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µ = one millionth · m = one thousandth · k = one thousand · M = one million.

Example: 680 µA ÷ 1,000 = 0.68 mA. Divide by 1,000 again to get 0.00068 A.

Check: Convert backwards to recover the original number. Capital M and lower-case m mean different things.

V = I × R · I = V ÷ R · R = V ÷ I.

Example: V = IR. Divide both sides by I: V ÷ I = R. With 12 V and 0.5 A, R = 12 ÷ 0.5 = 24 Ω.

Check: Put the answer back into the original formula: 0.5 × 24 = 12 V.

Rₜ = R₁ + R₂ + … The current is the same throughout the circuit.

Example: 10 Ω + 20 Ω = 30 Ω. Across 12 V, I = 12 ÷ 30 = 0.4 A. The drops are 4 V and 8 V.

Check: The voltage drops must add to the supply: 4 + 8 = 12 V.

For two resistors: Rₜ = (R₁ × R₂) ÷ (R₁ + R₂). Each branch has the same voltage.

Example: 10 Ω and 20 Ω across 12 V carry 1.2 A and 0.6 A. Total current = 1.8 A; total resistance ≈ 6.67 Ω.

Check: Total resistance must be below the smallest branch resistance. Branch currents add to supply current.

A current rating alone does not tell you a protective device’s operating time.

Example: A 40 A MCB marking is a current rating. To establish operating time, use the actual current and that device’s time–current curve.

Check: Keep the device’s actual operating time separate from the maximum time permitted for the circuit.

Read Schneider Electric’s breaker characteristics guide. Permitted disconnection times depend on the relevant circuit and earthing conditions; refer to your current course guidance.

For d.c. or purely resistive loads: P = VI; I = P/V; V = P/I. For resistors: P = I²R = V²/R. Power is measured in watts; 1 kW = 1,000 W.

Example: A 24 V load taking 2 A uses P = 24 × 2 = 48 W. A 0.025 Ω joint carrying 10 A dissipates P = 10² × 0.025 = 2.5 W.

Check: Use the voltage across the component being calculated. Add individual resistor powers to check against supply power.

R = ρL/A; L = RA/ρ; A = ρL/R. With ρ in Ω·m, use area in m²: 1 mm² = 0.000001 m². Alternatively use copper ρ = 0.0172 or aluminium ρ = 0.0265 Ω·mm²/m with area in mm².

Example: One 50 m copper conductor of 2.5 mm²: R = 0.0172 × 50 ÷ 2.5 = 0.344 Ω. A twin cable feeding a load 50 m away has 100 m of outgoing-and-return conductor, giving 0.688 Ω.

Check: At constant temperature, greater length increases resistance; greater area reduces it. Check whether the stated length is one conductor, one-way distance or the complete loop.