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11. Electricity

Text-book Questions

Text-book Questions

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.

11. Electricity

TextBook Exercise

Class 10 Science English Updated : 30 July 2026

NCERT Solutions Exercise

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Q1. A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R′, then the ratio R/R′ is –

(a) 1/25

(b) 1/5

(c) 5

(d) 25

Ans: (d) 25

Q2. Which of the following terms does not represent electrical power in a circuit?

(a) I 2R

(b) IR2

(c) VI

(d) V2/R

Ans: (b) IR2

P = VI = I2R = V2/R

Q3. An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be –

(a) 100 W

(b) 75 W

(c) 50 W

(d) 25 W

Ans: (d) 25 W

Q4. Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be –

(a) 1:2

(b) 2:1

(c) 1:4

(d) 4:1

Ans: (c) 1:4

Q5. How is a voltmeter connected in the circuit to measure the potential difference between two points?

Ans: Voltmeter is always connected in parallel.

Chapter 11: Electricity

NCERT Textbook Questions & Solutions

Q6. A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10^–8 Ω m. 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:

• Resistance, R = 10 Ω
• Resistivity, ρ = 1.6 × 10^–8 Ω m
• Diameter, d = 0.5 mm = 0.5 × 10^–3 m
• Radius, r = 0.25 × 10^–3 m

Cross-sectional area,

A = πr²

= 3.14 × (0.25 × 10^–3)²

= 1.963 × 10^–7 m²

Using,

R = ρL/A

L = RA/ρ

= (10 × 1.963 × 10^–7)/(1.6 × 10^–8)

L ≈ 122.7 m

If the diameter is doubled, the area becomes four times.

Since resistance is inversely proportional to area,

New resistance = 10/4 = 2.5 Ω

Answer:

• Length of the wire = 122.7 m
• New resistance = 2.5 Ω
• Decrease in resistance = 7.5 Ω

Q7. The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below. Plot a graph between V and I and calculate the resistance of that resistor.

Solution:

Plot the following points on a graph with current (I) on the X-axis and potential difference (V) on the Y-axis.

Current (A)	0.5	1.0	2.0	3.0	4.0
Voltage (V)	1.6	3.4	6.7	10.2	13.2

The graph obtained is approximately a straight line passing through the origin, showing that the resistor obeys Ohm's law.

Using any point,

R = V/I

= 13.2/4

R = 3.3 Ω

Answer: The resistance of the resistor is approximately 3.3 Ω.

Q8. 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.

Solution:

Given:

• Potential difference, V = 12 V
• Current, I = 2.5 mA = 0.0025 A

Using Ohm's law,

R = V/I

= 12/0.0025

R = 4800 Ω = 4.8 kΩ

Answer: The resistance of the resistor is 4800 Ω (4.8 kΩ).

Q9. 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?

Solution:

Total resistance,

R = 0.2 + 0.3 + 0.4 + 0.5 + 12

= 13.4 Ω

Using Ohm's law,

I = V/R

= 9/13.4

I ≈ 0.67 A

Since the resistors are connected in series, the same current flows through every resistor.

Answer: Current through the 12 Ω resistor is 0.67 A.

Q10. How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?

Solution:

Given:

• Voltage, V = 220 V
• Total current, I = 5 A
• Each resistor = 176 Ω

Equivalent resistance required,

R = V/I

= 220/5

R = 44 Ω

For n equal resistors in parallel,

R = 176/n

176/n = 44

n = 4

Answer: 4 resistors of 176 Ω connected in parallel are required.

Chapter 11: Electricity

NCERT Textbook Questions & Solutions

Q11. 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.

Solution:

(i) To obtain 9 Ω:

Connect two 6 Ω resistors in series.

Equivalent resistance = 6 + 6 = 12 Ω

Now connect this 12 Ω combination in parallel with the third 6 Ω resistor.

Equivalent resistance,

1/R = 1/12 + 1/6

= 1/12 + 2/12 = 3/12 = 1/4

R = 4 Ω

Since this gives 4 Ω, for 9 Ω:

Connect two 6 Ω resistors in parallel.

