In this post we are going to share the detailed solution for class 10th science chapter 12 . All the solutions are prepared by our esteemed teachers who are very well experienced in the teaching.
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Complete Science Solutions Chapter Wise : Class 10th NCERT Science Solutions – All Chapters
Chapter 12 Solutions
EXERCISE -1
QUE:-1 What does an electric circuit mean?
ANS:-
A continuous and closed path along which an electric current flows is called an electric circuit.
QUE:-2 Define the unit of current.
ANS:-
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.
I = \(\frac { Q }{ t }\) or 1 A = I C s-1
QUE:-3 Calculate the number of electrons constituting one coulomb of charge.
ANS:-
Charge on one electron, e = 1.6 x 10-19 C
Total charge, Q = 1 C
Number of electrons, n = \(\frac { Q }{ e }\) = \(\frac { 1C }{ 1.6x{ 10 }^{ -19 } }\) = 6.25 x 1018
EXERCISE -2
QUE:-1 Name a device that helps to maintain a potential difference across a conductor.
ANS:-
A battery.
QUE:-2 What is meant by saying that the potential difference between two points is 1 V?
ANS:-
The potential difference between two points is said to be 1 volt if 1 joule of work is done in moving 1 coulomb of electric charge from one point to the other.
QUE:-3 How much energy is given to each coulomb of charge passing through a 6 V battery?
ANS:-
Energy given by battery = charge x potential difference
or W = QV = 1C X 6V = 6J.
EXERCISE -3
QUE:-1 On what factors does the resistance of a conductor depend?
ANS:-
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
QUE:-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?
ANS:-
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. Therefore, smaller is the resistance of the conductor.
QUE:-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?
ANS:-
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.
QUE:-4 Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
ANS:-
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.
QUE:-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?
ANS:-
(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.
EXERCISE -4
QUE:-1 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
ANS:-
The required circuit diagram is shown below :
QUE:-2 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?
ANS:-
The required circuit diagram is shown on the right.
Total voltage, V = 3 x 2 = 6V
Total resistance, R = 5Ω + 8Ω + 12Ω = 25Ω
EXERCISE -5
QUE:-1 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.
ANS:-
When the resistances are connected in parallel, the equivalent resistance is smaller than the smallest individual resistance.
(i) Equivalent resistance < 1 Ω.
(ii) Equivalent resistance < 1 Ω.
QUE:-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?
ANS:-
Resistance of electric lamp, R1 = 100 Ω
Resistance of toaster, R2 = 50 Ω
Resistance of water filter, R3 = 500 Ω
Equivalent resistance Rp of the three appliances connected in parallel, is
Resistance of electric iron = Equivalent resistance of the three appliances connected in parallel = 31.25 Ω
Applied voltage, V = 220 V
Current, I = VR = 220V31.25Ω
QUE:-3 What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
ANS:-
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.
QUE:-4 How can three resistors of resistances 2 Ω, 3 Ω, and 6 Ω be connected togive a total resistance of
(a) 4 Ω, (b) 1 Ω?
ANS:-
(i) We can get a total resistance of 4Ω by connecting the 2Ω resistance in series with the parallel combination of 3Ω and 6Ω.
(ii) We can obtain a total resistance of 1Ω by connecting resistors of 2 Ω, 3 Ω and 6 Ω in parallel.
QUE:-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?
ANS:-
(i) Highest resistance can be obtained by connecting the four coils in series.
Then, R = 4Ω + 8Ω + 12Ω + 24Ω = 48Ω
(ii) Lowest resistance can be obtained by connecting the four coils in parallel.
EXERCISE -6
QUE:-1 Why does the cord of an electric heater not glow while the heating element does?
ANS:-
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.
QUE:-2 Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.
ANS:-
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
QUE:-3 An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s.
ANS:-
Here, R = 20 Ω, i = 5 A, t = 3s
Heat developed, H = I2 R t = 25 x 20 x 30 = 15,000 J = 1.5 x 104 J
EXERCISE -7
QUE:-1 What determines the rate at which energy is delivered by a current?
ANS:-
Resistance of the circuit determines the rate at which energy is delivered by a current.
QUE:-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.
ANS:-
Here, I = 5 A, V = 220 V, t = 2h = 7,200 s
Power, P = V I = 220 x 5 = 1100 W
Energy consumed = P x t = 100 W x 7200 s = 7,20,000 J = 7.2 x 105 J
EXERCISE -8 (CHEPTER END QUETION)
QUE:-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
ANS:-
(d) 25
QUE:-2 Which of the following terms does not represent electrical power in a circuit?
(a) I²R (b) IR² (c) VI (d) V²/R
ANS:-
(b) IR²
QUE:-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
ANS:-
(d) 25 W
QUE:-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
ANS:-
(c) 1 : 4
QUE:-5 How is a voltmeter connected in the circuit to measure the potential difference between two points?
ANS:-
A voltmeter is connected in parallel to measure the potential difference between two points.
QUE:-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?
ANS:-
If a wire of diameter doubled to it is taken, then area of cross-section becomes four times.
