Unit 1: Thermal, nuclear and electrical physics

QLDPhysicsSyllabus dot point

Topic 3: Electrical circuits

Define electric current, potential difference and resistance, and apply Ohm's law ($V = IR$) to simple resistive circuits

A focused answer to the QCE Physics Unit 1 dot point on Ohm's law. Defines current ($I = Q/t$), potential difference ($V = W/Q$) and resistance ($R = V/I$), distinguishes ohmic and non-ohmic conductors, and works the QCAA-style multi-resistor calculation from EA Paper 1.

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What this dot point is asking

QCAA wants you to define the three fundamental electrical quantities (current, potential difference, resistance), apply Ohm's law V=IRV = IR, and distinguish ohmic from non-ohmic conductors.

The three quantities

Electric current (II). The rate of flow of electric charge:

I=QtI = \frac{Q}{t}

SI unit: ampere (A). One ampere is one coulomb per second. By convention, current direction is the direction of positive charge flow (opposite to electron flow in a metal).

Potential difference (VV). The work done per unit charge moved between two points:

V=WQV = \frac{W}{Q}

SI unit: volt (V). One volt is one joule per coulomb. Potential difference is the cause of current flow in a circuit; it drives charge through resistors.

Resistance (RR). The opposition to current flow:

R=VIR = \frac{V}{I}

SI unit: ohm (Ω\Omega). One ohm is one volt per ampere. Resistance depends on the material, geometry and temperature of the conductor.

Ohm's law

For an ohmic conductor (most metals at constant temperature), resistance is constant and current is directly proportional to applied voltage:

V=IRV = I R

A current-voltage graph for an ohmic conductor is a straight line through the origin. For a non-ohmic conductor (a filament lamp, a diode, a thermistor), the line is curved or zero in some regions, and RR depends on the operating point.

Ohmic and non-ohmic examples

  • Ohmic: copper wire at constant temperature, carbon resistors.
  • Non-ohmic: filament lamp (resistance rises with temperature so the I-V curve flattens at high VV), semiconductor diode (zero current below the threshold, near-vertical above), thermistor (resistance falls with temperature).

For non-ohmic conductors, R=V/IR = V/I still gives the instantaneous resistance at the operating point, but RR is not constant.

Worked example

A heater draws 8.08.0 A at 240240 V. Find (a) the resistance and (b) the charge that passes through it in 1.01.0 minute of operation.

(a) R=V/I=240/8.0=30ΩR = V / I = 240 / 8.0 = 30 \,\Omega.

(b) Q=It=(8.0)(60)=480Q = I t = (8.0)(60) = 480 C.

Common traps

Calling current the speed of electrons. Drift speed of electrons in a wire is fractions of a millimetre per second. Current is the rate of charge flow, not the speed of charges.

Treating non-ohmic devices as ohmic. A filament lamp draws less than proportional current at high voltages because it heats up. The fixed RR assumption fails.

Mixing up volt and joule per coulomb when calculating energy. Energy per charge is VV, total energy is VQV Q, total energy per second is VIVI. Get the multiplication right.

Forgetting the direction convention. QCAA uses conventional current (positive charge direction). Electron flow is in the opposite direction.

How this appears in IA1 and EA

IA1. Often an ohm-meter reading or an unseen I-V graph asking for the resistance at a stated point and whether the device is ohmic.

EA Paper 1. Multiple choice on the units, the formula, and identifying ohmic from non-ohmic shapes.

EA Paper 2. Combined with the power-and-energy and series-and-parallel dot points to find unknown resistors or currents in a small circuit.

In one sentence

Electric current (I=Q/tI = Q/t) is the rate of charge flow in amperes, potential difference (V=W/QV = W/Q) is the energy delivered per coulomb in volts, resistance (R=V/IR = V/I) is the opposition to current flow in ohms, and ohmic conductors obey V=IRV = IR with constant RR, while non-ohmic conductors (lamps, diodes, thermistors) have RR that depends on the operating point.

Past exam questions, worked

Real questions from past QCAA papers on this dot point, with our answer explainer.

Year 11 SAC3 marksA circuit element carries a current of $0.40$ A when connected to a $12$ V supply. (a) Calculate its resistance. (b) Calculate the charge that flows in $5.0$ minutes. (c) State whether this device is ohmic if doubling $V$ to $24$ V gives a current of $1.2$ A.
Show worked answer →

(a) Resistance. R=V/I=12/0.40=30ΩR = V / I = 12 / 0.40 = 30\,\Omega.

(b) Charge. Q=It=(0.40)(5.0×60)=120Q = I t = (0.40)(5.0 \times 60) = 120 C.

(c) Ohmic test. If VV doubles and II would be ohmic, II should also double to 0.800.80 A. The observed 1.21.2 A is greater than 0.800.80 A, so the resistance is not constant. The device is non-ohmic.

Markers reward RR in ohms, the conversion of minutes to seconds, and the explicit ohmic test against the proportional prediction.

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