NCERT Solutions for Class 10 Science Chapter 11 – Electricity

Chapter 11 – Electricity

Intext Questions with Solutions of Class 10 Science Chapter 11 – Electricity

1.

Ans: A continuous, closed route of an electric current is called an electric circuit.
There are several electric components in the circuit such as: Conductors, Cell, Switch, Load.

2.

Ans: The unit of electric current is the ampere. If one coulomb of charge traverses a portion of a conductor in one second, the current is defined as one ampere.

3.

Ans:

4.

Ans: A battery.

5.

Ans: If one joule of work is required to move one coulomb of electric charge from one location to another, the potential difference between two points is said to be one volt.

6.

Ans: Energy given by battery = charge x potential difference
or W = QV = 1C X 6V = 6J.

7.

Ans: The following variables affect the conductor’s resistance:

1. The conductor’s temperature
2. Conductor cross-sectional area
3. The conductor’s length
4. The conductor’s material composition

8.

Ans: The relationship between resistance and cross-sectional area can be expressed as:

Resistance is inversely related to the cross-sectional area of the wire. As resistance diminishes, current amplifies.
A thicker wire allows less current to flow, while a thinner wire permits more current to pass.

9.

Ans: Through the use of Ohm’s Law, one is able to ascertain the change in the current that is passing through the electrical component.

Ohm’s Law states that the current is determined by the following:
I = V/R

From this point forward, the gap in potential is cut in half while maintaining the same level of resistance.
Assume that the new voltage V`=V/2.

The new resistance will be denoted by R’ = R, and the new quantity of current will be denoted by I’.
The following is an example of how Ohm’s law can be used to determine the change in the current:

Because of this, the amount of current that is going through the electrical component is cut in half.

10.

Ans: The heating elements of electric toasters, irons, and similar devices are constructed from an alloy instead of a pure metal due to

1. the significantly higher resistivity of alloys compared to pure metals, and
2. the resistance of alloys to oxidation (or combustion) at elevated temperatures, even when incandescent.

11.

Ans:

1. Because mercury has a higher resistance than iron, iron is a better conductor than mercury.
2. Silver is the best conductor of all the materials indicated in the table because it has the lowest resistance
   (1.60 × 10^–8).

12.

Ans: Three 2 V cells make into a battery with a potential of 6 V. Three resistors with resistances of 12 Ω, 8 Ω, and 5 Ω connected in series with a 6 V battery are depicted in the circuit diagram below.

13.

Ans: An ammeter must be connected in series with resistors, whereas a voltmeter should be connected in parallel to the resistor to measure the potential difference, as illustrated in the image below.

Ohm’s Law enables us to ascertain the readings of the ammeter and voltmeter.
The cumulative resistance of the circuit is 5 Ω + 8 Ω + 12 Ω = 25 Ω.
The circuit’s potential difference is 6 V; hence, the current traversing the circuit or the resistors can be computed as follows:

I = V/R = 6/25 = 0.24 A
Designate the potential difference across the 12 Ω resistor as V1.
The acquired current V1 can be computed as follows:
V1 = 0.24 A × 12 Ω = 2.88 V

Consequently, the ammeter will register 0.24 A, and the voltmeter will indicate 2.88 V.

14.

Ans:

15.

Ans:

16.

Ans: The benefits of connecting electrical equipment in parallel with the battery are:

1. In parallel circuits, if one electrical appliance stops working due to a malfunction, the other appliances continue to function normally.
2. In parallel circuits, each electrical appliance has its own switch, allowing it to be turned on and off separately without impacting other appliances.
3. In parallel circuits, all electrical appliances receive the same voltage (220 V) as the power supply line.
4. Parallel connection of electrical appliances reduces the overall resistance of the household circuit, resulting in a high current from the power source.

17.

Ans:

1. A total resistance of 4Ω can be achieved by connecting a 2Ω resistor in series with the parallel arrangement of 3Ω and 6Ω resistors.

1. A total resistance of 1Ω can be achieved by connecting 2 Ω, 3 Ω, and 6 Ω resistors in parallel.

18.