New resistance = \(\frac { 10 }{ 2 }\) = 2.5 Ω, Thus the new resistance will be \(\frac { 1 }{ 4 }\) times.
Decrease in resistance = (10 – 2.5) Ω = 7.5 Ω
QUE:-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) 0.5 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:-
The graph between V and I for the above data is given below.
The slope of the graph will give the value of resistance.
Let us consider two points P and Q on the graph.
and from P along Y-axis, which meet at point R.
Now, QR = 10.2V – 34V = 6.8V
And PR = 3 – 1 = 2 ampere
Thus, resistance, R = 3.4 Ω
QUE:-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.
ANS:-
Here, V = 12 V and I = 2.5 mA = 2.5 x 10-3 A
∴ Resistance, R = \(\frac { V }{ I }\) = \(\frac { 12V }{ 2.5\times { 10 }^{ 3 }A }\) = 4,800 Ω = 4.8 x 10-3 Ω
QUE:-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?
ANS:-
Total resistance, R = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω – 13.4 Ω
Potential difference, V = 9 V
Current through the series circuit, I = \(\frac { V }{ R }\) = \(\frac { 12V }{ 13.4\Omega }\) = 0.67 A
∵ There is no division of current in series. Therefore current through 12 Ω resistor = 0.67 A.
QUE:-10 How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?
ANS:-
Suppose n resistors of 176 Ω are connected in parallel.
Thus 4 resistors are needed to be connect.
QUE:-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.
ANS:-
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
(ii) When we connect a series combination of R1 and R2 in parallel with R3, as shown in Fig. (b), the equivalent resistance is
QUE:-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?
ANS:-
Here, current, I = 5 A, voltage, V = 220 V
∴ Maxium power, P = I x V = 5 x 220 = 1100W
Required no. of lamps
\( =\frac { Max.Power }{ Power\quad of\quad 1\quad lamp } \quad =\quad \frac { 1100 } { 10 } =110\)
∴ 110 lamps can be connected in parallel
QUE:-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?
ANS:-
(i) When the two coils A and B are used separately. R = 24 Ω, V = 220 V
(ii) When the two coils are connected in series,
(iii) When the two coils are connected in parallel.
QUE:-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.
ANS:-
(i) The circuit diagram is shown in figure.
Total resistance, R = 1Ω + 2Ω = 3Ω
Potential difference, V = 6 V
Power used in 2Ω resistor = I2R = (2)2 x 2 = 8 W
(ii) The circuit diagram for this case is shown :
Power used in 2 resistor = \(\frac { { v }^{ 2 } }{ R }\) =\(\frac { { 4 }^{ 2 } }{ 2 }\) = 8 W.
[ ∵ Current is different for different resistors in parallel combination.]
QUE:-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?
ANS:-
QUE:-16 Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?
ANS:-
Energy consumed by 250 W TV set in 1 h = 250 x 1 = 250 Wh.
Energy consumed by 1200 W toaster in 10 min = 1200 X 1/6 = 200 Wh.
∴ Energy consumed by TV set is more than the energy consumed by toaster in the given timings.
QUE:-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.
ANS:-
QUE:-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 aluminium wires usually employed for electricity transmission?
ANS:-
(a) It has high melting point and emits light at a high temperature.
(b) It has more resistivity and less temperature coefficient of resistance.
(c) (i) All appliances do not get same potential in series arrangement.
(ii) All appliances cannot be individually operated.
(d) R ∝ =1 / Area of cross – section.
(e) They are very good conductors of electricity.
SUMMARY OF THE CHEPTER
* A stream of electrons moving through a conductor constitutes an electric current.Conventionally, the direction of current is taken opposite to the direction of flow of electrons.
* The SI unit of electric current is ampere.
* To set the electrons in motion in an electric circuit, we use a cell or a battery. A cell generates a potential difference across its terminals. It is measured in volts (V).
* Resistance is a property that resists the flow of electrons in a conductor. It controls the magnitude of the current. The SI unit of resistance is ohm (Ω).
* Ohm’s law: The potential difference across the ends of a resistor is directly proportional to the current through it, provided its temperature remains the same.
* The resistance of a conductor depends directly on its length, inversely on its area of cross-section, and also on the material of the conductor.
* The equivalent resistance of several resistors in series is equal to the sum of their individual resistances.
* A set of resistors connected in parallel has an equivalent resistance Rp given by 1/Rp= 1/R1+1/R2+R3……
* The electrical energy dissipated in a resistor is given by
W = V × I × t
* The unit of power is watt (W). One watt of power is consumed when 1 A of current flows at a potential difference of N 1 V.
* The commercial unit of electrical energy is kilowatt hour (kWh).
1 kW h = 3,600,000 J = 3.6 × 106 J.
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- NCERT Class 10th Science Chapter 1 – Chemical Reactions and Equations Solutions
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- NCERT Class 10th Science Chapter – 3 Metals and Non Metals
- NCERT Class 10th Science Chapter – 4 Carbon and its Compounds
- NCERT Class 10th Science Chapter – 5 Periodic Classification of Elements
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Complete Science Solutions Chapter Wise : Class 10th NCERT Science Solutions – All Chapters
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