12. Electricity SEBA/ASSEB Class 10 Science– Important Questions and Answers | Assam HSLC

SEBA/ASSEB Class 10 Science Chapter 12: Electricity Important Questions and Answers | HSLC Assam

Chapter 12: Electricity

Multiple Question and Answer:

Question: Electrical resistivity of a given metallic wire depends upon

(a) its length (b) its thickness (c) its shape (d) nature of the material

Answer : (d) nature of the material.

Question : The unit of specific resistance of a material in S.I. system is : [SEBA 2015]

(a) (b) (c) (d)

Answer : (b)

Question: 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) (b) (c) (d)

Answer : (a)

[ Here , I = 1 A , t = 16 s

We have,

Again, ]

Question : Two wires have lengths , resistances , resistivities – each in the ratio 1:2 . what is the ratio of their diameters ?

(a) 1:2 (b) 1:4 (c) 4:1 (d)

Answer : (d)

[ We have,

For first wire :

For second wire :

So,

]

Question: What is the maximum resistance which can be made using five resistors each of Ω?

(a) Ω (b) 10 Ω (c) 5 Ω (d) 1 Ω

Answer : (d) 1 Ω

[Here ,

]

Question: Which of the following represents voltage?

(a)

(b) Work done × Charge

(c)

(d) Work done × Charge × Time

Answer : (a)

Question: A cylindrical conductor of length l and uniform area of cross- section A has resistance . Another conductor of length and resistance R of the same material has area of cross section

(a) (b) (c) 2A (d) 3A

Answer : (c) 2A .

[ We have,

Again, ]

Question: The resistivity does not change if

(a) the material is changed (b) the temperature is changed

Answer: (c) the shape of the resistor is changed.

Question: 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

Answer : (c) 20 J.

[Here, , volt

We have,

again,

Therefore, the heat dissipated by the 4 Ω resistor in 5 s is 20 J. ]

Question: 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

Answer : (d) 5 A.

[ We have ,

and considering safety margins, a 5 A fuse is suitable.]

Question: Unit of electric power may also be expressed as

(a) volt ampere (b) kilowatt hour (c) watt second (d) joule second

Answer : (a) Volt ampere .

Question : How many electrons will contribute in one second in causing on one micro-ampere of current ?[SEBA2017]

(a) (b) (c) (d)

Answer : (c) .

[ We have, microampere , second , coulomb .

We have,

]

Question : An electron bulb is rated 220V and 100W . When it is operated on 110 V, then the power consumed is :[SEBA 2018]

(a) 75 W (b) 25 W (c) 100 W (d) 50 W

Answer : (d) 50 W

[Here,

We have,

Again,

]

Question : The practical unit of electrical energy 1 kWh means : [SEBA 2019]

(a) J (b) J (c) J (d) J

Answer : (b) J

Question : The SI unit of electric charge is [SEBA 2021]

(a) Coulomb (b) volt (c) Watt (d) joule

Answer : (a) Coulomb .

Question : Which of the following is used to measure the potential difference of a circuit ? [SEBA 2023]

(a) Ammeter (b) Voltmeter (c) Galvanometer (d) Multimeter

Answer : (b) Voltmeter .

Question: Electrical appliances that have metal body are generally connected to the earthing wire. What is the reason to earth the wire ? [2025]

(a) To prevent the excess of current (b) To prevent the leakage of current

Answer: (b) To prevent the leakage of current

[ Earthing safely carries leaked current to the ground, preventing electric shock if the metal body becomes live due to a fault.]

Question: An electric bulb is connected to a 200 V generator. The current is 0.4 A. What is the power of the bulb ? [2025]

(a) 100 W (b) 90 W (c) 80 W (d) 110 W

Answer: (c) 80 W .

[ Here,

We have, ]

Question: Which of the following terms does not represent electrical power in a circuit [2025]

(a) I²R (b) V²/R (c) VI (d) IR²

Answer: (b) .

