ELECTRICAL QUANTITIES & CIRCUITS
Complete Exam Notes
Cambridge IGCSE Physics 0625 - Sections 4.2 & 4.3
Full Marks Revision Guide
⚡ For students aiming for 100% ⚡
📋 QUICK REFERENCE: KEY FORMULAS
Concept Formula Units
Current I = Q/t A (Amperes)
EMF ε = W/Q V (Volts)
Potential Difference V = W/Q V (Volts)
Resistance R = V/I Ω (Ohms)
Resistivity R = ρL/A Ω·m
Power P = IV or P = I²R or P = V²/R W (Watts)
Energy E = IVt or E = Pt J (Joules)
Series Resistance R = R₁ + R₂ + R₃... Ω
Parallel Resistance 1/R = 1/R₁ + 1/R₂ + 1/R₃... Ω
Efficiency η = (output/input) × 100% %
SECTION 4.2: ELECTRICAL QUANTITIES
4.2.1 ELECTRIC CHARGE
Definition: The property of matter responsible for electrical forces.
Key Concepts:
• Protons = positive charge; Electrons = negative charge
• Like charges repel; opposite charges attract
• Charge is conserved (cannot be created or destroyed)
• Formula: Q = ne where e = 1.6 × 10⁻¹⁹ C
Charging by Friction:
When two insulators rub together, electrons transfer from one to the other. The object losing electrons becomes
positive; the one gaining electrons becomes negative. ONLY electrons transfer (not protons).
EXAMPLE: An object has 5 × 10²⁰ electrons. Calculate charge.
Solution: Q = ne = 5 × 10²⁰ × 1.6 × 10⁻¹⁹ = 8000 C
4.2.2 ELECTRIC CURRENT
Definition: Electric current is the rate of flow of charge.
Formula: I = Q/t
Where: I = current (A), Q = charge (C), t = time (s)
Conventional Current Direction:
, • Flows from positive → negative terminal (outside cell)
• Electrons actually flow from negative → positive (opposite!)
• Examiners expect CONVENTIONAL current direction in answers
Ammeter Use:
• Always connected in SERIES with the component
• Has very LOW resistance (ideally 0Ω)
• Measures current flowing through it
EXAMPLE: Current 2.5A flows for 40 seconds. Calculate charge.
Solution: Q = I × t = 2.5 × 40 = 100 C
4.2.3 ELECTROMOTIVE FORCE (EMF) & POTENTIAL DIFFERENCE (P.D.)
CRUCIAL DISTINCTION - THIS IS HEAVILY TESTED!
Property EMF (ε) P.D. (V)
Definition Energy supplied per unit charge Energy converted per unit charge
by the cell by a component
What it measures Work done by the cell Work done against a load
Location INSIDE the cell ACROSS a component
Formula ε = W/Q V = W/Q
KEY RELATIONSHIP:
ε = V + Ir
Where: ε = EMF, V = Terminal P.D., I = Current, r = Internal resistance
When current flows, terminal P.D. is LESS than EMF because energy is lost in internal resistance!
EXAMPLE: Battery EMF 1.5V, when 2A flows, terminal P.D. is 1.4V. Find internal resistance.
Solution: ε = V + Ir → 1.5 = 1.4 + (2 × r) → 0.1 = 2r → r = 0.05 Ω
4.2.4 RESISTANCE
Definition: Resistance is the opposition to the flow of current.
Formula: R = V/I
Unit: Ohm (Ω) | This is Ohm's Law - V and I are directly proportional
Factors Affecting Resistance:
1. Length (L): R ∝ L - Longer wire = more resistance
2. Cross-sectional Area (A): R ∝ 1/A - Thicker wire = less resistance
3. Material (Resistivity ρ): Different materials have different resistivity
Combined Formula: R = ρL/A
Where: R = resistance (Ω), ρ = resistivity (Ω·m), L = length (m), A = cross-sectional area (m²)
Circuit Components:
• Fixed Resistor: Constant resistance
• Variable Resistor: Can change resistance (volume, brightness controls)
• LDR (Light-Dependent Resistor): Resistance ↓ when light ↑
• Thermistor (NTC): Resistance ↓ when temperature ↑
EXAMPLE: Wire length 4m, area 2×10⁻⁶ m², resistivity 1.7×10⁻⁸ Ω·m. Find R.
