IB Physics · Theme B The particulate nature of matter
B.5 Current and circuits
Summary to follow. 15 syllabus statements · 41 questions · about twenty minutes.
Compiled from the IB Physics guide (first assessment 2025, updated November 2023) and our question bank ·
Specialist review in progress
· How these pages are made
Assessed in Paper 1A (multiple choice), Paper 1B (data-based) and Paper 2 (short and extended response). IB Physics guide (first assessment 2025, updated November 2023).
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In preparation: 0 of 15 sections compiled and reviewed. The rest show key terms and common misconceptions from our question bank until they are.
Electromotive force (emf), ε
Electromotive force (emf), ε
The emf ε of a source such as a cell is the work done by the source (the energy it transfers from another form, such as chemical energy, into electrical energy) per unit charge that it moves around a complete circuit: ε = W/q. Despite its name, emf is not a force. SI unit: volt (V), where 1 V = 1 J C⁻¹. A cell of emf 1.5 V transfers 1.5 J to each coulomb of charge that passes through it.
Students often think A cell or battery is a container of charge (electricity) that flows out into the circuit, and a flat battery has run out of charge. In fact No. The charge carriers (free electrons in the wires) are present throughout the circuit before it is connected. The cell does work on them; it supplies energy, not charge.
Students often think Electromotive force is a force that pushes charges round the circuit, so a 1.5 V cell exerts 1.5 N on each coulomb. In fact No. Emf is energy transferred per unit charge, measured in volts (J C⁻¹), not a force measured in newtons.
Chemical cell
Chemical cell
A source of emf in which chemical reactions between the electrodes and the electrolyte transfer chemical energy into electrical energy. The cell does not store charge or 'electricity': the charge carriers that make up the current are already present throughout the circuit, and the cell does work on them. When the reactants are used up, the cell can no longer transfer energy; a rechargeable (secondary) cell can have its reactions reversed by driving a current through it the other way.
Solar (photovoltaic) cell
A source of emf that transfers the energy of incident light (electromagnetic radiation) directly into electrical energy. It stores no energy: its output exists only while light falls on it and increases with the intensity of that light, so a solar cell gives no output in darkness. It is not a solar heating panel; it does not work by using the Sun's heat.
Comparing chemical and solar cells as energy sources
Chemical cells: advantages — portable, and able to supply energy at any time, including in the dark and whatever the weather; disadvantages — they hold a finite amount of energy so must be replaced or recharged, contain chemicals that need careful disposal, and their internal resistance rises as they discharge. Solar cells: advantages — use a renewable source, need no fuel to be replaced, have low running costs and emit no greenhouse gases in use; disadvantages — output depends on light intensity and is zero at night, conversion efficiency is low so large areas are needed, the initial cost is high, and a storage device (often a rechargeable chemical cell) is needed for use when light is not available.
Students often think A solar cell soaks up energy from sunlight during the day and stores it, like a rechargeable battery, so it can supply energy at night. In fact No. A solar cell transfers light energy to electrical energy only while light falls on it; it stores nothing. Night-time use needs a separate store, usually a rechargeable chemical cell.
Students often think A battery is a store of electrical energy (electricity) that it simply lets out into the circuit. In fact No. A chemical cell stores chemical energy in its reactants; the reactions transfer this chemical energy to electrical energy only while there is a current.
Circuit diagram
Circuit diagram
A diagram that uses the standard symbols listed in the Physics data booklet (for example cell, battery, switch, lamp, resistor, ammeter, voltmeter, potentiometer, light-dependent resistor and thermistor) to show how components are connected. It shows the connections only — which components are in series and which in parallel — not the physical layout, positions or lengths of the wires. Connecting wires are taken to have negligible resistance. Two diagrams with the same connections represent the same circuit, however they are drawn.
Ammeter and voltmeter (ideal meters)
An ammeter measures the current through a component and is connected in series with it; an ideal ammeter has zero resistance, so it has no pd across it and does not change the current. A voltmeter measures the pd between two points and is connected in parallel with the component; an ideal voltmeter has infinite resistance, so no current passes through it. Unless otherwise stated, meters in IB questions are ideal; where a non-ideal meter is used, its resistance is constant.
Students often think A meter is a passive device that simply reads the circuit, so the circuit works in the same way and the readings are the same however the meter is connected. In fact No. An ammeter must be in series and a voltmeter in parallel with the component. A meter connected the wrong way changes the circuit: an ideal voltmeter in series stops the current, and an ammeter in parallel short-circuits the component.