Equivalent resistance = 3 Ω

Now connect this combination in series with the third 6 Ω resistor.

Total resistance = 3 + 6 = 9 Ω

(ii) To obtain 4 Ω:

Connect two 6 Ω resistors in series (12 Ω), then connect this combination in parallel with the third 6 Ω resistor.

Equivalent resistance = 4 Ω.

Answer:

• (i) 6 Ω || 6 Ω, then in series with 6 Ω → 9 Ω
• (ii) 6 Ω + 6 Ω, then in parallel with 6 Ω → 4 Ω

Q12. 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 across a 220 V line if the maximum allowable current is 5 A?

Solution:

Given:

• Power of each lamp = 10 W
• Voltage = 220 V
• Maximum current = 5 A

Current drawn by one lamp,

I = P/V = 10/220 = 0.045 A

Number of lamps,

n = 5/0.045

n ≈ 110

Answer: A maximum of 110 lamps can be connected in parallel.

Q13. 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?

Solution:

(i) When one coil is used:

R = 24 Ω

I = V/R = 220/24

I = 9.17 A

(ii) When both coils are connected in series:

R = 24 + 24 = 48 Ω

I = 220/48

I = 4.58 A

(iii) When both coils are connected in parallel:

Equivalent resistance,

R = (24 × 24)/(24 + 24) = 12 Ω

I = 220/12

I = 18.33 A

Answer:

• One coil: 9.17 A
• Series: 4.58 A
• Parallel: 18.33 A

Q14. Compare the power used in the 2 Ω resistor in each of the following circuits:

(i) A 6 V battery in series with 1 Ω and 2 Ω resistors.

(ii) A 4 V battery in parallel with 12 Ω and 2 Ω resistors.

Solution:

(i) Series circuit:

Total resistance = 1 + 2 = 3 Ω

Current,

I = 6/3 = 2 A

Power in 2 Ω resistor,

P = I²R

= 2² × 2

P = 8 W

(ii) Parallel circuit:

The 2 Ω resistor gets the full 4 V.

Power,

P = V²/R

= 4²/2

= 16/2

P = 8 W

Answer:

• Power in case (i) = 8 W
• Power in case (ii) = 8 W
• Hence, the power consumed by the 2 Ω resistor is the same in both circuits.

Chapter 11: Electricity

NCERT Textbook Questions & Solutions

Q15. 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?

Solution:

Given:

• Lamp 1: 100 W, 220 V
• Lamp 2: 60 W, 220 V
• Supply voltage = 220 V

Current drawn by the 100 W lamp,

I₁ = P/V = 100/220 = 0.455 A

Current drawn by the 60 W lamp,

I₂ = P/V = 60/220 = 0.273 A

Total current drawn from the mains,

I = I₁ + I₂

= 0.455 + 0.273 = 0.728 A

Answer: The current drawn from the line is approximately 0.73 A.

Q16. Which uses more energy, a 250 W TV set in 1 hour, or a 1200 W toaster in 10 minutes?

Solution:

Energy used by the TV:

Power = 250 W = 0.25 kW

Time = 1 hour

Energy = Power × Time

= 0.25 × 1

= 0.25 kWh

Energy used by the toaster:

Power = 1200 W = 1.2 kW

Time = 10 minutes = 1/6 hour

Energy = 1.2 × 1/6

= 0.20 kWh

Answer: The 250 W TV uses more energy (0.25 kWh) than the 1200 W toaster (0.20 kWh).

Q17. An electric heater of resistance 8 Ω draws 15 A from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater.

Solution:

Given:

• Resistance, R = 8 Ω
• Current, I = 15 A

The rate of heat developed is equal to the electrical power.

P = I²R

= 15² × 8

= 225 × 8

P = 1800 W

Answer: The rate at which heat is developed in the heater is 1800 W (1.8 kW).

Q18. Explain the following.

(a) Why is tungsten used almost exclusively for the filament of electric lamps?

Answer: Tungsten has a very high melting point (about 3400°C) and high resistivity. It can become white hot without melting, making it suitable for lamp filaments.