Question: Study Ohm’s law (V = IR) , a student prepared the following table :

V (in volt)	0.4	0.8	1.2	Y
I (in ampere )	0.1	0.2	X	0.5

The respective value of X and Y will be

(a) 0.3 and 2 (b) 0.1 and 0.4 (c) 0.3 and 1.6 (d) 0.4 and 1.8

Answer: (a) 0.3 and 2

[ Here,

We have,

For X : Here,

For Y : Here,

]

Very Short Question and Answer :

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

Answer: A voltmeter is connected in parallel to the two points between which the potential difference is to be measured.

Question: On what factors does the resistance of a conductor depend ?

Answer: The resistance of a conductor depends on the factors are length, area of the cross-section and the nature of the materials.

Question: Name two sources of direct current . [2025 M]

Answer: Two sources of direct current (DC) are electric cell and battery .

Question: What does an electric circuit mean ? [1M]

Answers: A continuous and closed path of an electric current is called an electric circuit.
Question: Define the unit of current. [1M]

Answers: One ampere (the unit of current) is constituted by the flow of one coulomb of charge per second .
Question: Calculate the number of electrons constituting one coulomb of charge . [1M]

Solution: Here, ,

The number of electron

Question: How much work is done in moving a charge of 2 C across two points having a potential difference 12 V ? [1M]
Solution : Here, Q = 2 Coulomb , V = 12 volt and W = ?
W = VQ
= 12 V × 2 C
= 24 J

Question: A current of 0.5 A is drawn by a filament of an electric bulb for 10 minutes. Find the amount of electric charge that flows through the circuit. [1M]
Solution : Here, I = 0.5 A , t = 10 min = 600 second and Q = ?
We know that , Q = It
Q = 0.5 A × 600 s
Q = 300 C

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

Solution: Here, V = 6 volt , Q = 1 Coulomb and W = ?

We know that , W = VQ = 6 × 1 J = 6 J

Question: What is the commercial unit of electrical energy?

Answer: The commercial unit of electrical energy is the kilowatt-hour (kWh).

Question: How the life of filament of an electric bulb is prolonged ? [1M 2017]

Answer: The bulbs are usually filled with chemically inactive nitrogen and argon gases to prolong the life of the filament.

Question: How much current will an electric bulb draw from a 220 V source, if the resistance of the bulb filament is 1200 Ω?

Solution : Here, V = 220 V; R = 1200 Ω.
Using Ohm’s, law , we have

Question: How much current will an electric heater coil draw from a 220 V source, if the resistance of the heater coil is 100 Ω?
Solution: Here , V = 220 V, R = 100 Ω and I = ?

Using Ohm’s, law , we have

Question: What is meant by saying that the potential difference between two points is 1 V ?
Answer: The potential difference between two points is 1 volt that mean a current carrying conductor when 1 joule of work is done to move a charge of 1 coulomb from one point to the other.

Question: 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: Current flows more easily through a thick wire than through a thin wire of the same material, when connected to the same source. This is because the resistance of a wire decreases as its cross-sectional area increases. Hence, a thick wire offers less resistance to the flow of current.

Question: Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
Answer: Coils of electric toasters and electric irons are made of an alloy rather than a pure metal because alloys have higher resistivity, high melting points, and do not oxidise easily at high temperatures. Hence, they can produce heat efficiently and withstand high temperatures without melting or burning.

Short Question and Answer :

Question: An electric bulb is connected to a 220 V generator. The current is 0.50 A. What is the power of the bulb?
Solution: Here, V = 220 volt , I = 0.50 A

We have,

Question: The potential difference between the terminals of an electric heater is 60 V, when it draws a current of 4 A from the source. What current will the heater draw if the potential difference is increased to 120 V ? [2025 2M]

Answer: Here,

According to Ohm's law,

Again,

Question: The potential difference between the terminals of an electric heater is 60 V when it draws a current of 4 A from the source. What current will the heater draw if the potential difference is increased to 120 V?
Answer : Here, ,
According to Ohm’s law,

When the potential difference is increased to 120 volt .

Again, , ,

The current through the heater becomes 8 A.

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

Answer: Given: Resistance of each resistor (R) = 176 Ω

Current (I) = 5 A Voltage (V) = 220 V

The formula for total resistance in a parallel circuit is given by:

Let N be the number of resistors required .