Students often think Local reasoning: a feature at one place (such as a zero-resistance path across a lamp, or a junction) determines the current there, whatever the rest of the circuit contains. In fact No. The current in any part of a circuit depends on the whole circuit (the emf and all the resistances in the loop), not only on the components at that point.
Electric current, I
Electric current, I
The rate of flow of charge past a point: I = Δq/Δt. Current is a scalar. In a series loop the current is the same at every point, because charge is conserved and does not accumulate: it is not used up by components. SI unit: ampere (A), where 1 A = 1 C s⁻¹. The number of charge carriers passing a point is the charge divided by the charge on each carrier (for electrons, e = 1.60 × 10⁻¹⁹ C).
Charge carriers and conventional current
Charge carriers are the mobile charged particles whose motion forms a current: free (delocalized) electrons in a metal, and positive and negative ions in an electrolyte or molten salt. Conventional current is taken to flow from the positive terminal of a source, around the external circuit, to the negative terminal; in a metal the electrons drift in the opposite direction. The drift of electrons is slow (typically well under 1 mm s⁻¹); a lamp lights almost at once because the electric field is set up throughout the circuit very quickly, so carriers everywhere start to drift together.
Direct current (dc)
A current whose direction does not change, such as the current supplied by a chemical cell or a solar cell. Only direct-current circuits are studied in this topic; alternating current (ac) circuits are not required.
Students often think A time given in minutes can be substituted directly into an equation that expects seconds (procedural error). In fact No. SI units must be used: times in seconds, so minutes must be multiplied by 60.
Students often think The charge that flows can be found by dividing the current by the time, Δq = I/Δt (procedural rearrangement error). In fact No. Multiplying both sides by Δt gives Δq = IΔt: the charge is the current multiplied by the time.
Electric potential difference, V
Electric potential difference, V
The work done per unit charge in moving a positive charge between two points along the path of the current: V = W/q. Across a resistor or lamp, it is the electrical energy transferred to other forms (such as internal energy and light) per unit charge passing through. A pd exists between two points whenever charge moved between them would have work done on or by it, even if no current is flowing (for example, across an open switch). SI unit: volt (V), where 1 V = 1 J C⁻¹.
Students often think A pd is a consequence of current: where there is no current there can be no pd. In fact Yes. A pd exists whenever work would be done on a charge moved between the points. Across an open switch in a circuit with a cell, the pd is the full emf even though the current is zero.
Students often think Voltage is the rate of energy transfer (or the energy itself), so V can be found by dividing the energy by the time. In fact No. Potential difference is energy transferred per unit charge (J C⁻¹). Energy transferred per second is power (J s⁻¹ = W).
Electrical conductor and insulator
Electrical conductor and insulator
A conductor contains a large number density of charge carriers that are free to move through it (mobile carriers): free electrons in a metal, ions in an electrolyte. An insulator also contains charged particles (electrons and nuclei), but its electrons are bound to their atoms or molecules and it has no ions free to move, so almost no charge carriers are mobile and the current is negligible for ordinary pds.
Students often think An insulator does not conduct because it contains no electrons or other charged particles. In fact No. Insulators contain protons and electrons, just as conductors do; the difference is that their electrons are bound to atoms or molecules, so there are almost no mobile charge carriers.
Students often think An electric current is always a flow of electrons, whatever the conductor: a conducting liquid must also carry its current by electrons. In fact No. A current is a flow of any mobile charge carriers. In a metal the carriers are electrons; in an electrolyte such as a salt solution they are positive and negative ions moving in opposite directions.
Origin of electrical resistance
Origin of electrical resistance
In a metal, free electrons accelerated by the electric field collide with the positive ions of the lattice, which are vibrating about fixed positions. In each collision energy is transferred from the electrons to the lattice, which opposes their drift and increases the internal energy of the metal. The rate of these collisions determines the resistance. When a metal's temperature rises, the ions vibrate with larger amplitude, collisions become more frequent, and the resistance increases.
Students often think Resistance is friction between the moving electrons and the surface of the wire, like water rubbing on the inside of a pipe. In fact No. Resistance arises inside the metal, from collisions of the free electrons with the vibrating positive ions of the lattice.
Students often think Resistance comes from the electrons bumping into each other as they crowd through the wire, so more current means more collisions and more resistance. In fact No. In the IB model, resistance arises from collisions of the free electrons with the lattice ions, which are vibrating about fixed positions; these collisions transfer energy to the lattice.
Electrical resistance, R
Electrical resistance, R
The ratio of the potential difference across a component to the current through it: R = V/I. This definition applies to every component at every operating point, ohmic or not; for a non-ohmic component R is found from V/I at the point in question, not from the gradient of the I–V graph. SI unit: ohm (Ω), where 1 Ω = 1 V A⁻¹.
