Unit 1: What ideas explain the physical world?

VICPhysicsSyllabus dot point

How are current, voltage and resistance related?

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

A focused answer to the VCE Physics Unit 1 dot point on electrical quantities and Ohm's law. Defines $I = Q/t$, $V = W/Q$ and $R = V/I$, distinguishes ohmic and non-ohmic conductors, and works the VCAA SAC-style I-V characteristic problem.

Generated by Claude OpusReviewed by Better Tuition Academy5 min answer

Have a quick question? Jump to the Q&A page

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.

Past exam questions, worked

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

Year 11 SAC4 marksA circuit element draws $0.20$ A at $4.0$ V. When the voltage is doubled to $8.0$ V, the current rises to $0.50$ 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=20ΩR = V/I = 4.0/0.20 = 20\,\Omega.

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

(b) Ohmic test. An ohmic conductor has constant RR. Here RR changed from 2020 to 16Ω16\,\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.

Related dot points