For N resistors connected in parallel, the total resistance is:

1Rp=NR

Rp=RN

Again,

A/Q,

Therefore, you would need 4 resistors of 176 Ω each, connected in parallel, to carry 5 A on a 220 V line.

Question: 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) Highest Total Resistance: Here, , , and

In a series connection, the total resistance is the sum of the individual resistances.

So, the highest total resistance that can be obtained is 48 Ω when all the coils are connected in series.

(b) Lowest Total Resistance:

So, the lowest total resistance that can be obtained is 2 Ω when all the coils are connected in parallel.

Question: 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: Here, V = 9 volt , , , , and

In a series circuit, the total resistance is the sum of the individual resistances:

Using Ohm's Law,

Therefore, 0.672 A of current would flow through the 12 Ω resistor in the given circuit.

Question: Draw a schematic diagram of an electric circuit consisting of a cell, an electric bulb, an ammeter and a plug key.

Answer: The diagram of an electric circuit is :

Long Question and Answer :

Question: 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: For the first lamp: Here, = 100 W , V = 220 volt

Using Ohm's law,

For the second lamp: Here, = 60 W , V = 220 volt

Therefore,

Therefore, the total current drawn from the line when the supply voltage is 220 V is 0.73 A.

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

Answer: For the TV set: Here, = 250 W , = 1 hour = 60 × 60 seconds = 3600 seconds

Energy for the TV set J

For the toaster: Here, = 1200 W , = 10 minutes = 10 × 60 seconds = 600 seconds

Energy for the toaster J

Therefore ,the toaster uses less energy compared to the TV set.

Question: 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.

Answer: (i) For a combination resistance of 9 Ω:

Here, , and

We have,

So,

(ii) Fotr a combination resistance of 4 Ω:

Again,

Question: 100 J of heat are produced each second in a 4 Ω resistance. Find the potential difference across the resistor. [3M 2019]

Answer: Here, H = 100 J, R = 4 Ω, t = 1 s, V = ?

We have,

Thus the potential difference across the resistor, V is

Question: State Joule’s law of heating. 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. [2018]

Answer: Joule’s law of heating: The heat produced in a resistor is directly proportional to the square of the current flowing through it, its resistance and the time for which the current flows.

Where is the heat produced, is the current, is the resistance and is the time.

Here, V = 12 volt , Current (I) = 2.5 mA = 0.0025 A

Using Ohm's Law,

Therefore, the value of the resistance of the unknown resistor is 4800 Ω .

Question: 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?

Solution: (a) Total Resistance of 4 Ω:

Here, , ,

We have,

(b) Total Resistance of 1 Ω :

We have,

Question: 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? [2025 3M]

Answer: Given, V = 220 volt , P = 10 W , = 5 A

We know that ,

For each lamp: P = 10 W , V = 220 volt

The number of lamp

Therefore, 110 lamps can be connected in parallel with each other across the two wires of the 220 V line, given a maximum allowable current of 5 A .

Question: Explain (a) Why is tungsten used almost exclusively for filament of electric lamps ? [2025 3M]

(b) Why are copper and aluminium wires usually employed for electricity transmission ?

Answer: (a) Tungsten is used for the filament of electric lamps because it has a very high melting point and high resistivity. It can be heated to a very high temperature and emit light without melting.

(b) Copper and aluminium wires are usually used for electricity transmission because they have low resistivity, which means they offer less resistance to the flow of electric current and reduce the loss of electrical energy as heat.

Question: Electric heaters designed to be used on a 220 V electric supply line are rated as 11 W. If the maximum supply current is 5A, how many electric heaters can be connected in parallel with each other across 220 V supply line ? How much total energy is consumed by these in 20 minutes ? [3M 2017]

Answer: Given, V = 220 volt , P = 11 Watt , I = 5 A , t = 20 minutes = 20 × 60 = 1200s

We have,

Let, n be the number of heaters connected in parallel.

The number of heaters

We have,

Question: What determines the rate at which energy is delivered by a current ? An electric heater of resistance 8 Ω draws 15A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater. [3M 2020]

Answer: The rate at which energy is delivered by an electric current is determined by the electric power.

