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Class X Physics - Chapter 3

Electricity

3.1 Introduction to Electricity

Electricity: Electricity is a form of energy resulting from the existence of charged particles (electrons and protons). It involves the flow of electric charge through a conductor.

Electric Charge:

Electric Charge: The fundamental property of matter that causes electric and magnetic effects. It exists in two types - positive and negative.

Properties of Electric Charge:

Example: When you rub a plastic comb with dry hair, electrons transfer from hair to comb, making the comb negatively charged and hair positively charged.

3.2 Electric Current

Electric Current: The rate of flow of electric charge through a conductor is called electric current.
I = Q/t
where I = current, Q = charge, t = time

Units and Measurements:

Quantity Symbol SI Unit Definition
Electric Current I Ampere (A) 1 A = 1 C/s
Electric Charge Q Coulomb (C) 1 C = 1 A × 1 s
One Ampere: If one coulomb of charge flows through a conductor in one second, the current is said to be one ampere.

Direction of Current:

Conventional Current: Direction of flow of positive charges (from positive to negative terminal)
Electron Flow: Direction of flow of electrons (from negative to positive terminal)
Note: Conventional current direction is opposite to electron flow direction
Example: If 60 C of charge flows through a wire in 20 seconds, find the current.

Given: Q = 60 C, t = 20 s
Solution: I = Q/t = 60/20 = 3 A

Answer: Current = 3 A
MCQ 1: If 120 C of charge flows through a conductor in 2 minutes, the current is:
(A) 1 A
(B) 2 A
(C) 60 A
(D) 240 A
Answer: (A) 1 A

3.3 Electric Potential and Potential Difference

Electric Potential: The electric potential at a point is the amount of work done in bringing a unit positive charge from infinity to that point.
Potential Difference: The potential difference between two points is the work done in moving a unit positive charge from one point to another.
V = W/Q
where V = potential difference, W = work done, Q = charge

Units:

Quantity Symbol SI Unit Definition
Potential Difference V Volt (V) 1 V = 1 J/C
Work Done W Joule (J) Energy transferred
One Volt: The potential difference between two points is one volt if one joule of work is done in moving one coulomb of charge from one point to another.
Example: If 50 J of work is done in moving 5 C of charge between two points, find the potential difference.

Given: W = 50 J, Q = 5 C
Solution: V = W/Q = 50/5 = 10 V

Answer: Potential difference = 10 V
MCQ 2: The work done in moving 2 C of charge through a potential difference of 6 V is:
(A) 3 J
(B) 8 J
(C) 12 J
(D) 4 J
Answer: (C) 12 J

3.4 Ohm's Law

Ohm's Law: The current flowing through a conductor is directly proportional to the potential difference across its ends, provided the temperature and other physical conditions remain constant.
V ∝ I
or V = IR
where V = potential difference, I = current, R = resistance

Mathematical Forms of Ohm's Law:

V = IR
I = V/R
R = V/I

Graphical Representation:

V-I Graph for Ohmic Conductor
            V (Volts)
                |
                |    /
                |   /
                |  /
                | /
                |/___________ I (Amperes)
                0
                
            Slope = R (Resistance)
Conditions for Ohm's Law:
Example: A conductor has a resistance of 5 Ω. If a current of 2 A flows through it, find the potential difference across its ends.

Given: R = 5 Ω, I = 2 A
Solution: V = IR = 2 × 5 = 10 V

Answer: Potential difference = 10 V
MCQ 3: If the potential difference across a resistor is 12 V and current through it is 3 A, its resistance is:
(A) 36 Ω
(B) 4 Ω
(C) 9 Ω
(D) 15 Ω
Answer: (B) 4 Ω

3.5 Resistance

Resistance: The property of a conductor to oppose the flow of current through it is called resistance.
R = ρl/A
where R = resistance, ρ = resistivity, l = length, A = cross-sectional area

Units:

Quantity Symbol SI Unit Definition
Resistance R Ohm (Ω) 1 Ω = 1 V/A
Resistivity ρ Ω⋅m Material property
One Ohm: The resistance of a conductor is one ohm if a current of one ampere flows through it when a potential difference of one volt is applied across its ends.

Factors Affecting Resistance:

Resistivity:

Resistivity: The resistance per unit length per unit cross-sectional area of a material is called resistivity. It is a characteristic property of the material.
Example: A copper wire of length 2 m and cross-sectional area 2 × 10⁻⁶ m² has resistivity 1.7 × 10⁻⁸ Ω⋅m. Find its resistance.

Given: l = 2 m, A = 2 × 10⁻⁶ m², ρ = 1.7 × 10⁻⁸ Ω⋅m
Solution: R = ρl/A = (1.7 × 10⁻⁸ × 2)/(2 × 10⁻⁶) = 0.017 Ω

Answer: Resistance = 0.017 Ω
MCQ 4: If the length of a wire is doubled keeping other factors constant, its resistance becomes:
(A) Half
(B) Double
(C) Four times
(D) One-fourth
Answer: (B) Double

3.6 Resistors in Series

Series Connection: When resistors are connected end to end in a single path, they are said to be connected in series.
Series Circuit
                ——R₁——R₂——R₃——
               |                |
               |                |
               +                -
                    Battery

Properties of Series Connection:

Rs = R₁ + R₂ + R₃ + ...
where Rs = equivalent series resistance

Key Points:

Example: Three resistors of 2 Ω, 3 Ω, and 5 Ω are connected in series. Find the equivalent resistance.

