VCE Systems Engineering (Year 12) · Unit 3 — Ethical Systems Design · 15m read

Electrical Principles — Ohm’s Law, Power and Series and Parallel Circuits

Electrotechnological subsystems run on these core relationships. Circuit calculations are a staple of the Systems Engineering exam.

Learning objective: Apply Ohm’s law, power equations, series and parallel resistance and voltage dividers to analyse DC circuits.

By the end of this lesson you can:

• I can use V = IR and P = VI.
• I can calculate total resistance in series and parallel.
• I can calculate a voltage divider output and a current-limiting resistor for an LED.

Voltage is pressure, current is flow. Like water in pipes: voltage pushes, current flows, and resistance is a narrow section that restricts flow.

Circuit formulas

• Ohm’s law: V = IR
• Power: P = VI = I^2R = \dfrac{V^2}{R}
• Series resistance: R_T = R_1 + R_2 + \dots
• Parallel resistance: \dfrac{1}{R_T} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dots
• Voltage divider: V_{out} = V_{in} \times \dfrac{R_2}{R_1 + R_2}
• LED resistor: R = \dfrac{V_s - V_f}{I}
• Energy: E = Pt

Series vs parallel

SERIES — one path. Same current everywhere; voltages add up to the supply; total resistance is the sum. If one component fails, the circuit breaks.

PARALLEL — multiple paths. Same voltage across each branch; branch currents add to the total; total resistance is LESS than the smallest branch. Branches work independently.

Worked example (Medium)

A red LED (forward voltage 2.0 V, current 20 mA) is powered from a 9 V battery. Find the series resistor and the power it dissipates.

1. Resistor. Voltage across resistor ÷ current. R = (9 - 2.0) / 0.020 = 350\ \Omega
2. Choose a standard value. Use the next higher preferred value. 390\ \Omega
3. Power. P = I²R (using 350 Ω). P = 0.020^2 \times 350 = 0.14\ \text{W}

Answer: About 350 Ω (use 390 Ω), dissipating about 0.14 W — a 0.25 W resistor is suitable.

Worked example (Medium)

Find the total resistance of 6 Ω and 3 Ω in parallel, then the current drawn from a 12 V supply.

1. Parallel. 1/RT = 1/6 + 1/3 = 1/2. R_T = 2\ \Omega
2. Current. I = V/R. 12 / 2 = 6\ \text{A}

Answer: RT = 2 Ω and I = 6 A.

Common mistakes

• Forgetting to convert mA to A Units must be consistent. Fix: 20 mA = 0.020 A.
• Adding parallel resistances Parallel uses reciprocals. Fix: Remember RT in parallel is always less than the smallest resistor.

Practice questions

1. Two 100 Ω resistors in series have a total resistance of:

   1. 50 Ω
   2. 100 Ω
   3. 200 Ω
   4. 10 000 Ω

   (1 marks)

2. A voltage divider has R1 = 10 kΩ and R2 = 5 kΩ across 12 V. Find Vout across R2.

   (2 marks)

3. A 12 V motor draws 1.5 A for 2 hours. Find the power and energy used.

   (2 marks)

Summary

• V = IR; P = VI = I²R = V²/R; E = Pt.
• Series adds; parallel uses reciprocals.
• Voltage dividers scale a voltage with two resistors.
• LED resistor: R = (Vs − Vf)/I.