Here, R = 8 Ω , I = 15 A and t = 2 hours = 2 × 60 × 60 seconds = 7200 seconds
We know that,

Question: Write the S.I. unit of resistivity . An electric motor takes 5 A from a 220V source. Determine the power of the motor and energy consumed in 2 hours. [3M 2020]

Answer: The S.I. unit of resistivity is .

Here,

We have,

Again,

Question: The resistance of a metal wire of length 1 m is 26 Ω at 20°C. If the diameter of the wire is 0.3 mm, what will be the resistivity of the metal at that temperature? [3M 2019]

Answer: Here, , ,

We have,

The resistivity of the metal at 20°C is .

Question: An electric iron consumes energy at a rate of 840 W when heating is at the maximum rate and 360 W when the heating is at the minimum. The voltage is 220 V. What are the current and the resistance in each case?

Solution: (a) When heating is at the maximum rate : Here, P = 840 watt , V = 220 volt

We know that ,

and the resistance of the electric iron is

(b) When heating is at the minimum rate : Here, P = 360 watt , V = 220 volt

and the resistance of the electric iron is .

Question: A copper wire has diameter 0.5 mm and resistivity of . 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?

Solution: Given, , ,

We know that ,

Therefore, the length of the copper wire required to have a resistance of 10 Ω is 122.5 meters.

The resistance changes if the diameter of the wire is doubled.

Here, d'=2d and

We have,

Now,

Therefore, the resistance is time when the diameter of the wire is doubled.

Question: 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?
Solution: Here, , , and volt

We have,

The total current (I) drawn by these three appliances when connected in parallel to the 220 V source.

Using Ohm's law ,

So, the resistance of the electric iron is 31.25 Ω, and the current through it is 7.04 A.

Question: An electric lamp, whose resistance is 20 Ω, and a conductor of 4 Ω resistance are connected to a 6 V battery .

Calculate (a) the total resistance of the circuit, (b) the current through the circuit, and (c) the potential difference across the electric lamp and conductor.

Solution: (a) Here, , and

The total resistance

(b) We have,

For the conductor : Here,

Question: In the circuit diagram given in Fig. 12.10, suppose the resistors , and have the values 5 Ω, 10 Ω, 30 Ω, respectively, which have been connected to a battery of 12 V. Calculate (a) the current through each resistor, (b) the total current in the circuit, and (c) the total circuit resistance.

Answer: Here, , , and

(a) Using Ohm’s law ,

, and

(b) The total current in the circuit

Question: In given figure , ,,, , and a 12 V battery is connected to the arrangement.

Calculate (a) the total resistance in the circuit, and (b) the total current flowing in the circuit.

Answer: Here, ,,, , and

(a) The total resistance in the circuit : For resistance and :

For resistance , and :

(b) The total current flowing in the circuit : Here,

We have,

Question: 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: Here, , , and volt

We have,

The total current (I) drawn by these three appliances when connected in parallel to the 220 V source.

Using Ohm's law ,

So, the resistance of the electric iron is 31.25 Ω, and the current through it is 7.04 A.

Question: 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: Given, V = 220 volt ,

Resistance (R) of each coil (A and B) = 24 Ω

Case 1: Using coil A : Here, = 24 Ω , V = 220 V

Applying Ohm's Law,

The current flowing through coil A when used separately is 9.17 A.

Case 2: Using coils A and B :

Total resistance () , V = 220 volt

Applying Ohm's Law,

The current flowing through the series combination of coil A and coil B is 4.58 A.

Case 3: Using coils A and B in parallel :

Total resistance () , Voltage (V) = 220 V

Applying Ohm's Law,

The current flowing through the parallel combination of coil A and coil B is 18.33 A.

Posted 2 weeks ago

12. Electricity SEBA/ASSEB Class 10 Science– Important Questions and Answers | Assam HSLC

SEBA/ASSEB Class 10 Science Chapter 12: Electricity Important Questions and Answers | HSLC Assam

Chapter 12: Electricity

Multiple Question and Answer:

Very Short Question and Answer :

Short Question and Answer :

Long Question and Answer :

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