NCERT In-Text Solutions For Class 10 Science Chapter 12 Electricity
CBSE Class 10 Physics Electricity In-Text Solution will help students to score good in class tests as well as in the CBSE Class 10 Board Exam. The solution provided below will also help students to clear their concepts about electricity.
Before getting into the in-text questions, let’s have a look on the topics covered in this chapter under CBSE Class 10 Science Electricity
Topics covered in electricity are:
- Electric Charge and Electric Current
- Electric Potential and Potential Difference
- Circuit Diagram
- Ohm’s Law
- Factors on Which Resistance of A Conductor Depends
- Combination of resistors
- Heating Effect of Electric Current
- Electric Power
Don’t forget to check the important links for CBSE Class 10 Electricity Given at the end of this page.
INTEXT QUESTIONS PAGE NO. 200
1. What does an electric circuit mean?
AnswerAn electric circuit consists of electric devices, switching devices, source of electricity, etc. that are connected by conducting wires.
2. Define the unit of current.
AnswerThe unit of electric current is ampere (A).
1 A is defined as the flow of 1 C of charge through a wire in 1 s.
3.Calculate the number of electrons constituting one coulomb of charge.
Answer6 x 1018 electrons
INTEXT QUESTIONS PAGE NO. 202
1.Name a device that helps to maintain a potential difference across a conductor.
AnswerA source of electricity such as cell, battery, power supply, etc. helps to maintain a potential difference across a conductor.
2.What is meant by saying that the potential difference between two points is 1 V?
AnswerIf 1 J of work is required to move a charge of amount 1 C from one point to another, then it is said that the potential difference between the two points is 1 V.
3.How much energy is given to each coulomb of charge passing through a 6 V battery?
AnswerWork done = Potential difference x Charge
Where, Charge = 1 C and Potential difference = 6 V
Work done = 6 x 1 = 6 J
Therefore, 6 J of energy is given to each coulomb of charge passing through a battery of 6 V.
INTEXT QUESTIONS PAGE NO. 209
1.On what factors does the resistance of a conductor depend?
AnswerThe resistance of a conductor depends upon the following factors:
(a) Length of the conductor
(b) Cross-sectional area of the conductor
(c) Material of the conductor
(d) Temperature of the conductor
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?
AnswerResistance of wire,
ρ=Resistivity of the material of the wire
l = Length of the wire
A = Area of cross-section of the wire
Resistance is inversely proportional to the area of cross-section of the wire. Thicker the wire, lower is the resistance of the wire and vice-versa. Therefore, current can flow more easily through a thick wire than a thin wire.
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?
AnswerThe amount of current flowing through the electrical component is reduced by half.
4.Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
AnswerThe resistivity of an alloy is higher than the pure metal. Moreover, at high temperatures, the alloys do not melt readily. Hence, the coils of heating appliances such as electric toasters and electric irons are made of an alloy rather than a pure metal.
5.(a) Which among iron and mercury is a better conductor?
AnswerResistivity of mercury is more than that of iron. This implies that iron is a better conductor than mercury.
(b) Which material is the best conductor?
INTEXT QUESTIONS PAGE NO. 213
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.
AnswerThree cells of potential 2 V, each connected in series, is equivalent to a battery of potential 2 V + 2 V + 2 V = 6V. The following circuit diagram shows three resistors of resistances 5 Ω, 8 Ω and 12 Ω respectively connected in series and a battery of potential 6 V.
2.Redraw the circuit of question 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure potential difference across the 12 Ω resistor. What would be the readings in the ammeter and the voltmeter?
AnswerTo measure the current flowing through the resistors, an ammeter should be connected in the circuit in series with the resistors. To measure the potential difference across the 12 Ω resistor, a voltmeter should be connected parallel to this resistor, as shown in the following figure.,
INTEXT QUESTIONS PAGE NO. 216
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.
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?
3.What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
AnswerThere is no division of voltage among the appliances when connected in parallel. The potential difference across each appliance is equal to the supplied voltage. The total effective resistance of the circuit can be reduced by connecting electrical appliances in parallel.
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?
AnswerThere are three resistors of resistances 2 Ω, 3 Ω, and 6 Ω respectively.
(a) The following circuit diagram shows the connection of the three resistors. Here, 6 Ω and 3 Ω resistors are connected in parallel.
Therefore, their equivalent resistance will be given by
This equivalent resistor of resistance 2Ω is connected to a 2Ω resistor in series.
Therefore, equivalent resistance of the circuit of the circuit = 2Ω + 2Ω = 4Ω.
Hence, the total resistance of the circuit is 4 Ω.
The following circuit diagram shows the connection of the three resistors.
All the resistors are connected in series. Therefore, their equivalent resistance will be given as Therefore, the total resistance of the circuit is 1 Ω
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?
AnswerThere are four coils of resistances 4 Ω, 8 Ω, 12 Ω and 24 Ω respectively
(a) If these coils are connected in series, then the equivalent resistance will be the highest, given by the sum 4 + 8 + 12 + 24 = 48 Ω
(b) If these coils are connected in parallel, then the equivalent resistance will be the lowest, given by
Therefore, 2Ω is the lowest total resistance.
INTEXT QUESTIONS PAGE NO. 218
1.Why does the cord of an electric heater not glow while the heating element does?
AnswerThe heating element of an electric heater is a resistor. The amount of heat produced by it is proportional to its resistance. The resistance of the element of an electric heater is very high. As current flows through the heating element, it becomes too hot and glows red. On the other hand, the resistance of the cord is low. It does not become red when current flows through it.
2.Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.
AnswerAns. The amount of heat (H) produced is given by the Joule’s law of heating as H=VIt
Where, Voltage, V = 50 V, Time, t = 1 h = 1 × 60 × 60 s
3. An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s.
AnswerThe amount of heat (H) produced is given by the Joule’s law of heating as H=VIt
Current, I = 5 A
Time, t = 30 s
Voltage, V = Current × Resistance = 5 × 20 = 100 V
H = 100 x 5 x 30 = 1.5 x 104 J
Therefore, the amount of heat developed in the electric iron is 1.5 x 104 J
INTEXT QUESTIONS PAGE NO. 220
1.What determines the rate at which energy is delivered by a current?
AnswerThe 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.
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.
AnswerPower (P) is given by the expression, P = VI
Voltage, V = 220 V
Current, I = 5 A
P = 220 x 5 = 1100 W
Energy consumed by the motor = Pt
Time, t = 2 h = 2 × 60 × 60 = 7200 s
Therefore, P = 1100 × 7200 = 7.92 × 106 J
Therefore, power of the motor = 1100 W
Energy consumed by the motor = 7.92 × 106 J
Important links on CBSE Class 10 Science Chapter 12 Electricity
- MCQs on CBSE Class 10 Science Chapter 1
- MCQs on CBSE Class 10 Science Chapter 2
- MCQs on CBSE Class 10 Science Chapter 3
- MCQs on CBSE Class 10 Science Chapter 4
- MCQs on CBSE Class 10 Science Chapter 5
- MCQs on CBSE Class 10 Science Chapter 6
- MCQs on CBSE Class 10 Science Chapter 7
- MCQs on CBSE Class 10 Science Chapter 8
- MCQs on CBSE Class 10 Science Chapter 9
- MCQs on CBSE Class 10 Science Chapter 10
- MCQs on CBSE Class 10 Science Chapter 11
- MCQs on CBSE Class 10 Science Chapter 13
- MCQs on CBSE Class 10 Science Chapter 14
- MCQs on CBSE Class 10 Science Chapter 15