FORMULA CHAPTER
Chapter 08 – Current Electricity
Electromotive Force
Worked example
Given: W = 12 J, Q = 2 C Calculation: EMF = 12 / 2 Answer: EMF = 6 V
Potential Difference
Worked example
Given: W = 12 J, Q = 3 C Calculation: V1 − V2 = 12/3 Answer: Potential difference = 4 V
Electric Current
Worked example
Given: Q = 10 C, t = 2 s Calculation: I = 10/2 Answer: I = 5 A
Ohm’s Law
Worked example
Given: I = 2 A, R = 5 ohm Calculation: V = 2 × 5 Answer: V = 10 V
Terminal Voltage of a Cell
Worked example
Given: E = 12 V, I = 2 A, r = 0.5 Ω Calculation: V = 12 − 2 × 0.5 Answer: V = 11 V
Resistance of a Conductor
Worked example
Given: ρ = 2 × 10⁻⁸ Ω m, l = 2 m, A = 1 × 10⁻⁶ m² Calculation: R = (2 × 10⁻⁸ × 2)/(10⁻⁶) Answer: R = 0.04 Ω
Electrical Conductivity
Worked example
Given: ρ = 2 × 10⁻⁸ Ω m Calculation: σ = 1/(2 × 10⁻⁸) Answer: σ = 5 × 10⁷ S/m
Resistors in Series
Worked example
Given: R₁ = 2 Ω, R₂ = 3 Ω, R₃ = 5 Ω Calculation: R = 2 + 3 + 5 Answer: R = 10 Ω
Resistors in Parallel
Worked example
Given: R₁ = 6 Ω, R₂ = 3 Ω Calculation: 1/R = 1/6 + 1/3 = 1/2 Answer: R = 2 Ω
Cells Connected in Series
Worked example
Given: n = 3, E = 2 V, R = 5 Ω, r = 0.5 Ω Calculation: I = 6/(5 + 1.5) Answer: I ≈ 0.92 A
Cells Connected in Parallel
Worked example
Given: n = 4, E = 2 V, R = 1 Ω, r = 0.5 Ω Calculation: I = 8/(4 + 0.5) Answer: I ≈ 1.78 A
Joule’s Law of Heating
Worked example
Given: I = 2 A, R = 5 Ω, t = 10 s Calculation: H = 2² × 5 × 10 Answer: H = 200 J
Electrical Energy
Worked example
Given: V = 12 V, I = 2 A, t = 10 s Calculation: W = 12 × 2 × 10 Answer: W = 240 J
Power from Work and Time
Worked example
Given: W = 600 J, t = 3 s Calculation: P = 600 / 3 Answer: P = 200 W
Electrical Power from Voltage and Current
Worked example
Given: V = 230 V, I = 2 A Calculation: P = 230 × 2 Answer: P = 460 W
Electrical Work from Voltage and Charge
Worked example
Given: V = 6 V, Q = 3 C Calculation: W = 6 × 3 Answer: W = 18 J