Electricity Notes And Questions Answers | Chapter 12 | Class 10 | Science 2024

Last updated on July 14th, 2024 at 05:07 pm

Electricity Notes And Questions Answers

Below are some of the very important NCERT Class 10 Science chapter 12 Electricity Notes And Questions Answers. These Class 10 Electricity Notes And Questions Answers have been prepared by expert teachers and subject experts based on the latest syllabus and pattern of term 2. Questions with Answers to help students understand the concept.

These Questions for Class 10 Science Electricity Notes And Questions Answers are very important for the latest CBSE term 2 pattern. These class 10 notes and Q and A are very important for students who want to score high in CBSE Board.

We have put together these NCERT  Questions of Class 10 Science chapter 12 Electricity Notes And Questions Answers for practice on a regular basis to score high in exams. Refer to these Questions with Answers here along with a detailed explanation.

Electricity is one of the most convenient and widely used  forms of energy in today’s world. It is a controllable and  convenient form of energy. 

Ohm’s Law 

It gives a relationship between current I, flowing in a metallic wire and potential difference V, across its terminals. According to this law, the electric current flowing through a  conductor is directly proportional to the potential difference applied across its ends, providing the physical conditions  (such as temperature) remain unchanged. 

If V is the potential difference applied across the ends of a conductor through which current I flows, then according to Ohm’s law,

     V  ∝ I   

 or,   V  =  1R

or,     I = V / R

Where, R is the constant of proportionality called resistance of the conductor at a given temperature.

The conductors which obey Ohm’s law are called ohmic conductors while the conductors which do not obey Ohm’s law are called non-ohmic conductors. 

V-l Graph

The graph between the potential difference V and the corresponding current I is found to be a straight line passing through the origin for ohmic (metallic) conductors.

Resistance

It is that property of a conductor by virtue of which it opposes/resists the flow of charges/current through it, Its SI  unit is ohm and it is represented by Ω

Resistance of a conductor is given by, 

R = V/1

It is said to be 1 ohm, if a potential difference of 1 volt across  the ends of the conductor makes a current of 1 ampere to flow through it.

Factors on which the Resistance of a Conductor Depends

The electrical resistance of a conductor depends on the following factors 

(i) Length of the conductor  The resistance of conductor h is directly proportional to the length l

i.e. R ∝ l                ——   (i)

(ii) Area of cross section of the conductor The resistance of a conductor R is inversely proportional to its area of cross-section A

i.e. R ∝ l / A           —–     (ii)

(iii) Nature of the material of the conductor The resistance of a conductor depends on the nature of the material of which it is made. Some material have low resistance,  whereas others have high resistance

Therefore,  from Eqs. (i) and (ii), we can write 

        R ∝ l / A 

Or,   R ∝ ρ l / A

Where, ρ is the constant of proportionality and is called electric resistivity or specific resistance of the material of the conductor

Resistivity

It is defined as the resistance of a conductor of unit length and unit area of cross-section. Its SI unit is ohm-meter (Ω-m).

• The resistivity of a material does not depend on its length or thickness but depends on the nature of the substance and temperature. It is a characteristic property of the material of the conductor and varies only if its temperature changes.
• Insulators such as glass, rubber, ebonite, etc., have a very  high resistivity (1012 to 1017 Ω -m), while conductors have a very low resistivity (10-8 to 10-6 Ω-m).
• Alloys have higher resistivity than that of their constituents in metals. They do not oxidize easily at high temperatures, this is why they are used to make heating elements of devices such as electric iron, heaters, etc.  
• Tungsten is almost exclusively used for filaments of electric bulbs, whereas copper and aluminum are generally used for electrical transmission lines.

Combination of Resistors

There are two methods of joining the resistors which are as given below  

1. Series Combination

When two or more resistors are connected end to end to each other, then they are said to be connected in series.

i.e.,     

The equivalent resistance is thus greater than the resistances of either resistor. This is also Known as maximum effective resistance. 

