Electric current, voltage, resistance, Ohm's law $V = IR$, series and parallel circuits, electric power $P = VI$, and household electricity

What this dot point is asking

QCAA wants Year 11 students to apply Ohm's law and energy conservation to analyse electric circuits.

Charge, current, voltage, resistance

Charge $q$ (coulombs C). Elementary charge $e = 1.6 \times 10^{-19}$ C.

Current $I$ ($I = q/t$, amperes A). Rate of charge flow.

Voltage / potential difference $V$ ($V = E/q$, volts V). Energy per unit charge.

Resistance $R$ (ohms $\Omega$). Opposition to current.

Ohm's law

For ohmic conductors. Equivalent: $I = V/R$, $R = V/I$.

Series circuits

Same current, voltages add.

$R_{\text{total}} = R_1 + R_2 + R_3 + \ldots$

Parallel circuits

Same voltage, currents add.

$\frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + \ldots$

For two: $R_p = R_1 R_2 / (R_1 + R_2)$.

Mixed circuits

Simplify step by step: identify parallel combinations, replace with equivalent; identify series sums; apply Ohm's law. The key physical principles that let you do this are conservation of charge (the current into a junction equals the current out) and conservation of energy (the sum of voltage drops around any loop equals the source voltage). These are Kirchhoff's two laws, and although QCE Unit 1 does not name them formally, every mixed-circuit reduction relies on them.

Electric power

Watts (W) = J/s.

Energy

$E = Pt$ (joules or kWh). Household billing in kWh.

1 kWh = 3.6 MJ.

Real cells: EMF and internal resistance

An ideal source delivers a fixed voltage, but a real battery has a small internal resistance $r$. Its electromotive force (EMF, $\varepsilon$) is the energy per coulomb the chemical reaction supplies, while the terminal voltage measured across the cell is $V = \varepsilon - Ir$. As the current drawn rises, more of the EMF is lost across $r$ and the terminal voltage sags, which is why a torch dims as its battery ages and $r$ grows. QCE problems test this as a difference between the labelled cell voltage and the voltage actually delivered to a load.

Household electricity

The Australian household supply is $230\ \text{V}$ AC at $50\ \text{Hz}$. The quoted $230\ \text{V}$ is the root-mean-square (RMS) value, the steady DC voltage that would deliver the same average power; the peak voltage is $V_{\text{peak}} = \sqrt{2} \times 230 = 325\ \text{V}$. Power is distributed at high voltage and stepped down so that, for a given delivered power $P = VI$, the current and hence the $I^2 R$ transmission loss are minimised.

Safety devices each address a different hazard. Fuses and thermal-magnetic circuit breakers open the circuit when current exceeds a rating, preventing $I^2 R$ overheating of the wiring. The earth wire provides a low-resistance fault path so that a fault to a metal casing draws a large current that trips the breaker rather than leaving the case live. Residual current devices (RCDs) compare the current in the active and neutral conductors and trip within milliseconds if they differ by about $30\ \text{mA}$, which is the signature of current leaking to earth through a person.

Examples in context

Example 1. A Townsville household installs a $5.0 \text{ kW}$ rooftop solar inverter feeding a $230 \text{ V}$ switchboard. The peak current is $I = P/V = 5000/230 = 21.7 \text{ A}$, sized by the network operator to the $32 \text{ A}$ residential feed-in limit. Each circuit's protective device is rated against $I^2 R$ heating in the copper. When a tradesperson adds a $2.4 \text{ kW}$ pool pump on the same final sub-circuit, total current rises to $32 \text{ A}$ and the thermal-magnetic breaker trips, illustrating the household-scale link between Ohm's law, power dissipation and electrical safety.

Example 2. Cairns light-rail substations rectify the $33 \text{ kV}$ Ergon distribution feed to $750 \text{ V}$ DC for traction motors. A train drawing $1.5 \text{ MW}$ pulls $I = 1.5 \times 10^6 / 750 = 2000 \text{ A}$ from the overhead. Engineers minimise the conductor resistance to about $0.05 \text{ ohms per km}$, capping $I^2 R$ losses to $200 \text{ kW per km}$. The same Ohm's-law accounting that the QCAA Unit 1 dot point asks Year 11 students to apply to a torch circuit is what determines feeder spacing and copper cross-section on the actual line.

Try this

Q1. Define electric current in terms of charge, and calculate the current when $90 \text{ C}$ flows through a wire in $30 \text{ s}$. [2 marks]

• Cue. $I = Q/t$, so $I = 90/30 = 3.0 \text{ A}$.

Q2. A heater of resistance $24 \text{ ohms}$ is connected to a $240 \text{ V}$ supply. Calculate the current drawn and the power dissipated, and state the total energy dissipated in $5.0$ minutes. [4 marks]

• Cue. $I = V/R = 10 \text{ A}$; $P = VI = 2400 \text{ W}$; $W = Pt = 7.2 \times 10^5 \text{ J}$.

Q3. A student wires three $6.0 \text{ ohm}$ resistors. (a) Find the equivalent resistance in series and in parallel. (b) Across a $12 \text{ V}$ supply, find the supply current in each case. (c) Identify which configuration dissipates more power and explain why in terms of $P = V^2/R$. [3+2+2 marks; ISMG: Knowledge and conceptual understanding, Analysis and interpretation]

• Cue. (a) $18 \text{ ohms}$ vs $2.0 \text{ ohms}$; (b) $0.67 \text{ A}$ vs $6.0 \text{ A}$; (c) parallel dissipates more because lower $R$ at fixed $V$.