Students often think The resistance at a point equals the inverse gradient ΔV/ΔI of the tangent to the I–V graph at that point. In fact No. Resistance is R = V/I, the ratio of the coordinates at the point. The gradient of the tangent gives how the current changes with pd, which is not the resistance unless the graph is a straight line through the origin.
Students often think The gradient of any graph of current and pd gives the resistance directly, whichever quantity is on which axis. In fact No. With I plotted against V, the gradient has units A V⁻¹: it is a conductance-like quantity, not a resistance. In any case resistance is found from V/I at the point.
Resistivity, ρ
Resistivity, ρ
A property of a material (at a given temperature) defined by ρ = RA/L, where R is the resistance of a uniform sample of length L and cross-sectional area A. Resistance depends on the sample's dimensions (R = ρL/A, so R ∝ L and R ∝ 1/A); resistivity does not. For a wire of diameter d, A = π(d/2)². SI unit: ohm metre (Ω m).
Students often think The diameter of a wire can be substituted for r in A = πr² (procedural error). In fact No. The radius is half the diameter; using the diameter makes the area four times too large and the resistance four times too small.
Students often think Areas convert between mm² and m² by the same factor, 10⁻³, as lengths (procedural error). In fact No. 1 mm = 10⁻³ m, so 1 mm² = (10⁻³ m)² = 10⁻⁶ m².
Ohm's law
Ohm's law
The current in a conductor is directly proportional to the potential difference across it, provided the physical conditions, such as temperature, remain constant. A metal conductor at constant temperature is taken to obey Ohm's law. Ohm's law is not the equation R = V/I: that equation defines resistance for any component, whereas Ohm's law is the experimental finding that, for certain conductors, V/I is constant.
Students often think Ohm's law is V = IR, and since every conductor has a resistance, every conductor obeys Ohm's law. In fact No. V = IR (R = V/I) is the definition of resistance and applies to every component. Ohm's law is the statement that, for some conductors at constant temperature, the current is proportional to the pd.
Students often think Because R = V/I, the resistance of a conductor is proportional to the pd across it (doubling the pd doubles its resistance), and this is taken to be what Ohm's law states. In fact No. R = V/I is how resistance is measured. For an ohmic conductor, increasing V increases I in proportion, so R is unchanged; R depends on the material, dimensions and temperature.
Ohmic and non-ohmic conductors
Ohmic and non-ohmic conductors
An ohmic conductor obeys Ohm's law: its I–V graph is a straight line through the origin, so V/I is the same at every pd (for example, a metal wire kept at constant temperature). A non-ohmic conductor has a V/I that changes with the pd: its I–V graph is curved or does not pass through the origin. A filament lamp is non-ohmic: its I–V graph (I on the vertical axis) is a curve through the origin whose gradient decreases as the pd increases, because the filament heats up and its resistance rises.
Heating effect of a current
When there is a current in a resistor, electrical energy is transferred to the internal energy of the resistor (collisions of charge carriers with the lattice), which may raise its temperature. The rate of this transfer is the power dissipated, P = I²R. In a metal, a rise in temperature increases the resistance, which is why a filament lamp is non-ohmic.
Students often think A component is ohmic whenever its I–V graph is a straight line, wherever the line crosses the axes. In fact No. An ohmic component has an I–V graph that is a straight line through the origin (I ∝ V). A straight line that does not pass through the origin means V/I changes with V, so the component is non-ohmic.
Students often think A filament's resistance increases because it expands when it heats up, and the extra length increases R = ρL/A. In fact No. Expansion increases the length and the area by similar fractions, and the change is tiny (about 1% for a tungsten filament). The large rise in resistance is due to more frequent electron–ion collisions, which increase the resistivity.
Electrical power, P
Electrical power, P
The rate at which energy is transferred: P = IV. For a resistor, substituting V = IR gives P = I²R = V²/R. The energy transferred in a time t at constant power is E = Pt. Use P = I²R when the current is known or common (series components) and P = V²/R when the pd is known or common (parallel components). SI unit: watt (W), where 1 W = 1 J s⁻¹.
Students often think Power dissipated can be found as P = IR or P = V/R, without squaring (procedural error). In fact No. The power depends on the square of the current (or pd): doubling the current through a resistor quadruples the power.
Students often think Power and energy are the same thing, so the power of a device, in watts, is the energy it transfers, in joules. In fact No. Power is the rate of energy transfer (J s⁻¹ = W). The energy transferred in a time t is E = Pt.