Ans:

1. The resistance of the four resistors connected in series will be the largest since it is the sum of their individual resistances. The resistors linked in series will have a total equivalent resistance of 4 Ω + 8 Ω + 12 Ω + 24 Ω = 48 Ω.
2. The resistors’ equivalent resistances will be the lowest if they are linked in parallel.
   When connected in parallel, their corresponding resistance is

19.

Ans: The heat generated by a system is directly proportional to its resistance.

The thermal element generates an increased amount of heat, which results in its glowing appearance due to its increased resistance.

Additionally, the resistance of the wire is reduced, which results in a reduction in the amount of heat generated. Therefore, the heating element emits light, while the filament remains unlit.

20.

Ans:

21.

Ans: Joule’s law of heating, which is provided by the equation, can be used to determine how much heat is produced.
H = Vit   
When we change the values in the equation above, we obtain
H =100 × 5 × 30 = 1.5 × 10⁴ J
The electric iron produced 1.5 × 10⁴ J of heat in 30 seconds.

22.

Ans: The power of an appliance is the rate at which energy is delivered by a current or consumed by the device.

23.

Ans:

Exercise Questions with Solutions of Class 10 Science Chapter 11 – Electricity

1.

Ans: (d) 25

2.

Ans: (b) IR²

3.

Ans: (d) 25 W

4.

Ans: (c) 1 : 4

5.

Ans: To measure the potential difference between two places, the voltmeter needs to be linked in parallel.
This is due to the voltmeter’s ability to measure the voltage across the parallel circuit.
However, voltmeters cannot be utilized for a series circuit. In a series circuit, an ammeter is utilized instead.

6.

Ans: The resistance of a copper wire, characterized by its length in meters and cross-sectional area in square meters, is expressed by the formula.

7.

Ans: The IV characteristic is the graph that shows the relationship between current and voltage. On one side, we see the voltage, and on the other, we see the current. The table provides the various current levels for various voltage values. Below are the I V graph plot for the properties of the provided resistor.

8.

Ans:

9.

Ans: There is no present division in a series connection. Every resistor has the same amount of current flowing through it.
We apply Ohm’s law to determine how much current is passing through the resistors.
However, let’s first determine the equivalent resistance in this way.
R = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω = 13.4 Ω

Now, using Ohm’s law,

The current flowing across the 12 Ω is 0.671 A.

10.

Ans:

11.

Ans: Here, R₁ = R₂ = R₃ = 6 Ω.

1. When R1 is connected in series with the parallel arrangement of R2 and R3, as illustrated in Fig. (a).The corresponding resistance is

2. When a series combination of R1 and R2 is connected in parallel with R3, as illustrated in Fig. (b), the corresponding resistance is

12.

Ans:

13.

Ans:

14.

Ans:

15.

Ans:

16.

Ans:

17.

Ans:

18.

Ans:

1. Tungsten is predominantly utilized for the filaments of electric lamps due to its exceptionally high melting point of 3300°C. When electricity is passed through a tungsten filament, its temperature increases to 2700°C, producing heat and light energy without melting.
2. The conductors of electric heating devices, such as bread toasters and electric irons, are composed of an alloy rather than a pure metal. This is due to the higher resistivity of alloys compared to pure metals, as well as their resistance to oxidation and burning at elevated temperatures.
3. The series arrangement is not utilized in domestic circuits because, in a series circuit, the failure of one electrical appliance due to a defect results in the cessation of operation of all other appliances, as the entire circuit becomes interrupted.
4. The resistance of a wire is inversely proportional to its cross-sectional area, expressed as R ∝ (1/πr²). An increase in the cross-sectional area of a conductor of fixed length results in a decrease in resistance, as this allows for a greater number of free electrons to facilitate movement within the conductor.
5. Copper and aluminium wires are commonly utilized for electricity transmission due to their low resistances. Thus, they do not experience excessive heating when conducting electric current.

Related Study Resources of Chapter 11 – Electricity

Students can use the links below to get extra study materials for Class 10 Science Chapter 11: Electricity.

Sl No.	Related Links
1	Class 10 Science Chapter 11 Electricity – Important Questions
2	Class 10 Science Chapter 11 NCERT Textbook

Download Electricity NCERT Solutions PDF

You can download the PDF from the link below for offline study