The current through each resistor is the same. The potential  difference across each resistor is different. 

2. Parallel Combination 

When two or more resistors are connected simultaneously  between two points to each other, then they are said to be  connected in  parallel combination.

The equivalent resistance is less than the resistance of either resistor, This is also known as minimum effective resistance.

The current from the source is greater than the current through either resistor. The potential difference across each resistor is the same.

Heating Effect of Electric Current

When an electric current is passed through a high resistance wire like nichrome wire, then the wire becomes very hot and produces heat.

In purely resistive circuits, the source of energy continuously gets dissipated entirely in the form of heat. This is called the  heating effect of current.

This is obtained by the transformation of electrical energy into heat energy. e.g. electric heater, electric iron, etc.  Heat produced is expressed as, H = I² x R x t 

It is known as Joule’s law of heating.

This law implies that heat produced in a resistor is  

(i) directly proportional to the square of current for a given resistance. 

(ii) directly proportional to the resistance for a given current. 

(iii) directly proportional to the time for which the current  flows through the resistor.

Practical Applications of Heating Effect of Electric Current

There are two applications of heating effect of electric current which are given below  

1. Electric Bulb

It has a filament made of tungsten. So, most of the power  consumed by this, is dissipated in the form of heat and some part is converted into light because it has high resistivity and high melting point,

The filament is thermally isolated and the bulb is filled with chemically inactive nitrogen and argon gas to prolong the life of filament.

2. Electric Fuse

It is used as a safety device in household circuits. It protects the circuits, by stopping the flow of any unduly high electric current. It is connected in series with the mains supply. It  consists of an alloy of lead and tin which has an appropriate melting point.

When the current flowing through the circuit exceeds the safe limit, then the fuse wire melts and breaks the circuit. This helps to protect the other circuit elements from heavy current. Fuses are always rated for different current values such as 1 A, 2 A, 5 A, 10 A, 15 A, etc.

Electric Power

It is defined as the amount of electric energy consumed in a circuit per unit time.

Electric power is expressed as, 

P = VI 

or, P = V² / R

The SI unit of electric power is watt (W). 

It is said to be 1 watt, if 1 ampere current flows through a circuit having 1 volt potential difference.

i.e   1 watt = 1 volt x 1 ampere = 1 VA

Commercial units of electrical energy are kilowatt-hour.

1 kWh =  3.6 x 10⁶ J

Frequently Asked Questions

Short Answer Type Questions  

1.  State the law which gives the relationship between current and potential difference. Define the unit of resistance.

Ans. Ohm’s gives the relationship between current I flowing in a metallic wire and potential difference V, across its terminals. According to this law, the electric current flowing through a conductor is directly proportional to the potential difference applied across its ends, providing physical conditions remain unchanged.

i.e.     V ∝  I or V = lR

where R is constant of proportionality called resistance of conductor at a given temperature.

Unit of resistance is ohm. It is said to be 1 ohm, if the potential difference of 1 volt across the ends of the conductor makes a current of I A to flow through it.  

i.e.  1 ohm (Ω) = 1 V / 1 A = VA-1

2. Define resistance. Give its SI unit.    (CBSE 2019)

Ans. Resistance is the property of a conductor by virtue of which it opposes/resist the flow of charges / flow of current through it. It’s SI unit is ohm and is represented by the Greek letter  Ω (ohm). Resistance of a conductor is given by R  = V / I   

3. The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are as given below

V (volts)      0.5  1.0  1.5  2.0  2.5  3.0  4.0 5.0

I (amperes) 0.1  0.2  0.3  0.4  0.5  0.6  0.8 1.0

Plot a graph between V and I and also calculate the resistance of that resistor.    (CBSE 2018)

