Basic Electricity and Electronic Components
Basic Electricity and Electronic Components is examined through circuit calculations, component identification and practical rating choices.
What the exam tests
This domain checks whether you can apply Ohm's law and power relations to series, parallel and mixed circuits, identify components from markings and packages, and choose correct ratings. Expect resistor colour-code reading, capacitor and inductor behaviour, diode and BJT terminal identification, and simple soldering judgement.
Core rules
- Ohm's law: V = IR; power P = VI = I²R = V²/R.
- Series: R_total = R1 + R2 + ...; same current; voltages add. Parallel: 1/R_total = 1/R1 + 1/R2 + ...; same voltage; branch currents add.
- Four-band resistor: digit, digit, multiplier, tolerance. Gold = ±5%, silver = ±10%, brown = ±1%. A 220 Ω ±5% resistor reads red-red-brown-gold.
- Capacitor: C = Q/V; energy = ½CV². Capacitors in parallel add; in series use the reciprocal rule. Electrolytics are polarised: observe polarity.
- Silicon diode drops about 0.7 V forward; germanium about 0.3 V. BJT: β = Ic/Ib; emitter arrow shows conventional current direction (NPN out, PNP in).
Worked example
A 12 V supply feeds 220 Ω and 330 Ω in series. Total resistance = 550 Ω. Current I = 12/550 = 0.0218 A (21.8 mA). Voltage across 220 Ω = 0.0218 × 220 = 4.8 V; across 330 Ω = 7.2 V. Check: 4.8 + 7.2 = 12 V. Power in 330 Ω = I²R = (0.0218)² × 330 ≈ 0.157 W, so a 0.25 W resistor is adequate but a 0.125 W part is not.
Traps that cost marks
Reading the multiplier band as a digit; using the parallel formula for capacitors; forgetting that an LED needs a series current-limiting resistor; choosing a resistor wattage from voltage alone; reversing electrolytic capacitor polarity; assuming a BJT's base current equals collector current.