11. Electricity - Science Class 10 In English

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:

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.