Ans. Scale : At x-axis, 1 div (1 cm) = 0.1 A

At y-axis, I div (1 cm) = 0.5 V

The graph between current and potential difference is shown below

Resistance (R) of the resistor is determined by the slope of V- I graph

 Resistance  = R = Slope of graph = V = I

  R = y2 – y1 / x2 – x1 = 1.5 – 1.0 / 0.3 – 0.2

     = 0.5 / 0.1  = 5 Ω

4. Why are coils of electric toasters and electric irons made of alloy rather than a pure metal?    (NCERT)

Ans. Alloys have a higher resistibility than their constituent metals. They do not oxidize or burn at higher temperatures as they have a high melting point. Thus, they are used to make coils of electric toaster and electric irons rather than pure metals

5. (i) List the factors on which the resistance of a conductor in the shape of a wire depends.

(ii) Why are metals good conductors of electricity, whereas glass is a bad conductor of electricity? Give a reason. 

(iii) What type of material is used in electrical heating devices? Give a reason. (CBSE 2018)  

Ans. 

(i) Following factors on which resistance of a wire depends 

a. length of wire = R ∝ 1

b.  Area of cross-section of wire : R ∝ I / A

c.   Resistivity of material of wire: R ∝ ρ

      R = ρ l / A

(ii) Metals are good conductor as their resistivity is very low whereas glass is a bad conductor having high resistivity 

(iii) Alloys are used as heating elements as their resistivity and melting points both are very high

6. 

A student has two resistors 2 Ω and 3 Ω. She has to put one of them in place of R2 as shown in the circuit. The current that she needs in the entire circuit is exactly 9 A.

Show by calculation which of the two resistors she should choose,

7. A metal wire has a diameter of 0.25 mm and electrical resistivity of 0.8 x 10 -8 Ω-m.

(i) What will be the length of this wire to make a resistance of 5 Ω?

(ii) How much will the resistance change, if the diameter of the wire is doubled?

8. Redraw the given circuits putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference 12 Ω resistor. What would be the readings in ammeter and voltmeter? (NCERT)

Ans. 7 and 8

Equivalent resistance of the circuit

R = R1 + R2 + R3 = 5 + 8 + 12 = 2.88 Ω

In series combination, current flowing through all the resistance is same and equal to the total current flowing through the circuit.

 ∴ Current in the resistors

     l = V / R = 6 / 25 = 0.24 A

 ∴ Ammeter reading = 0.24 A

Potential across 12      resistance 

V =  lR =  0.24 × 12 =  2.88 V 

∴ Voltmeter reading is 2.88 v

9. What is the reason for connecting electrical appliances in parallel in household circuits? 

Ans. Parallel combination of resistance is highly useful in circuits used in daily life, as the circuits used have components of different resistance that require different amounts of current.

The type of combination in a circuit divides the current among the component electrical gadgets so that they can have the necessary amount of current to operate properly. This is the reason for connecting electrical appliances in parallel in household circuits.

10. What is (i) the highest and (ii) the lowest total resistance which can be secured by combinations of four coils of resistances 

4Ω, 8 Ω, 12 Ω and 24 Ω? (NCERT) 

Ans: (i) resistance is maximum when resistors are connected in series.

11. Consider the circuit diagram as given below

12. With the help of suitable circuit diagrams prove that the reciprocal of the equivalent resistance of a group of resistances joined in parallel is equal to the sum of the reciprocals of the individual resistances (CBSE 2019) 

Ans. 11 and 12

13. A battery E is connected to three identical lamps P, Q and R as shown in figure. Initially, the switch S is kept open and the lamp P and Q are observed to glow with the same brightness. Then, S is closed. 

How will the brightness of the glow of bulbs P and Q change? Justify your answer.

Ans. The brightness of glow bulb P will increase and brightness of bulb Q  will decrease.

This is because, on closing S, bulbs Q and R will be in parallel and the combination will be in series with bulb P. Hence, the total resistance of the circuit will decrease and the current flowing in the circuit will increase because the glow of bulb P will increase.

Also bulbs Q and R will be in parallel in this case.  So, the current gets divided and lesser current flows through Q.  Hence, the bulb Q decreases.

