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QLDPhysicsSyllabus dot point

Topic 3: Electrical circuits

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

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

Generated by Claude Opus 4.87 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. The three quantities
  3. Ohm's law
  4. Ohmic and non-ohmic examples
  5. How this appears in IA1 and EA
  6. Examples in context
  7. Try this

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.

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.

Examples in context

Example 1. A Townsville workshop 48 V48 \text{ V} welder uses a copper lead with R=0.020 ohmsR = 0.020 \text{ ohms} at 200 A200 \text{ A}. Voltage drop along the lead is V=IR=4.0 VV = IR = 4.0 \text{ V}, dissipating I2R=800 WI^2 R = 800 \text{ W} as heat. Switching to twice the cross-section halves RR and quarters the dissipation. The same Ohm's-law accounting tells the apprentice why old or undersized leads scorch; QCAA Unit 1 EA Paper 1 typically sets the same sums on a smaller domestic appliance.

Example 2. A Bundaberg cane farmer pumps water through a 4.0 kW4.0 \text{ kW} submersible motor connected by 80 m80 \text{ m} of 4 mm24 \text{ mm}^2 copper cable (R0.36 ohmsR \approx 0.36 \text{ ohms}). At rated current I=P/V=17 AI = P/V = 17 \text{ A} on 230 V230 \text{ V}, cable voltage drop is V=IR=6.1 VV = IR = 6.1 \text{ V} (2.72.7 per cent of supply). Doubling cable length doubles the drop and quadruples the I2RI^2 R heat. Australian Standard AS/NZS30003000 caps voltage drop at 55 per cent, so the farmer must upsize cable if a second pump is added.

Try this

Q1. Define electric current and potential difference, and state Ohm's law. [3 marks]

  • Cue. I=Q/tI = Q/t; V=W/QV = W/Q; V=IRV = IR for an ohmic conductor.

Q2. A 230 V230 \text{ V} kettle draws 9.0 A9.0 \text{ A}. Calculate the resistance of the heating element and the charge through it in 4.0 minutes4.0 \text{ minutes}. [3 marks]

  • Cue. R=V/I=25.6 ohmsR = V/I = 25.6 \text{ ohms}; Q=It=2160 CQ = It = 2160 \text{ C}.

Q3. A student measures VV across a filament lamp at six currents and finds VV vs II curves upward. (a) Define an ohmic conductor. (b) Explain why the lamp is non-ohmic. (c) Calculate the resistance at the operating point V=240 VV = 240 \text{ V}, I=0.42 AI = 0.42 \text{ A} and compare with the cold value Rc=50 ohmsR_c = 50 \text{ ohms}. [2+3+3 marks; ISMG: Knowledge and conceptual understanding, Analysis and interpretation]

  • Cue. (a) Constant RR at fixed TT; (b) heating raises filament RR; (c) R=571 ohmsR = 571 \text{ ohms}, factor about eleven above cold.

Exam-style practice questions

Practice questions written in the style of QCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Year 11 SAC3 marksA circuit element carries a current of 0.400.40 A when connected to a 1212 V supply. (a) Calculate its resistance. (b) Calculate the charge that flows in 5.05.0 minutes. (c) State whether this device is ohmic if doubling VV to 2424 V gives a current of 1.21.2 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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