(b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?

Answer: Alloys have high electrical resistance, high melting point and do not oxidise easily even at high temperatures. Therefore, they are ideal for heating elements.

(c) Why is the series arrangement not used for domestic circuits?

Answer: In a series circuit, the same current flows through all appliances. If one appliance fails or is switched off, the entire circuit breaks and all other appliances stop working. Hence, domestic wiring uses parallel connections.

(d) How does the resistance of a wire vary with its area of cross-section?

Answer: The resistance of a wire is inversely proportional to its cross-sectional area. As the area increases, the resistance decreases.

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

Answer: Copper and aluminium have very low resistivity and are good conductors of electricity. They allow current to flow with minimum energy loss and are therefore widely used for electrical transmission.

11. Electricity

Additional Questions With Solutions

Class 10 Science English Updated : 30 July 2026

Chapter-12. Electricity

Q1: What is an electric circuit?

Ans: A continuous and closed path of an electric current is called an electric circuit.

Q2: Name the instrument which measures electric current in a circuit.

Ans: Ammeter.

Q3: In which order, an ammeter is connected in a circuit?

Ans: In series.

Q4: In which direction does the electric current flow in a circuit?

Ans: the electric current flows in the circuit from positive terminal of the cell to negative terminal of the cell.

Q5: Name the device which produces potential difference in a wire?

Ans: Cell or battery.

Q6: What is potential difference?

Ans: The electrons move only if there is a difference of electric pressure in a conducting metallic wire. This difference of electric pressure is called potential difference.

Q7. Two bulbs of 100 W and 25 W are connected in series to 200 V AC mains. Which bulb glows more brightly? Explain your observation.

Solution:

Given:

• Power of first bulb, P₁ = 100 W
• Power of second bulb, P₂ = 25 W
• Supply voltage = 200 V

Using the relation,

R = V²/P

Resistance of the 100 W bulb,

R₁ = 200²/100 = 400 Ω

Resistance of the 25 W bulb,

R₂ = 200²/25 = 1600 Ω

Since the bulbs are connected in series, the same current flows through both bulbs.

The heat (or power) produced in each bulb is given by:

P = I²R

As the 25 W bulb has a much higher resistance (1600 Ω), it dissipates more power in the series circuit and therefore glows more brightly.

Answer: The 25 W bulb glows more brightly because it has higher resistance than the 100 W bulb. Since the same current flows through both bulbs in series, the bulb with greater resistance produces more heat and light.

Q8: Why does the element of a room heater become red hot but the lead wires remain cold?

Ans:  The element of a room heater has high resistance. According to jule’s law of heating , increasing in resistance cause increasing in heat. While the device wire has not so highly resistance.  

Q9: Calculate the number of electrons constituting one coulomb of charge.

Ans: 6 x10¹⁸ electrons.

Q10: State the Ohm’s Law.

Ans: The electric current flowing through a metallic wire is directly proportional to the potential difference V, across its ends provided its temperature remains the same. This is called Ohm’s law.

11. Electricity

Long Additional Questions With Solutions

Class 10 Science English Updated : 30 July 2026

Chapter 11: Electricity

Long Additional Questions with Solutions

Q1. State Ohm's Law. Describe an activity to verify Ohm's Law with a labelled circuit diagram.

Solution:

Ohm's Law states that the current flowing through a conductor is directly proportional to the potential difference across its ends, provided the temperature remains constant.

V = IR

Activity:

• Connect a battery, resistor, key and ammeter in series.
• Connect a voltmeter across the resistor.
• Close the key and note the current and potential difference.
• Change the voltage and record different readings.
• Plot a graph between V and I.

The graph is a straight line passing through the origin, verifying Ohm's Law.

Conclusion: The ratio V/I remains constant and is equal to the resistance of the conductor.

Q2. Explain the factors affecting the resistance of a conductor. Derive the relation for resistance.

Solution:

The resistance of a conductor depends on:

• Length (L) – Resistance increases with increase in length.
• Cross-sectional area (A) – Resistance decreases as the area increases.
• Nature of material – Different materials have different resistivities.
• Temperature – For metallic conductors, resistance generally increases with temperature.