14. (i) Write Joule’s law of heating.

(ii) Two lamps, one rated 100 W, 220 V and the other 60W, 220 V are connected in parallel to the electric main supply.

Find the current drawn by two bulbs from the line, if the supply voltage is 220 V. (CBSE 2018)

15. In an electrical circuit, two resistors of 2Ω and 4 Ω are connected in series to a 6 V battery. Find the heat dissipated by the 4 Ω resistor in 5 s. (NC,ER’I’ Exemplar)

Ans. 14 and 15

And 14.

(i) According to joule’s law of heating amount of heat produced in a resistor is 

1. Directly proportional to square of current flowing through the resistor 

H  ∝ l²

2. Directly proportional to the resistance of the resistor 

H ∝R

3. Directly proportional to time for which the current flows through the resistor

H ∝ t

Hence, H = l² Rt

(ii)

16. Why does the cord of an electric heater not glow while the heating element does?    (NCERT)

Ans. The heating element of the heater is made up of an alloy that has very high resistance. So, when the current flows through it, it becomes very hot and glows red. But the resistance of the cord is less because it is made up of copper or aluminum, so it does not glow.

17. Why are fuses used in electrical circuits?

Ans. Electric fuse is used to protect the electrical circuit from overloading and short circuits. When the current flowing through a circuit exceeds the safe limit, the temperature of fuse wire increases and due to heating effect, it melts and breaks the circuit.

18. Define the Sl unit of electric power, What is the commercial unit of electrical energy?

Ans. The SI unit of electric power is watt (W). It is the power consumed by a device that carries l A of current when operated at a potential difference of 1 V. 

Thus, 1 W = I volt x 1 ampere = 1 VA 

The commercial unit of electrical energy is kilowatt hour (kWh), commonly known as ‘unit’. 

1 kWh = 3.6 x 10⁶ J

19. The electric power consumed by a device may be calculated by using either of the two expressions

P = I²R or P = V² / R

The first expression indicates that the power is directly proportional to R, whereas the second  expression indicates inverse proportionality. How can seemingly different dependence of P on R in these expressions be explained?

Ans. P =  l²R is used when current flowing in every component of the circuit is constant. This is the case of a series combination of the devices in the circuit.

P = V² / R is used when potential difference (V) across every component of the circuit is constant. This expression is used in case of parallel combination in the circuit in series combination R is greater than the value of R in parallel combination

20. Three 2 Ω resistors, A, B and C are connected as  shown in figure. Each of them dissipates energy and can withstand a maximum power of 18 W without melting. Find the maximum current that can flow through the three resistors? (NCERT Exemplar)

21. An electric geyser rated 1500 W, 250 V is connected to a 250 V line mains. Solve 

(i) the electric current drawn by it.

(ii) energy consumed by it in 50 h.

(iii) cost of energy consumed, if each unit costs ₹6.

Long Answer Type Questions 

22. In the given circuit, A, B, C and D are four lamps connected with a battery of 60 V.

Analyze the circuit to answer the following questions. 

(i) What kind of combination are the lamps arranged in (series or parallel)?

(ii) Explain with reference to your above answer, what are the advantages (any two) of this combination of lamps?

(iii) Explain with proper calculations which lamp glows the brightest.

(iv) Find out the total resistance of the circuit.

Ans. 20, 21, and 22

(i) In the circuit all the lamps have seen voltage i.e. 60 V but each lamp is having different current. So, the lamps are arranged in parallel combination 

(ii) The two advantage of lamps in parallel combination are 

1. If one lamp get faulty, it will not affect the working of other lamps 
2. In a parallel combination of lamps, each lamp will use the full potential of the battery.

(iii) The lamp with the highest power will grow the brightest 

As, power =  voltage ×  current 

In this case, all the lamps have the same voltage of 60 volt.

