11. Electricity Class 10 Science [LATEST] Solutions Text-book Questions in english - CBSE Study

Text-book Questions

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Page no. 200 

Q1. What does an electric circuit mean?

Ans: A continuous and closed path of an electric current is called an electric circuit. In which various electric components are arranged in series or parrallel. 

Q2. Define the unit of current.

Ans: S.I unit of current is Ampere, which denoted by Letter 'A'. Ampere is defined as "When one coulomb of charge flows in one second it is callled one Ampere of current".

Q3. Calculate the number of electrons constituting one coulomb of charge.

Ans: One electron possesses a charge of 1.6 × 10^-19 C, 

Therefore, Charge on 1 electron = 1.6 × 10^-19 C

Total Charge = 1 coulomb = 1C (given) 

The number of elecrons = ? 

Total Charge = number of electrons × Charge on 1 electron

1C = number of electrons × 1.6 × 10^-19 C

The number of electrons constituting one coulomb of charge is 6 × 10¹⁸

Page no. 202 

Q1. Name a device that helps to maintain a potential difference across a conductor.

Ans: Cell or Battery is a device that helps to maintain a potential difference across a conductor. 

Q2. What is meant by saying that the potential difference between two points is 1 V?

Ans: If 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 point is 1 V.  

Q3. How much energy is given to each coulomb of charge passing through a 6 V battery?

Ans: The energy given to each coulomb of charge is equal to the amount of work done to move it. 

V = W/Q

Chapter 11: Electricity

NCERT Textbook Questions & Solutions

Page 209

Q1. On what factors does the resistance of a conductor depend?

Solution:

The resistance of a conductor depends on the following factors:

• The length of the conductor – Resistance increases with an increase in length.
• The cross-sectional area – Resistance decreases as the area increases.
• The nature of the material – Different materials have different resistivities.
• The temperature of the conductor – For metallic conductors, resistance generally increases with temperature.

Answer: Resistance depends on the length, cross-sectional area, nature of the material, and temperature of the conductor.

Q2. Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?

Solution:

A thick wire has a larger cross-sectional area than a thin wire. Since resistance is inversely proportional to the cross-sectional area, a thick wire offers less resistance to the flow of current.

Answer: Current flows more easily through a thick wire because it has lower resistance.

Q3. Let the resistance of an electrical component remain constant while the potential difference across its ends decreases to half of its former value. What change will occur in the current through it?

Solution:

According to Ohm's law,

V = IR

Since the resistance remains constant, the current is directly proportional to the potential difference.

If the potential difference becomes half, the current also becomes half.

Answer: The current through the component will become half of its original value.

Q4. Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?

Solution:

Alloys such as nichrome are used because they have:

• High electrical resistance.
• High melting point.
• Do not oxidise or burn easily even at high temperatures.

These properties make alloys suitable for heating elements.

Answer: Heating coils are made of alloys because they have high resistance, high melting point, and resist oxidation.

Q5. Use the data in Table 12.2 to answer the following:

(a) Which among iron and mercury is a better conductor?

Solution:

A material with lower resistivity is a better conductor.

The resistivity of iron is lower than that of mercury.

Answer: Iron is a better conductor than mercury.

(b) Which material is the best conductor?

Solution:

Among the materials listed in Table 12.2, silver has the lowest resistivity.

Answer: Silver is the best conductor of electricity.

Page 213

Q1. 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, a 12 Ω resistor, and a plug key, all connected in series.

Solution:

The required circuit consists of:

• A battery of three 2 V cells connected in series (Total voltage = 6 V).
• A plug key (K).
• A 5 Ω resistor.
• An 8 Ω resistor.
• A 12 Ω resistor.

All the components are connected in series.

 (+) ── K ── [5Ω] ── [8Ω] ── [12Ω] ── (−)
          │
      Battery (3 × 2 V)

Q2. 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?

Solution:

Given:

• Total voltage = 6 V
• Resistors = 5 Ω, 8 Ω and 12 Ω (in series)

Total resistance,

R = 5 + 8 + 12 = 25 Ω

Using Ohm's law,

I = V/R = 6/25 = 0.24 A

Therefore, the ammeter reads 0.24 A.

Potential difference across the 12 Ω resistor,

V = IR = 0.24 × 12 = 2.88 V

Answer:

• Ammeter reading = 0.24 A
• Voltmeter reading = 2.88 V

Chapter 11: Electricity

NCERT Textbook Questions & Solutions

Page 216

Q1. Judge the equivalent resistance when the following are connected in parallel – (a) 1 Ω and 10⁶ Ω, (b) 1 Ω, 10³ Ω and 10⁶ Ω.

Solution:

(a) For 1 Ω and 10⁶ Ω in parallel:

Using the formula,

1/R = 1/R₁ + 1/R₂

1/R = 1/1 + 1/10⁶

Since 1/10⁶ is extremely small compared to 1,

R ≈ 1 Ω

(b) For 1 Ω, 10³ Ω and 10⁶ Ω in parallel:

1/R = 1/1 + 1/10³ + 1/10⁶

The additional terms are very small compared to 1.