The relation is:

R = ρL/A

where ρ is the resistivity of the material.

Conclusion: Resistance is directly proportional to length and inversely proportional to cross-sectional area.

Q3. Explain the equivalent resistance of resistors connected in series and derive the expression for it.

Solution:

In a series combination, resistors are connected one after another, and the same current flows through each resistor.

Let the resistances be R₁, R₂ and R₃.

Total potential difference,

V = V₁ + V₂ + V₃

Using Ohm's Law,

V = IR₁ + IR₂ + IR₃

V = I(R₁ + R₂ + R₃)

Therefore,

R = R₁ + R₂ + R₃

Conclusion: The equivalent resistance in series is the sum of all individual resistances.

Q4. Explain the equivalent resistance of resistors connected in parallel and derive the expression for it.

Solution:

In a parallel combination, all resistors have the same potential difference across them.

Total current,

I = I₁ + I₂ + I₃

Using Ohm's Law,

I = V/R₁ + V/R₂ + V/R₃

Dividing by V,

1/R = 1/R₁ + 1/R₂ + 1/R₃

Conclusion: The reciprocal of the equivalent resistance is equal to the sum of the reciprocals of the individual resistances.

Q5. Compare series and parallel combinations of resistors.

Solution:

Series Combination	Parallel Combination
Same current flows through all resistors.	Same potential difference across each resistor.
Equivalent resistance increases.	Equivalent resistance decreases.
If one resistor fails, the entire circuit breaks.	Failure of one resistor does not affect the others.
Used in decorative light strings.	Used in household electrical wiring.

Q6. Explain Joule's Law of Heating. Derive the expression for heat produced in a conductor.

Solution:

According to Joule's Law, the heat produced in a conductor is directly proportional to the square of the current, the resistance of the conductor and the time for which the current flows.

Electrical work done,

W = VIt

Using Ohm's Law, V = IR

W = I²Rt

Since electrical work is converted into heat,

H = I²Rt

This is known as Joule's Law of Heating.

Q7. Explain electric power and derive the different expressions for electric power.

Solution:

Electric power is the rate at which electrical energy is consumed or converted into other forms of energy.

Power is given by,

P = W/t

Since W = VIt,

P = VI

Using Ohm's Law:

P = I²R

Also,

P = V²/R

Conclusion: Electric power can be calculated using any of the above expressions depending on the known quantities.

Q8. Explain the heating effect of electric current. Mention its applications and disadvantages.

Solution:

When electric current passes through a conductor, electrical energy is converted into heat energy. This is called the heating effect of electric current.

Applications:

• Electric iron
• Electric heater
• Electric kettle
• Toaster
• Electric fuse

Disadvantages:

• Causes energy loss in transmission lines.
• May damage electrical appliances due to overheating.
• Can lead to short circuits and fire hazards.

Q9. Explain the construction and working of an electric fuse. Why is it important in domestic circuits?

Solution:

An electric fuse is a safety device made of a thin wire of low melting point connected in series with an electrical circuit.

When excessive current flows due to overloading or short circuit, the fuse wire melts and breaks the circuit.

Importance:

• Protects electrical appliances.
• Prevents overheating.
• Reduces the risk of electrical fires.
• Protects users from electrical hazards.

Q10. Explain the causes of overloading and short-circuiting in domestic electric circuits. Mention the precautions to prevent them.

Solution:

Causes of Overloading:

• Using many appliances on the same socket.
• Operating high-power appliances simultaneously.
• Faulty electrical wiring.

Causes of Short Circuit:

• Damage to insulation of wires.
• Direct contact between live and neutral wires.
• Loose or faulty electrical connections.

Precautions:

• Use proper-rated fuse or MCB.
• Avoid overloading electrical circuits.
• Replace damaged wires immediately.
• Ensure proper earthing.
• Use good-quality electrical appliances and wiring.

Conclusion: Proper wiring, suitable protective devices and careful use of electrical appliances help prevent electrical accidents.

📘

Science

Class 10 (English Medium)

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