For lamp A, current  = 3 A   

                Power = 60 × 3 = 180w

For lamp B, current  = 4 A   

                Power = 60 × 4 = 240 w

For lamp C,  current  = 5 A   

                Power = 60 × 5 = 300 w

For lamp D, current  = 3 A   

                Power = 60 × 3 = 180w

(iv) Let R  be the total resistance of the circuit 

Total current in the circuit, l = 3 + 4 + 5 + 3 = 15 A

Voltage. V = 60 V

Using Ohm’s Law, V = IR

        R = V / l = 60 / 15 = 4 Ω

23. Find the equivalent resistance of the following circuit, find the current and potential at each resistor.

Ans. In the given circuit R2, R3 and R 4 are in parallel combination. As, current through R2, R3 and R4 are different so their equivalent resistance R is

24.(i) State Ohm’s  

(ii) How is an ammeter connected in an electric circuit?

(iii) The power of a lamp is 100 W. Find the energy consumed by it in 1 min.     

(iv) A wire of resistance 5 Ω is bent in the form of a closed circle. Find the resistance between two  points at the ends of any diameter of the circle.

Ans. 

(i) A state that the electric current flowing through a conductor directly proportional to the potential difference applied across its end provided the physical conditions such as temperature remains unchanged 

(ii) An ammeter is connected in series in a circuit

(iii) Given, P = 100w and time t = 1 minute = 60s

As, Energy E = Pt = 100 × 60 = 6000j

(iv) Given, a wire with resistance 5 ohm

Now, wire is converted into a ring as shown below

25. An electric lamp of resistance 20 Ω and a conductor  of resistance 4 Ω are connected to a 6 V battery as shown in the circuit given below.

Calculate

(i) the total resistance of the circuit.

(ii) the current through the circuit. 

(iii) the potential difference across the (a) electric lamp and (b) conductor.

(iv) the power of the lamp,    (CBSE 2019) 

Ans. Given, resistance of lamp (R1) = 20 Ω

Resistance of conductor (R2) = 4 Ω

Potential difference of battery (V) = 6 V

(i) Total resistance, 

R = R1 + R2 = 20 + 4 = 24 Ω

(ii)  Current through the circuit 

     l = V / R = 6 / 24 = 0.25 A

(iii)  (a) Potential difference across electric lamp = lR1

(b) Potential difference across a conductor 

      = lR2 = 0.25 × 4 = 1 V

(iv) Power of lamp = l²R1

  = (0.25)² × 20

  = 0.0625 × 20 = 1.25 W

26. (i) How should two resistors, with resistances R1(Ω) and R2 (Ω) be connected to a battery of e.m.f. V volts so that the electrical power consumed is minimum?

(ii) In a house, 3 bulbs of 100 watt each are lighted for 5 hours daily, 2 fans of 50 watt each are used for 10 hours daily and an electric heater of 1.00 kWh is used for half an hour daily. Calculate the total energy consumed in a month of 31 days and its cost at the rate of ₹ 3.60 per kWh.

Ans.

27. (a) Define power and state its SI unit,

(b) A torch bulb is rated 5V and 500 mA, Calculate its

(i) power

(ii) resistances

(iii) energy consumed when it is lit for 2½ hours.

Ans. 

(a) Power is defined as the rate at which electric energy is dissipated for consumed in an electric circuit 

  P = VI = I²R = V² / R

The SI unit of electric power is watt (W)

It is the power consumed by the device that carries l A of current when operated at a potential difference of 1V 

  1 W = 1 volt × 1 ampere = 1 VA

(b) Given, voltage rating V = 5 V  and current rating I = 500 mA

(i) As, we know power of bulb

     P = Vl = 5 × 500 × 10-3 = 2.5 W 

        = 2.5 × 10-3 kW

(ii) Resistance of bulb, 

      R = V / I

      R = 5 / 500 × 10-3 = 10 Ω

(iii) Energy consumed in 2½ hour

       E = P.t = 2.5 × 2.5 / 1000

          = 6.25 / 1000 

          = 0.00625 kWh

Case Based Questions

28. Read the following and answer the questions from (i) to (v) given below 

Several resistors may be combined to form a network. The combination should have two endpoints to connect it with battery or other circuit elements. When the resistors are connected in series, the current in each resistor is the same but the potential difference is different in each resistor. 

