A 4.0\ resistor carries a current of 1.5\ A. Find the power dissipated. [2 marks]

SEAB Original-style practice: A copper wire of length 2.0\ m and cross-sectional area 1.5 × 10^-7\ m^2 has resistivity 1.7 × 10^-8\ \ m. (a) Find its resistance. (b) A potential difference of 0.50\ V is applied. Find the current. (c) Find the power dissipated.

(a) Resistance: R = LA = (1.7 × 10^-8)(2.0)1.5 × 10^-7 = 3.4 × 10^-81.5 × 10^-7 = 0.227\. (b) Ohm's law: I = VR = 0.500.227 = 2.2\ A. (c) Power: P = VI = 0.50 × 2.2 = 1.1\ W (equivalently P = I^2 R or V^2/R). Markers reward R = L / A, Ohm's law for the current, and the power from P = VI, each with units.

SEAB Original-style practice: (a) Define electric current and the coulomb. (b) A current of 0.80\ A flows for 5.0 minutes. Calculate the charge transferred and the number of electrons involved (e = 1.60 × 10^-19\ C).

(a) Electric current is the rate of flow of electric charge, I = Qt. One coulomb is the charge passing a point when a current of one ampere flows for one second. (b) Charge: Q = It = 0.80 × (5.0 × 60) = 0.80 × 300 = 240\ C. Number of electrons: n = Qe = 2401.60 × 10^-19 = 1.5 × 10^21. Markers reward current as rate of flow of charge, the coulomb defined via the ampere and second, the charge from Q = It (with time in seconds), and the electron count from Q/e.

SingaporePhysicsSyllabus dot point

What is electric current as a flow of charge, and how do resistance and resistivity describe how a conductor opposes that flow?

Define electric current, potential difference and resistance, apply Ohm's law and resistivity, and relate electrical power to current and voltage

A focused answer to the H2 Physics learning outcome on current and resistance. Current as the rate of flow of charge, potential difference, Ohm's law, resistivity, ohmic and non-ohmic behaviour, and electrical power.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-06

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

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to define current, potential difference and resistance, to apply Ohm's law and the resistivity relation, to distinguish ohmic from non-ohmic conductors, and to relate electrical power to current and voltage. These are the building blocks for circuit analysis.

The answer

Current and charge

Electric current is the rate of flow of electric charge:

I = \frac{Q}{t}

measured in amperes (A). Conventional current is taken as the flow of positive charge, opposite to the actual drift of electrons in a metal. One coulomb is the charge transported by a current of one ampere in one second.

Potential difference

The potential difference (voltage) between two points is the work done per unit charge in moving charge between them:

V = \frac{W}{Q}

measured in volts (V), equivalent to joules per coulomb. It drives current through a component.

Resistance and Ohm's law

Resistance is the opposition to current flow:

R = \frac{V}{I}

measured in ohms ( $\Omega$ ). Ohm's law states that for a metallic conductor at constant temperature, the current is proportional to the potential difference, so $R$ is constant. A component obeying this is ohmic (a straight-line, through-the-origin $I$ - $V$ graph). Non-ohmic components (a filament lamp, a diode) have a curved $I$ - $V$ characteristic.

Resistivity

Resistance depends on the conductor's dimensions and material:

R = \frac{\rho L}{A}

where $L$ is the length, $A$ the cross-sectional area, and $\rho$ the resistivity (a material property, units $\Omega\ \text{m}$ ). A longer or thinner wire has more resistance; a lower-resistivity material conducts better.

Electrical power

The power dissipated in a component is:

P = VI = I^2 R = \frac{V^2}{R}

Use whichever form matches the quantities you know. The energy transferred in time $t$ is $E = Pt = VIt$ .

Worked example

A filament lamp is rated $12\ \text{V}$ , $24\ \text{W}$ . Find (a) the current through it at this rating, (b) its resistance at operating temperature, and (c) explain why its resistance is lower when first switched on.

Step 1: Find the operating current

P = VI \implies I = \frac{P}{V} = \frac{24}{12} = 2.0\ \text{A}

Step 2: Find the operating resistance

R = \frac{V}{I} = \frac{12}{2.0} = 6.0\ \Omega

(Equivalently $R = V^2/P = 144/24 = 6.0\ \Omega$ .)

Step 3: Consider the cold filament

When first switched on, the filament is cold. The resistivity of a metal increases with temperature, so a cold filament has a lower resistivity and hence a lower resistance.

Step 4: Explain the consequence

A lower cold resistance means a large surge current flows at switch-on (by $I = V/R$ ), which is why filament lamps most often fail at the moment they are turned on. This also makes the lamp non-ohmic: its $I$ - $V$ graph curves as the filament heats.

Examples in context

Example 1. Why power lines run at high voltage. Transmitting power $P = VI$ at a high voltage means a small current for the same power. Since the heating loss in the cables is $I^2 R$ , a smaller current drastically reduces the energy wasted as heat, which is why the grid steps voltage up for long-distance transmission.

Example 2. Choosing a fuse. A fuse is a thin wire that melts when the current exceeds a rating, breaking the circuit. Its resistance and cross-section are chosen so that the $I^2 R$ heating melts it at the intended current, protecting the appliance from dangerous overcurrents.

Try this

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

Cue. Resistance $R = V/I$ , the ratio of potential difference to current; unit ohm ( $\Omega$ ).

Q2. A $4.0\ \Omega$ resistor carries a current of $1.5\ \text{A}$ . Find the power dissipated. [2 marks]

Cue. $P = I^2 R = (1.5)^2 \times 4.0 = 9.0\ \text{W}$ .

Q3. Two wires of the same material have the same length, but one has twice the cross-sectional area. Compare their resistances. [2 marks]

Cue. $R = \rho L / A$ , so the thicker wire (double the area) has half the resistance.

Exam-style practice questions

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

Original5 marksA copper wire of length

2.0\ \text{m}

and cross-sectional area

1.5 \times 10^{-7}\ \text{m}^2

has resistivity

1.7 \times 10^{-8}\ \Omega\ \text{m}

. (a) Find its resistance. (b) A potential difference of

0.50\ \text{V}

is applied. Find the current. (c) Find the power dissipated.

Show worked answer →

(a) Resistance: $R = \dfrac{\rho L}{A} = \dfrac{(1.7 \times 10^{-8})(2.0)}{1.5 \times 10^{-7}} = \dfrac{3.4 \times 10^{-8}}{1.5 \times 10^{-7}} = 0.227\ \Omega$ .

(b) Ohm's law: $I = \dfrac{V}{R} = \dfrac{0.50}{0.227} = 2.2\ \text{A}$ .

Markers reward $R = \rho L / A$ , Ohm's law for the current, and the power from $P = VI$ , each with units.

Original4 marks(a) Define electric current and the coulomb. (b) A current of

0.80\ \text{A}

flows for

5.0

minutes. Calculate the charge transferred and the number of electrons involved (

e = 1.60 \times 10^{-19}\ \text{C}

Show worked answer →

(a) Electric current is the rate of flow of electric charge, $I = \dfrac{Q}{t}$ . One coulomb is the charge passing a point when a current of one ampere flows for one second.

(b) Charge: $Q = It = 0.80 \times (5.0 \times 60) = 0.80 \times 300 = 240\ \text{C}$ .

Number of electrons: $n = \dfrac{Q}{e} = \dfrac{240}{1.60 \times 10^{-19}} = 1.5 \times 10^{21}$ .

Markers reward current as rate of flow of charge, the coulomb defined via the ampere and second, the charge from $Q = It$ (with time in seconds), and the electron count from $Q/e$ .