Electricity

Electricity – Complete Solutions

3.1 Electric Current

Definition:

Electric current is the rate of flow of electric charge through a conductor. It is the amount of charge flowing per unit time.

Formula: I = Q/t

Where: I = Current (in Amperes, A)

Q = Charge (in Coulombs, C)

t = Time (in seconds, s)

Unit:

Ampere (A) – SI unit of electric current

1 A = 1 Coulomb / 1 second

Direction of Current:

Conventional current flows from positive terminal to negative terminal (direction of flow of positive charges).

Electron flow is in opposite direction (from negative to positive).

Example: If 60 Coulombs of charge flows through a cross-section in 5 seconds:

I = Q/t = 60/5 = 12 Amperes

3.2 Potential Difference (Voltage)

Definition:

Electric potential difference between two points is the work done per unit positive charge to move the charge from one point to another.

Formula: V = W/Q

Where: V = Potential difference (in Volts, V)

W = Work done (in Joules, J)

Q = Charge (in Coulombs, C)

Unit:

Volt (V) – SI unit of potential difference

1 Volt = 1 Joule / 1 Coulomb

Example: Work done in moving 2 Coulombs of charge is 10 Joules:

V = W/Q = 10/2 = 5 Volts

3.3 Resistance

Definition:

Resistance is the opposition offered by a material to the flow of electric current through it. It depends on the nature of the material and its dimensions.

Formula: R = ρL/A

Where: R = Resistance (in Ohms, Ω)

ρ (rho) = Resistivity of the material

L = Length of the conductor

A = Cross-sectional area of the conductor

Unit:

Ohm (Ω) – SI unit of resistance

1 Ohm = 1 Volt / 1 Ampere

Resistivity:

Resistivity is the property of a material to resist the flow of electric current. It depends on the nature of the material, not on its dimensions.

Good Conductors (Low Resistivity):

• Silver (ρ = 1.6 × 10⁻⁸ Ω·m)
• Copper (ρ = 1.7 × 10⁻⁸ Ω·m)
• Aluminum (ρ = 2.7 × 10⁻⁸ Ω·m)

Poor Conductors/Insulators (High Resistivity):

• Rubber (ρ ≈ 10¹⁵ Ω·m)
• Glass (ρ ≈ 10¹¹ Ω·m)
• Wood (ρ ≈ 10⁸ Ω·m)

3.4 Ohms Law

Statement:

Ohms Law states that the electric current flowing through a conductor is directly proportional to the potential difference applied across it, provided the temperature and other physical conditions remain constant.

Formula: V = IR

Where: V = Potential difference (Volts)

I = Current (Amperes)

R = Resistance (Ohms)

Alternative forms:

I = V/R

R = V/I

Limitations of Ohms Law:

• It applies only to ohmic conductors (like metals at constant temperature)
• It does not apply to non-ohmic conductors (like semiconductor diodes)
• Temperature must remain constant

Example: A bulb has resistance 100 Ω and is connected to 220 V supply:

Current I = V/R = 220/100 = 2.2 A

3.5 Electric Power and Energy

Power:

Power is the rate of doing work or the rate of consumption of electrical energy.

Formula: P = VI = I²R = V²/R

Where: P = Power (in Watts, W)

Unit: Watt (W) = 1 Joule/second

Other units:

• 1 kW = 1000 W
• 1 MW = 10⁶ W

Energy:

Electrical energy consumed in a circuit is given by:

Formula: E = Pt = VIt = I²Rt

Where: E = Energy (in Joules, J)

Commercial unit: Kilowatt-hour (kWh)

1 kWh = 3.6 × 10⁶ J

Example: A 60 W bulb is used for 8 hours:

Energy = Pt = 60 × 8 × 3600 = 1,728,000 J = 0.48 kWh

3.6 Series and Parallel Circuits

Series Circuit:

Characteristics:

• Components connected end-to-end in a single path
• Same current flows through all components
• Voltage divides among components
• Total resistance increases

Formulas:

I = I₁ = I₂ = I₃ (same current)

V = V₁ + V₂ + V₃ (voltage divides)

R_total = R₁ + R₂ + R₃

Application: Christmas lights, flashlight bulbs

Parallel Circuit:

Characteristics:

• Components connected with common starting and ending points
• Same voltage across all components
• Current divides among components
• Total resistance decreases

Formulas:

V = V₁ = V₂ = V₃ (same voltage)

I = I₁ + I₂ + I₃ (current divides)

1/R_total = 1/R₁ + 1/R₂ + 1/R₃

Application: Domestic wiring, appliances in homes

3.7 Solved Examples

Example 1: A wire has resistance 10 Ω at 0°C. Find its resistance at 50°C if coefficient of resistance is 0.005 per °C

Solution:

R_t = R₀[1 + α(t – t₀)]

R_50 = 10[1 + 0.005(50 – 0)]

R_50 = 10[1 + 0.25]

R_50 = 10 × 1.25 = 12.5 Ω

Example 2: Calculate the current drawn by a 60 W bulb at 220 V

Solution:

P = VI

60 = 220 × I

I = 60/220 = 0.27 A

Example 3: Two resistors of 4 Ω and 6 Ω are connected in series and then in parallel. Compare total resistance in both cases

Solution:

Series: R_total = 4 + 6 = 10 Ω

Parallel: 1/R_total = 1/4 + 1/6 = 3/12 + 2/12 = 5/12

R_total = 12/5 = 2.4 Ω

Series resistance (10 Ω) > Parallel resistance (2.4 Ω)