When the resistors are connected in parallel, the voltage drop across each resistor is same but the current is different in each resistor. 

(i) What do you mean by complex circuit? 

(ii) In the graph, it shows the resistance in series and parallel for two identical wires. Which of the  following denotes series combination and parallel combination, individually?

(iv) What is minimum effective resistance?

(v) When three resistors of resistances R, 2R and 3R are connected in series then, how will the value of current get affected in each resistor by applying a voltage V across the circuit?

Ans. (i) The electrical circuit in which some resistance are connected in series combination and some in parallel combination this type of combination is called Complex circuit.

(ii) Resistance in series is always greater than resistance in parallel. The slope of V-l graph gives resistance. Hence, line B denotes resistance in series combination and line A denotes resistance in parallel combination.

(iii) Equivalent circuit diagram is

(iv) When the equivalent resistance is less than the resistance of least resistor. This is known as minimum effective resistance.

(v) In series combination, the value of current is same through each resistor.

So, it will not be affected at all.

29. Read the following and answer the questions from (i) to (v) given below

A cell or a battery is the source of electrical energy. Due to the chemical reactions inside them a potential difference is setup which is responsible for the flow of current through any electrical circuit. So, to maintain this flow, the source continuously has to provide the energy. But only a part of this energy helps in maintaining the current consumed into useful work.

Rest of it may be consumed in the form of heat by raising the temperature of the appliances.

(i) How heat produced in a resistor is related to current flowing in that resistor?

(ii) Give two practical applications of the heating effect of current.

(iii) Which type of energy is transformed into heat energy? 

(iv) An electric iron of resistance 25 Ω takes a current of 7 A. Calculate the heat developed in 0.5 min.

(v) 200 J of heat is produced in 10 s in a 5 Ω resistance.  Find the potential difference across the resistor.

Ans.

(i)  According to joule’s law of heating, the heat produced in a resistor is directly proportional to the square of current flowing through that resistor 

(ii) Practical application of heating effect of current is electric heater, electric iron etc. 

(iii) Electrical energy is transformed into heat energy

(iv) Given, resistance  R = 25 Ω, 

      current l =7A

   Time t = 05 min = 05 × 60 = 30s Heat H = ?

    We know that, Heat H = l²Rt

      H = (7)² × 25 × 30 = 36750j

         = 3.68 × 10⁴ J

So, the heat developed is 3.68 × 10⁴ J

30. Read the following and answer questions from (i) to (v) given below

In resistance for a system of the resistor, there are two methods of joining the resistors together as shown below

It showed an electric circuit in which 3 resistors having resistor R1, R2 and R3 respectively are joined end to end i.e in series.

While the combination of the resistors in which 3 resistors connected together which point X and Y are said to be parallel.

(i) Calculate the potential difference across a series combination of resistors. 

(ii) What is the value of current in a series combination? 

(iii) Write the formula of electrical energy  dissipated in the resistor. 

(iv) If 200 resistors, each of value 0.2 Ω are connected in series, what will be the resultant resistance?  

(v) What will be the effective resistance shown in figure below? 

Ans. 

(i) An applied potential V produces current l in the resistors and R1, R2 and R3 causing a potential drop V1, V2 and V3 respectively, through each resistor 

Total potential, V = V1 +  V2 + V3 

(ii) In series combination of resistors, the current is same at every point of the circuit, i.e. the current through each resistor is same

Final Words

From the above article, you have practiced Class 10 chapter 12 Electricity Notes And Questions Answers. We hope that the above-mentioned notes and Q & A for term 2 will surely help you in your exam. 

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