Question 1

What is Ohm's Law and when does it apply?

Accepted Answer

Ohm's Law states that voltage (V) equals current (I) times resistance (R): V = I × R. It applies to linear (ohmic) circuit elements where resistance remains constant regardless of voltage or current. Examples of ohmic devices: resistors, metal conductors at constant temperature, resistive heating elements. Non-ohmic devices where it doesn't directly apply: diodes, LEDs, BJTs, MOSFETs, capacitors, inductors, and power supplies. In AC circuits, use impedance Z instead of resistance R: V = I × Z, where Z = √(R² + (X_L − X_C)²).

Question 2

How do I calculate electrical power in a circuit?

Accepted Answer

Power can be calculated three ways depending on known quantities: P = V × I (know voltage and current — most common), P = I² × R (know current and resistance — for resistor heat dissipation), P = V² / R (know voltage and resistance — for load calculations). In AC circuits: Real power P = V × I × cos(φ) in watts (W). Apparent power S = V × I in volt-amperes (VA). Reactive power Q = V × I × sin(φ) in VAR. Power factor PF = P / S = cos(φ). South African standard mains supply: 230 V single-phase (SANS 1019), 400 V three-phase.

Question 3

How do series and parallel resistor circuits differ?

Accepted Answer

Series resistors: Total resistance R_total = R1 + R2 + R3 + ... Current is the same through all resistors. Voltage divides across each in proportion to its resistance. Parallel resistors: 1/R_total = 1/R1 + 1/R2 + 1/R3 + ... Voltage is the same across all branches. Current divides in inverse proportion to resistance. Practical tip: Wiring light bulbs in series → one fails, all go out (old Christmas lights). Wiring in parallel → each operates independently (modern installations). For two resistors in parallel: R = (R1 × R2) / (R1 + R2) — the product-over-sum shortcut.

Question 4

How do I select the correct resistor wattage rating?

Accepted Answer

Calculate power dissipated in the resistor: P = I² × R or P = V² / R. Then apply a safety margin: Minimum wattage rating ≥ calculated power × 2 (derate to 50%). Common resistor wattage ratings: 1/8 W (0.125W), 1/4 W (0.25W), 1/2 W (0.5W), 1W, 2W, 5W, 10W, 25W, 50W. Example: A 100Ω resistor with 5V across it: P = 5² / 100 = 0.25W → use a 0.5W or 1W resistor. Surface-mount (SMD) resistors: 0402 = 0.0625W, 0603 = 0.1W, 0805 = 0.125W, 1206 = 0.25W. In South Africa, common brands available: Yageo, Vishay, Bourns from local electronics distributors.

Question 5

What is the voltage divider rule and how do I use it?

Accepted Answer

A voltage divider uses two series resistors to produce a fraction of the supply voltage: V_out = V_in × R2 / (R1 + R2). Common uses: Setting reference voltages, level shifting (e.g., 5V Arduino input to 3.3V MCU), biasing transistors, potentiometers (variable voltage divider). Design tip: Choose R1 + R2 significantly smaller than the load impedance to prevent the load from affecting the output voltage. Rule of thumb: R_load > 10 × (R1 + R2). Example: 12V supply → 5V output: ratio = 5/12. If R1 = 7kΩ and R2 = 5kΩ → V_out = 12 × 5000/12000 = 5V.

Question 6

How do LED current-limiting resistors work?

Accepted Answer

LEDs require a current-limiting resistor to prevent overdrive and failure: R = (V_supply − V_f) / I_f. Where V_f = LED forward voltage, I_f = LED forward current. Typical LED values: Red/Yellow/Green LED: V_f ≈ 1.8–2.2V, I_f = 20mA. Blue/White LED: V_f ≈ 3.0–3.5V, I_f = 20mA. High-power LED: I_f = 350mA–1A. Example (5V supply, red LED): R = (5 − 2.0) / 0.020 = 150Ω → use 150Ω or 180Ω resistor. Power in resistor: P = I² × R = 0.02² × 150 = 60mW → 1/4W resistor is adequate. South African mains LED drivers: typically 230V input, constant current output.

Question 7

What is the difference between AC and DC resistance calculations?

Accepted Answer

DC circuits: Resistance R is constant. V = I × R, P = I × V. AC circuits: Resistors still behave the same (impedance = resistance). But inductors add inductive reactance: X_L = 2πfL (Ω) — increases with frequency. Capacitors add capacitive reactance: X_C = 1/(2πfC) (Ω) — decreases with frequency. Total impedance: Z = √(R² + (X_L − X_C)²). At resonance (X_L = X_C): Z = R minimum, circuit draws maximum current. South Africa mains frequency: 50 Hz (SANS 1019). In electronic design, use ohms for DC/low-frequency and impedance at 50 Hz or signal frequency for audio/power work.

Question 8

How do I calculate voltage drop in electrical cables?

Accepted Answer

Voltage drop in a cable conductor: V_drop = I × R_cable. Where R_cable = ρ × L / A (resistivity × length / cross-section area). Copper resistivity ρ ≈ 0.0172 Ω·mm²/m at 20°C; aluminium ≈ 0.0282 Ω·mm²/m. Example: 10A through 6mm² copper cable, 20m run (40m total): R = 0.0172 × 40 / 6 = 0.115Ω. V_drop = 10 × 0.115 = 1.15V. SANS 10142-1 (South African wiring code) limits voltage drop to 5% of nominal supply voltage for most circuits. Maximum drop example: 230V supply × 5% = 11.5V max allowable drop. For motors: limit to 3% under-load to prevent overheating.

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