Four electrical quantities underpin all circuit analysis and design: voltage, current, resistance, and power. Mastering their definitions, units, interrelationships, and calculation methods is fundamental to VCE Systems Engineering.
KEY TAKEAWAY: Voltage drives current through resistance; power is the rate at which energy is converted. All four quantities are linked through Ohm’s Law and the power equations.
Voltage (potential difference) is the electrical energy transferred per unit charge between two points:
$$V = \frac{W}{Q}$$
| Symbol | Quantity | Unit |
|---|---|---|
| $V$ | Voltage | Volt (V) |
| $W$ | Energy | Joule (J) |
| $Q$ | Charge | Coulomb (C) |
EXAM TIP: Voltage is measured with a voltmeter connected in parallel. Current is measured with an ammeter connected in series. Swapping these connections is a common error.
Current is the rate of flow of electric charge:
$$I = \frac{Q}{t}$$
| Symbol | Quantity | Unit |
|---|---|---|
| $I$ | Current | Ampere (A) |
| $Q$ | Charge | Coulomb (C) |
| $t$ | Time | second (s) |
Worked example: A 9 V battery drives 180 mA through a resistor. How much charge passes in 10 s?
$$Q = I \times t = 0.18 \times 10 = 1.8 \text{ C}$$
Resistance is the opposition to current flow, arising from collisions between charge carriers and conductor atoms:
$$R = \frac{V}{I}$$
| Symbol | Quantity | Unit |
|---|---|---|
| $R$ | Resistance | Ohm ($\Omega$) |
Factors affecting resistance:
- Material (copper low, nichrome high)
- Length: longer wire → higher resistance
- Cross-sectional area: thicker wire → lower resistance
- Temperature: most conductors increase resistance with temperature
REMEMBER: Ohm’s Law ($V = IR$) holds for ohmic conductors at constant temperature. Non-ohmic devices (diodes, thermistors) do not follow a linear $V$–$I$ relationship.
Power is the rate at which electrical energy is converted to other forms:
$$P = VI = I^2 R = \frac{V^2}{R}$$
| Form | Use when |
|---|---|
| $P = VI$ | Both $V$ and $I$ are known |
| $P = I^2 R$ | $I$ and $R$ are known |
| $P = V^2 / R$ | $V$ and $R$ are known |
Worked example 1: A 12 V lamp draws 2 A.
$$P = VI = 12 \times 2 = 24 \text{ W}$$
$$R = \frac{V}{I} = \frac{12}{2} = 6 \text{ } \Omega$$
Worked example 2: A 100 $\Omega$ resistor has 5 V across it.
$$I = \frac{V}{R} = \frac{5}{100} = 0.05 \text{ A} = 50 \text{ mA}$$
$$P = I^2 R = (0.05)^2 \times 100 = 0.25 \text{ W}$$
VCAA FOCUS: Power calculations in circuits are examined frequently. Know all three power formula forms and select the most efficient one for the given information.
$$W = Pt = VIt$$
Worked example: A 500 W element runs for 3 minutes.
$$W = 500 \times 180 = 90{,}000 \text{ J} = 90 \text{ kJ}$$
A 24 V supply connected to R1 = 40 $\Omega$ in series with R2 = 80 $\Omega$:
$$R_T = 40 + 80 = 120 \text{ } \Omega$$
$$I = \frac{24}{120} = 0.2 \text{ A}$$
$$V_1 = 0.2 \times 40 = 8 \text{ V}, \quad V_2 = 0.2 \times 80 = 16 \text{ V}$$
$$P_1 = (0.2)^2 \times 40 = 1.6 \text{ W}, \quad P_2 = (0.2)^2 \times 80 = 3.2 \text{ W}$$
$$P_T = 24 \times 0.2 = 4.8 \text{ W} = P_1 + P_2 \checkmark$$
COMMON MISTAKE: When using $P = V^2/R$ in a series circuit, use the voltage across that component, not the supply voltage. Using the supply voltage gives the wrong answer for individual components.
| Quantity | Symbol | Unit | Key equation |
|---|---|---|---|
| Voltage | $V$ | Volt (V) | $V = IR$ |
| Current | $I$ | Ampere (A) | $I = V/R$ |
| Resistance | $R$ | Ohm ($\Omega$) | $R = V/I$ |
| Power | $P$ | Watt (W) | $P = VI = I^2R = V^2/R$ |
| Energy | $W$ | Joule (J) | $W = Pt$ |
APPLICATION: Selecting a current-limiting resistor for an LED requires $R = (V_{supply} - V_{LED}) / I_{LED}$, then checking $P = I^2 R$ is within the resistor’s power rating — applying all four quantities in one real design step.
