class 10 Science Chapter 12 Electricity

    The resistances are connected in series.

    Ohm’s law can be used to obtain the readings of ammeter and voltmeter. According to Ohm’s law,

    V = IR,

    Where,

    Potential difference, V = 6 V

    Current flowing through the circuit/resistors = I

    Resistance of the circuit, R = 5 + 8 + 12 = 25Ω

    I = V/R = 6/25 = 0.24 A

    Potential difference across 12 Ω resistor = V1

    Current flowing through the 12 Ω resistor, I = 0.24 A

    Therefore, using Ohm’s law, we obtain

    V1 = IR = 0.24 x 12 = 2.88 V

    Therefore, the reading of the ammeter will be 0.24 A.

    The reading of the voltmeter will be 2.88 V.

    Page Number: 216

    Question 1

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

     (a) 1 Ω and 106Ω, (b) 1 Ω and 103Ω and 106Ω.

    Answer:

    (a) When 1 Ω and 106 Ω are connected in parallel:

    Let R be the equivalent resistance.

    NCERT Solutions for Class 10 Science Chapter 12 Electricity

    Therefore, equivalent resistance ≈ 1 Ω

    (b) When 1Ω, 10³ Ω and 10³ Ω are connected in parallel:

    Let R be the equivalent resistance.

    NCERT Solutions for Class 10 Science Chapter 12 Electricity

    Therefore, equivalent resistance = 0.999 Ω

    Question 2

    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 ?

    Answer:

    The electric lamp, the toaster and the water filter connected in parallel to a 220 V source can be shown as using a circuit diagram as follows:

    NCERT Solutions for Class 10 Science Chapter 12 Electricity

    The equivalent resistance of the resistors can be calculated as follows:

    NCERT Solutions for Class 10 Science Chapter 12 Electricity

    The resistance of the electric iron box is 31.25 Ω.

    Question 3 

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

    Answer:

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

    • In parallel circuits, if an electrical appliance stops working due to some defect, then all other appliances keep working normally.
    • In parallel circuits, each electrical appliance has its own switch due to which it can be turned on turned off independently, without affecting other appliances.
    • In parallel circuits, each electrical appliance gets the same voltage (220 V) as that of the power supply line.
    • In the parallel connection of electrical appliances, the overall resistance of the household circuit is reduced due to which the current from the power supply is high.

    Question 4

    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?

    Answer:

    (a) The circuit diagram below shows the connection of three resistors

    NCERT Solutions for Class 10 Science Chapter 12 Electricity

    From the circuit above, it is understood that 3 Ω and 6 Ω are connected in parallel. Hence, their equivalent resistance is given by

    NCERT Solutions for Class 10 Science Chapter 12 Electricity

    The equivalent resistor 2 Ω is in series with the 2 Ω resistor. Now the equivalent resistance can be calculated as follows:

    Req= 2 Ω +2 Ω = 4 Ω

    Hence, the total resistance of the circuit is 4 Ω.

    (b) The circuit diagram below, shows the connection of three resistors.

    NCERT Solutions for Class 10 Science Chapter 12 ElectricityNCERT Solutions for Class 10 Science Chapter 12 Electricity

    From the circuit, it is understood that all the resistors are connected in parallel. Therefore, their equivalent resistance can be calculated as follows:

    NCERT Solutions for Class 10 Science Chapter 12 Electricity

    The total resistance of the circuit is 1 Ω.

    Question 5

    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?

    Answer:

    (a) If the four resistors are connected in series, their total resistance will be the sum of their individual resistances and it will be the highest. The total equivalent resistance of the resistors connected in series will be 4 Ω + 8 Ω + 12 Ω + 24 Ω = 48 Ω.

    (b) If the resistors are connected in parallel, then their equivalent resistances will be the lowest.

    Their equivalent resistance connected in parallel is

    NCERT Solutions for Class 10 Science Chapter 12 Electricity

    Hence, the lowest total resistance is 2 Ω.

    Page Number: 218

    Question 1

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

    Answer:

    Heat generated in a circuit is given by I²R t. The heating element of an electric heater made of nichrome glows because it becomes red-hot due to the large amount of heat produced on passing current because of its high resistance, but the cord of the electric heater made of copper does not glow because negligible heat is produced in it by passing current because of its extremely low resistance.

    Question 2

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

    Answer:

    Here, 

    Q = 96,000 C, 

    t =1 hour = 1 x 60 x 60 sec = 3,600 s, 

    V = 50 V

    Heat generated, 

    H = VQ = 50Vx 96,000 C = 48,00,000 J = 4.8 x 10⁶ J

    Question 3

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

    Answer:

    Here, 

    R = 20 Ω, 

    I = 5 A, 

    t = 3s

    Heat developed, 

    H = I²R t = 25 x 20 x 30 = 15,000 J = 1.5 x 10⁴ J

    Page Number: 220

    Question 1

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

    Answer:

    The rate of consumption of electric energy in an electric appliance is called electric power. Hence, the rate at which energy is delivered by a current is the power of the appliance.

    Question 2

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

    Answer:

    Here, 

    I = 5 A, 

    V = 220 V, 

    t = 2h = 7,200 s

    Power, P = VI = 220 x 5 = 1100 W

    Energy consumed = P x t = 100 W x 7200 s = 7,20,000 J = 7.2 x 10⁵ J

    Back Exercise Questions

    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

    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

    Question 5

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

    Answer:

    A voltmeter is connected in parallel to measure the potential difference between two points.