R ≈ 1 Ω

Answer:

• (a) Equivalent resistance = 1 Ω (approximately)
• (b) Equivalent resistance = 1 Ω (approximately)

Q2. 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:

Current through the lamp:

I₁ = 220/100 = 2.2 A

Current through the toaster:

I₂ = 220/50 = 4.4 A

Current through the water filter:

I₃ = 220/500 = 0.44 A

Total current:

I = 2.2 + 4.4 + 0.44 = 7.04 A

Let the resistance of the electric iron be R.

Using Ohm's law,

R = V/I = 220/7.04 = 31.25 Ω

Answer:

• Resistance of the electric iron = 31.25 Ω
• Current through the iron = 7.04 A

Q3. What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?

Solution:

In household wiring, electrical appliances are connected in parallel because:

• Each appliance receives the full supply voltage.
• Each appliance can be switched on or off independently.
• If one appliance stops working, the others continue to operate.
• Each appliance draws the required current according to its resistance.

Answer: Parallel connection ensures proper voltage, independent operation, and uninterrupted functioning of other appliances.

Q4. 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 = 4 Ω

First connect 3 Ω and 6 Ω in parallel.

Equivalent resistance,

1/R = 1/3 + 1/6 = 3/6 = 1/2

R = 2 Ω

Now connect this 2 Ω in series with the remaining 2 Ω resistor.

Total resistance = 2 + 2 = 4 Ω

(b) Total resistance = 1 Ω

Connect all three resistors in parallel.

1/R = 1/2 + 1/3 + 1/6

= 3/6 + 2/6 + 1/6 = 6/6 = 1

Therefore,

R = 1 Ω

Answer:

• (a) Connect 3 Ω and 6 Ω in parallel, then connect the combination in series with 2 Ω.
• (b) Connect all three resistors in parallel.

Q5. What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistances 4 Ω, 8 Ω, 12 Ω and 24 Ω?

Solution:

(a) Highest resistance:

The highest resistance is obtained when all resistors are connected in series.

R = 4 + 8 + 12 + 24 = 48 Ω

(b) Lowest resistance:

Connect all resistors in parallel.

1/R = 1/4 + 1/8 + 1/12 + 1/24

= 6/24 + 3/24 + 2/24 + 1/24

= 12/24 = 1/2

Therefore,

R = 2 Ω

Answer:

• (a) Highest resistance = 48 Ω
• (b) Lowest resistance = 2 Ω

Chapter 11: Electricity

NCERT Textbook Questions & Solutions

Page 218

Q1. Why does the cord of an electric heater not glow while the heating element does?

Solution:

The heating element of an electric heater is made of an alloy such as nichrome, which has a high electrical resistance. Due to its high resistance, a large amount of heat is produced when current passes through it, causing it to become red hot and glow.

On the other hand, the connecting cord is made of copper, which has very low resistance. Therefore, very little heat is produced in the cord, so it does not become hot enough to glow.

Answer: The heating element glows because it has high resistance and produces more heat, whereas the copper cord has low resistance and does not produce enough heat to glow.

Q2. Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.

Solution:

Given:

• Charge, Q = 96000 C
• Potential difference, V = 50 V

The heat (electrical energy) produced is given by:

H = V × Q

H = 50 × 96000

H = 4,800,000 J

Answer: The heat generated is 4.8 × 10⁶ J.

Q3. An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s.

Solution:

Given:

• Resistance, R = 20 Ω
• Current, I = 5 A
• Time, t = 30 s

According to Joule's law of heating,

H = I²Rt

H = (5)² × 20 × 30

H = 25 × 20 × 30

H = 15000 J

Answer: The heat developed is 15,000 J.

Page 220

Q1. What determines the rate at which energy is delivered by a current?

Solution:

The rate at which electrical energy is consumed or delivered is called electric power.

Electric power depends on:

• The potential difference across the device.
• The current flowing through the device.

Mathematically,

P = VI

Thus, the greater the current or potential difference, the greater is the rate of energy transfer.

Answer: The rate at which energy is delivered by a current is determined by the electric power (P = VI).

Q2. An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.

Solution:

Given:

• Potential difference, V = 220 V
• Current, I = 5 A
• Time, t = 2 h

Step 1: Calculate the power

P = VI

P = 220 × 5

P = 1100 W = 1.1 kW

Step 2: Calculate the energy consumed

Energy = Power × Time

= 1.1 kW × 2 h

Energy = 2.2 kWh

In SI units,

Energy = 1100 × 7200 = 7.92 × 10⁶ J

Answer:

• Power of the motor = 1100 W (1.1 kW)
• Energy consumed in 2 h = 2.2 kWh or 7.92 × 10⁶ J.