Given: R₁ = 2 Ω, R₂ = 3 Ω, R₃ = 5 Ω
Solution: Rs = R₁ + R₂ + R₃ = 2 + 3 + 5 = 10 Ω

Answer: Equivalent resistance = 10 Ω
MCQ 5: Two resistors of 4 Ω and 6 Ω are connected in series. The equivalent resistance is:
(A) 2.4 Ω
(B) 5 Ω
(C) 10 Ω
(D) 24 Ω
Answer: (C) 10 Ω

3.7 Resistors in Parallel

Parallel Connection: When resistors are connected between the same two points, providing multiple paths for current, they are said to be connected in parallel.
Parallel Circuit
                    ——R₁——
                   |      |
                ——— ——R₂—— ———
               |   |      |   |
               |    ——R₃——    |
               |              |
               +              -
                   Battery

Properties of Parallel Connection:

1/Rp = 1/R₁ + 1/R₂ + 1/R₃ + ...
where Rp = equivalent parallel resistance

For Two Resistors in Parallel:

Rp = (R₁ × R₂)/(R₁ + R₂)

Key Points:

Example: Two resistors of 6 Ω and 3 Ω are connected in parallel. Find the equivalent resistance.

Given: R₁ = 6 Ω, R₂ = 3 Ω
Solution: Rp = (R₁ × R₂)/(R₁ + R₂) = (6 × 3)/(6 + 3) = 18/9 = 2 Ω

Answer: Equivalent resistance = 2 Ω
MCQ 6: Three resistors each of 9 Ω are connected in parallel. The equivalent resistance is:
(A) 27 Ω
(B) 9 Ω
(C) 3 Ω
(D) 1 Ω
Answer: (C) 3 Ω

3.8 Electric Power

Electric Power: The rate of consumption of electrical energy or the rate at which electrical work is done is called electric power.
P = W/t = VI
where P = power, W = work done, t = time, V = voltage, I = current

Alternative Forms of Power Formula:

P = VI = I²R = V²/R

Units:

Quantity Symbol SI Unit Commercial Unit
Power P Watt (W) Kilowatt (kW)
Energy E Joule (J) Kilowatt-hour (kWh)
One Watt: The power consumed is one watt when one ampere of current flows through a device at a potential difference of one volt.

Derivation of Power Formulas:

Starting from P = VI:
Using Ohm's law V = IR:
P = VI = (IR)I = I²R

Also, using I = V/R:
P = VI = V(V/R) = V²/R
Example: A bulb operates at 220 V and draws a current of 0.5 A. Find the power consumed.

Given: V = 220 V, I = 0.5 A
Solution: P = VI = 220 × 0.5 = 110 W

Answer: Power consumed = 110 W
MCQ 7: The power consumed by a device of resistance 8 Ω when 2 A current flows through it is:
(A) 16 W
(B) 32 W
(C) 4 W
(D) 10 W
Answer: (B) 32 W

3.9 Electric Energy

Electric Energy: The total amount of work done by an electric current or the total electrical energy consumed by a device is called electric energy.
E = Pt = VIt = I²Rt = V²t/R
where E = energy, P = power, t = time

Commercial Unit of Energy:

Kilowatt-hour (kWh): The energy consumed when a device of power 1 kW operates for 1 hour.
Relation: 1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J
Energy Conversion:
1 kWh = 3.6 × 10⁶ J = 3.6 MJ
1 unit of electricity = 1 kWh

Electricity Bill Calculation:

Cost = Energy consumed (kWh) × Rate per unit
Example: A 100 W bulb is used for 5 hours daily for 30 days. If the rate is ₹5 per kWh, find the monthly electricity bill.

Given: P = 100 W = 0.1 kW, t = 5 × 30 = 150 hours, Rate = ₹5/kWh
Solution:
Energy = P × t = 0.1 × 150 = 15 kWh
Cost = 15 × 5 = ₹75

Answer: Monthly bill = ₹75
MCQ 8: A device of 2 kW operates for 3 hours. The energy consumed is:
(A) 6 kWh
(B) 5 kWh
(C) 1.5 kWh
(D) 6 kW
Answer: (A) 6 kWh

3.10 Heating Effect of Electric Current

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

Joule's Law:

Joule's Law: The heat produced in a conductor is directly proportional to the square of current, resistance of the conductor, and time for which current flows.
H = I²Rt
where H = heat produced, I = current, R = resistance, t = time

Alternative Forms:

H = VIt = V²t/R
All forms give heat in Joules

End of Chapter 1

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