Solution: R = ρL/A = (1.7×10⁻⁸ × 4)/(2×10⁻⁶) = 0.034 Ω
Complete Exam Notes
Cambridge IGCSE Physics 0625 - Sections 4.2 & 4.3
Full Marks Revision Guide
⚡ For students aiming for 100% ⚡
📋 QUICK REFERENCE: KEY FORMULAS
Concept Formula Units
Current I = Q/t A (Amperes)
EMF ε = W/Q V (Volts)
Potential Difference V = W/Q V (Volts)
Resistance R = V/I Ω (Ohms)
Resistivity R = ρL/A Ω·m
Power P = IV or P = I²R or P = V²/R W (Watts)
Energy E = IVt or E = Pt J (Joules)
Series Resistance R = R₁ + R₂ + R₃... Ω
Parallel Resistance 1/R = 1/R₁ + 1/R₂ + 1/R₃... Ω
Efficiency η = (output/input) × 100% %
SECTION 4.2: ELECTRICAL QUANTITIES
4.2.1 ELECTRIC CHARGE
Definition: The property of matter responsible for electrical forces.
Key Concepts:
• Protons = positive charge; Electrons = negative charge
• Like charges repel; opposite charges attract
• Charge is conserved (cannot be created or destroyed)
• Formula: Q = ne where e = 1.6 × 10⁻¹⁹ C
Charging by Friction:
When two insulators rub together, electrons transfer from one to the other. The object losing electrons becomes
positive; the one gaining electrons becomes negative. ONLY electrons transfer (not protons).
EXAMPLE: An object has 5 × 10²⁰ electrons. Calculate charge.
Solution: Q = ne = 5 × 10²⁰ × 1.6 × 10⁻¹⁹ = 8000 C
4.2.2 ELECTRIC CURRENT
Definition: Electric current is the rate of flow of charge.
Formula: I = Q/t
Where: I = current (A), Q = charge (C), t = time (s)
Conventional Current Direction:
, • Flows from positive → negative terminal (outside cell)
• Electrons actually flow from negative → positive (opposite!)
• Examiners expect CONVENTIONAL current direction in answers
Ammeter Use:
• Always connected in SERIES with the component
• Has very LOW resistance (ideally 0Ω)
• Measures current flowing through it
EXAMPLE: Current 2.5A flows for 40 seconds. Calculate charge.
Solution: Q = I × t = 2.5 × 40 = 100 C
4.2.3 ELECTROMOTIVE FORCE (EMF) & POTENTIAL DIFFERENCE (P.D.)
CRUCIAL DISTINCTION - THIS IS HEAVILY TESTED!
Property EMF (ε) P.D. (V)
Definition Energy supplied per unit charge Energy converted per unit charge
by the cell by a component
What it measures Work done by the cell Work done against a load
Location INSIDE the cell ACROSS a component
Formula ε = W/Q V = W/Q
KEY RELATIONSHIP:
ε = V + Ir
Where: ε = EMF, V = Terminal P.D., I = Current, r = Internal resistance
When current flows, terminal P.D. is LESS than EMF because energy is lost in internal resistance!
EXAMPLE: Battery EMF 1.5V, when 2A flows, terminal P.D. is 1.4V. Find internal resistance.
Solution: ε = V + Ir → 1.5 = 1.4 + (2 × r) → 0.1 = 2r → r = 0.05 Ω
4.2.4 RESISTANCE
Definition: Resistance is the opposition to the flow of current.
Formula: R = V/I
Unit: Ohm (Ω) | This is Ohm's Law - V and I are directly proportional
Factors Affecting Resistance:
1. Length (L): R ∝ L - Longer wire = more resistance
2. Cross-sectional Area (A): R ∝ 1/A - Thicker wire = less resistance
3. Material (Resistivity ρ): Different materials have different resistivity
Combined Formula: R = ρL/A
Where: R = resistance (Ω), ρ = resistivity (Ω·m), L = length (m), A = cross-sectional area (m²)
Circuit Components:
• Fixed Resistor: Constant resistance
• Variable Resistor: Can change resistance (volume, brightness controls)
• LDR (Light-Dependent Resistor): Resistance ↓ when light ↑
• Thermistor (NTC): Resistance ↓ when temperature ↑
EXAMPLE: Wire length 4m, area 2×10⁻⁶ m², resistivity 1.7×10⁻⁸ Ω·m. Find R.
Solution: R = ρL/A = (1.7×10⁻⁸ × 4)/(2×10⁻⁶) = 0.034 Ω
