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Electric current, potential difference and Ohm's law (VCE Physics Unit 1)

Syllabus dot point

“Define electric current, potential difference and resistance, and apply Ohm's law (V=IRV = IR) to ohmic and non-ohmic conductors, including filament lamps and diodes”

7 min read

Quick answer

Current is charge flow rate (I=Q/tI=Q/t), voltage is energy per charge (V=W/QV=W/Q) and resistance is R=V/IR=V/I. Ohmic conductors obey V=IRV=IR with constant RR (straight-line I-V graph); non-ohmic devices like filament lamps and diodes have a resistance that changes with the operating point.

Jump to a section
  1. What this dot point is asking
  2. Electric current
  3. Potential difference
  4. Resistance
  5. Ohm's law
  6. Non-ohmic devices
  7. VCAA exam style
  8. Common traps
  9. In one sentence
  10. Examples in context
  11. Try this
  12. Exam-style questions

What this dot point is asking

VCAA wants you to define current, voltage and resistance, apply Ohm's law to ohmic devices, and distinguish non-ohmic devices (filament lamps, diodes, thermistors) where resistance varies with operating conditions.

Electric current

The rate of flow of charge:

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

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

Potential difference

Work done per unit charge:

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

SI unit: volt (V) = joule per coulomb. Drives current through circuit elements.

Resistance

Opposition to current:

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

SI unit: ohm (Ω\Omega) = volt per ampere. Depends on material, geometry and temperature.

Ohm's law

For an ohmic conductor (constant temperature, most metals at low currents):

V=IRV = IR

with RR constant. The I-V graph is a straight line through the origin.

For non-ohmic devices, R=V/IR = V/I still gives the instantaneous resistance at the operating point, but RR varies with VV or II.

Non-ohmic devices

Filament lamp
As current flows, the filament heats up and its resistance rises. I-V graph curves so that the slope decreases at higher VV.
Semiconductor diode
Zero current below the threshold voltage (about 0.70.7 V for silicon). Near-vertical above. Strongly direction-dependent (only conducts one way).
Thermistor
A semiconductor whose resistance falls steeply with temperature. Used in temperature sensors. The I-V curve is concave up.

VCAA exam style

Year 11 SAC tasks typically include:

  • Calculating RR at two operating points and identifying whether the device is ohmic.
  • Reading an I-V graph to find RR at a stated voltage.
  • Distinguishing ohmic and non-ohmic shapes.

Common traps

Calling drift velocity "current"
Electron drift velocity in a wire is typically fractions of a millimetre per second. Current is the rate of charge flow.
Treating R=V/IR = V/I as a definition that proves Ohm's law
R=V/IR = V/I defines instantaneous resistance for any device. Ohm's law is the additional claim that RR is constant for ohmic devices.
Forgetting direction in diodes
A diode conducts only in one direction. Reversing the voltage usually drops the current to near zero.
Mixing units of charge and current
Ampere-second is coulomb; ampere-hour is 36003600 C.

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 devices (lamps, diodes, thermistors) have RR that varies with the operating point.

Examples in context

Example 1. Grid battery cell characterisation (simplified model). Grid batteries such as the Hornsdale Power Reserve (owned by Neoen) are built from many cells. Take a made-up lithium-iron-phosphate cell with a nominal terminal potential difference of 3.23.2 V. When delivering 130130 A during a frequency-control event, internal resistance of approximately 0.00050.0005 Ω\Omega causes a voltage drop ΔV=IR=130×0.0005=0.065\Delta V = IR = 130 \times 0.0005 = 0.065 V, so the terminal voltage falls to 3.1353.135 V. This is ohmic behaviour: voltage scales linearly with current. The cell heats at rate P=I2R=1302×0.0005=8.5P = I^2 R = 130^2 \times 0.0005 = 8.5 W, which the liquid cooling loop must remove to keep the cell below 45∘45^\circC and prolong life.

Example 2. Filament lamp non-ohmic behaviour in heritage Melbourne street lights. Some Carlton heritage lamps still use 150150 W tungsten filaments. At room temperature (20∘20^\circC), the cold filament resistance is about 3030 Ω\Omega; at operating temperature (2500∘2500^\circC), it rises to roughly 360360 Ω\Omega, a factor of 1212. Switching the lamp on, the inrush current is I=230/30≈7.7I = 230/30 \approx 7.7 A, but steady-state current is only I=230/360≈0.64I = 230/360 \approx 0.64 A. A VV-II plot is therefore strongly non-linear: doubling the voltage less than doubles the current because the filament heats. This is why filament lamps are classed as non-ohmic, in contrast to nichrome heating elements.

Try this

Q1. Define electric current and state its SI unit. [2 marks]

  • Cue. Rate of flow of charge, I=Q/tI = Q/t, measured in amperes (11 A = 11 C s−1^{-1}).

Q2. A 1212 V battery delivers a current of 1.51.5 A through an unknown resistor for 55 minutes. Calculate (a) the resistance, and (b) the total charge that has flowed. [2+2 marks]

  • Cue. (a) R=V/I=8R = V/I = 8 Ω\Omega. (b) Q=It=1.5×300=450Q = It = 1.5 \times 300 = 450 C.

Q3. A filament lamp shows a non-linear voltage-current graph. (a) Define an ohmic conductor. (b) Explain why a tungsten lamp behaves non-ohmically. (c) Sketch the VV versus II graph and indicate the region where the lamp behaves approximately ohmically. [2+2+2 marks]

  • Cue. (a) Constant ratio V/IV/I at fixed temperature. (b) Filament resistance rises with temperature. (c) Curve concave to the VV axis; near origin (cold) the slope is roughly constant.

Exam-style questions

Questions in the style of VCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Year 11 SAC-style4 marks
A circuit element draws 0.200.20 A at 4.04.0 V. When the voltage is doubled to 8.08.0 V, the current rises to 0.500.50 A. (a) Find the resistance at each operating point. (b) Is the element ohmic? Justify.
Show worked answer →

(a) Resistance at each point.

At 4.04.0 V: R=V/I=4.0/0.20=20 ΩR = V/I = 4.0/0.20 = 20\,\Omega.

At 8.08.0 V: R=V/I=8.0/0.50=16 ΩR = V/I = 8.0/0.50 = 16\,\Omega.

(b) Ohmic test. An ohmic conductor has constant RR. Here RR changed from 2020 to 16 Ω16\,\Omega when VV doubled. So the element is non-ohmic.

An ohmic prediction would have been I=0.40I = 0.40 A at 8.08.0 V (proportional); the observed 0.500.50 A exceeds this, suggesting a device where RR falls with increasing VV or TT (a thermistor with negative temperature coefficient, for example).

Markers reward RR in ohms at each point, the ohmic test against the proportional prediction, and named non-ohmic examples.

Practise this

Sources & how we know this

ExamExplained