    Question 6

    A copper wire has diameter 0.5 mm and resistivity of 1.6 x 10-8 Ωm. 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:

    The resistance of the copper wire of length in meters and area of cross-section m² is given by the formula

    NCERT Solutions for Class 10 Science Chapter 12 Electricity

    The length of the wire is 122.72 m and the new resistance is 2.5 Ω.

    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 :

    NCERT Solutions for Class 10 Science Chapter 12 Electricity

    Answer:

    The plot between voltage and current is known as IV characteristic. The current is plotted in the y-axis while the voltage is plotted in the x-axis. The different values of current for different values of voltage are given in the table. The I-V characteristics for the given resistor is shown below:

    NCERT Solutions for Class 10 Science Chapter 12 Electricity

    The slope of the line gives the value of resistance.
    The slope can be calculated as follows:
    Slope = 1/R = PR/RQ = 2/6.8
    To calculate R,
    R = 6.8/2 = 3.4 Ω
    The resistance of the resistor is 3.4 Ω.

    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:

    Here, 

    V = 12 V and 

    I = 2.5 mA =  2.5 x 10-3 A

    ∴ Resistance, R = V/I = 12V/2.5×10-3A = 4,800 Ω = 4.8 x 10-3 Ω

    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:

    In series connection, there is no division of current. The current flowing across all the resistors is the same.

    To calculate the amount of current flowing across the resistors, we use Ohm’s law.

    But first, let us find out the equivalent resistance as follows:

    R = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω = 13.4 Ω

    Now, using Ohm’s law,

    NCERT Solutions for Class 10 Science Chapter 12 Electricity

    The current flowing across the 12 Ω is 0.671 A.

    Question 10 

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

    Answer:

    Let us consider the number of resistors required as ‘x'.

    The equivalent resistance of the parallel combination of resistor R is given by

    NCERT Solutions for Class 10 Science Chapter 12 Electricity

    The number of resistors required 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) 4Ω

    Answer:

    Here, R1 = R2 = R3 = 6 Ω.

    (i) When we connect R1 in series with the parallel combination of R2 and R3 as shown in Fig. (a).

    The equivalent resistance is

    NCERT Solutions for Class 10 Science Chapter 12 Electricity

    (ii) When we connect a series combination of R1 and R2 in parallel with R3, as shown in Fig. (b), the equivalent resistance is

    NCERT Solutions for Class 10 Science Chapter 12 Electricity

    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:

    The resistance of the bulb can be calculated using the expression

    P1 = V²/R1

    R1 = V²/P1

    Substituting the values, we get

    NCERT Solutions for Class 10 Science Chapter 12 Electricity
    Hence, 110 lamps can be connected in parallel.

    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:

    (i) When the two coils A and B are used separately. R = 24 Ω, V = 220 V

    NCERT Solutions for Class 10 Science Chapter 12 Electricity

    (ii) When the two coils are connected in series,

    NCERT Solutions for Class 10 Science Chapter 12 Electricity

    (iii) When the two coils are connected in parallel.

    NCERT Solutions for Class 10 Science Chapter 12 Electricity

    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:

    Potential difference, V = 6 V

    1 Ω and 2 Ω resistors are connected in series. Therefore, equivalent resistance of the circuit, R = 1 + 2 = 3 Ω

    According to Ohm’s law,

    V = IR

    Where,

    I is the current through the circuit

    I = 6/3 = 2 A

    This current will flow through each component of the circuit because there is no division of current in series circuits. Hence, current flowing through the 2 Ω resistor is 2 A. Power is given by the expression,

    P = (I)²R = (2)² x 2 = 8 W

    (ii) Potential difference, V = 4 V

    12 Ω and 2 Ω resistors are connected in parallel. The voltage across each component of a parallel circuit remains the same. Hence, the voltage across 2 Ω resistor will be 4 V.

    Power consumed by 2 Ω resistor is given by

    P= V²/R = 4²/2 = 8 W

    Therefore, the power used by 2 Ω resistor is 8 W.

    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:

    Power of first lamp (P1) = 100 W

    Potential difference (V) = 220 V

    NCERT Solutions for Class 10 Science Chapter 12 Electricity

    Question 16

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

    Answer:

    The energy consumed by electrical appliances is given by the equation

    H = Pt, where P is the power of the appliance and t is the time

    Using this formula, the energy consumed by a TV of power ration 250 W, can be calculated as follows:

    H = 250 W × 3600 seconds = 9 × 10⁵ J

    Similarly, the energy consumed by a toaster of power rating 1200 W is

    H = 1200 W × 600 s = 7.2 × 10⁵ J

    From the calculations, it can be said that the energy consumed by the TV is greater than the toaster.

    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:

    The rate at which the heat develops in the heater can be calculated using the following formula

    P = I²R

    Substituting the values in the equation, we get

    P = (15A)² × 8 Ω = 1800 W

    The electric heater produces heat at the rate of 1800 watt.

    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 copper and aluminium wires are usually employed for electricity transmission?

    Answer:

    a. The resistivity and melting point of tungsten is very high. Due to this property, it doesn’t burn readily when heated. Electric lamps operate at high temperature. Hence, tungsten is a choice of metal for the filament of electric lamps.

    b. The conductors of electric heating devices are alloys because of their high resistivity. Due to its high resistivity it produces large amount of heat.

    c. The voltage is divided in series circuit as result each component in the circuit receives a small voltage because of which the amount of current decreases and the device gets hot and does not work properly. This is the reason why series circuits are not used in domestic circuits.

    d. Resistance is inversely proportional to the area of cross section. When the area of cross section increases the resistance decreases and vice versa.

    e. Copper and aluminium are good conductors of electricity and have low resistivity because of which they are usually employed for electricity transmission.









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