Resistors in series
Resistors in series
Components connected one after another in a single path. The current is the same in each: I = I₁ = I₂ = …. The pds add to the total: V = V₁ + V₂ + …, and the pd divides in proportion to resistance. The total resistance is R_s = R₁ + R₂ + …, so adding a resistor in series increases the total resistance and decreases the current.
Resistors in parallel
Components connected across the same two points, each on its own branch. Each has the same pd: V = V₁ = V₂ = …. The branch currents add to the total: I = I₁ + I₂ + …, and the current divides in inverse proportion to the branch resistances. The total resistance is given by 1/R_p = 1/R₁ + 1/R₂ + …, so it is less than the smallest branch resistance, and adding a branch decreases the total resistance.
Students often think Every component in a circuit has the whole supply pd across it, because the pd is 'what the cell gives'. In fact No. In series, the pds across the components add up to the supply pd, so each has only part of it. Only components in parallel directly across the supply each have the full supply pd.
Students often think In a series circuit the supply pd is always shared equally among the components, whatever their resistances. In fact No. In series the pd divides in proportion to resistance: V₁/V₂ = R₁/R₂. It is shared equally only between identical components.
Internal resistance, r, and terminal potential difference
Internal resistance, r, and terminal potential difference
A real cell has a resistance r of its own (the internal resistance), so energy is dissipated inside it when there is a current. For a cell of emf ε connected to an external resistance R, ε = I(R + r). The terminal pd, V = IR = ε − Ir, is less than the emf by the pd across the internal resistance, Ir (sometimes called the 'lost volts'). The terminal pd equals ε only when no current is drawn; it falls as the current increases.
Students often think The pd across the terminals of a cell is always equal to its emf, so the internal resistance can be ignored. In fact No. Terminal pd V = ε − Ir. It equals the emf only when no current flows; it falls as the current increases.
Students often think The current can be found from I = ε/R, using the external resistance alone, and the internal resistance used only afterwards (procedural error). In fact No. The total resistance of the loop includes the internal resistance: I = ε/(R + r).
Thermistor
Thermistor
A resistor made of a semiconducting material whose resistance varies with temperature. For the negative temperature coefficient (NTC) thermistors commonly used, resistance decreases as temperature increases, because more charge carriers are released. Used as a temperature sensor, often in a potential divider.
Light-dependent resistor (LDR)
A resistor made of a semiconducting material whose resistance decreases as the intensity of the light falling on it increases, because the light releases more charge carriers. Its resistance is high in darkness. Used as a light sensor, often in a potential divider.
Potentiometer (potential divider)
A resistor with a sliding contact. Connected across a supply and used as a potential divider, it gives an output pd between one end of the track and the slider that can be varied continuously from zero to the full supply pd. For a uniform track and an ideal voltmeter, the output is the supply pd multiplied by the fraction of the track's resistance between that end and the slider, because the pd across series resistances divides in proportion to their resistances.
Students often think Every resistor behaves like a metal: a hotter thermistor has a higher resistance. In fact No. For the NTC thermistors used in IB circuits, resistance decreases as temperature increases, because more charge carriers are released in the semiconductor.
Students often think An LDR 'resists light', so the brighter the light, the greater its resistance. In fact No. An LDR's resistance decreases as light intensity increases, because the light releases more charge carriers; it is highest in darkness.
Diagnostic a bearings check, not a test
10 questions, one per part of the topic where we can. Answer them, then see which statements you own and which to read.
1 A cell has an emf of 1.5 V. What does this mean?
Answer and reasoning
The cell holds a store of 1.5 C of charge, which it releases into the circuit. — A student who pictures a cell as a container of charge picks this. The volt is a joule per coulomb, not a quantity of charge, and a cell stores no charge: the carriers are already in the circuit and the cell does work on them.
The cell pushes each coulomb of charge around the circuit with a force of 1.5 N. — A student who takes the name 'electromotive force' literally picks this. Emf is energy per unit charge, measured in J C⁻¹ (volts), not a force measured in newtons.
For each coulomb passing through the cell, 1.5 J is transferred to electrical energy. — Emf is the work done by the source per unit charge, ε = W/q, and 1 V = 1 J C⁻¹. So 1.5 J of chemical energy is transferred to electrical energy for every coulomb that the cell moves around the circuit.
The cell drives a current of 1.5 A through any circuit connected to it. — A student who thinks a cell supplies a fixed current picks this. The emf is in volts, not amperes, and the current the cell drives depends on the total resistance of the circuit: I = ε/(R + r).
Working ε = W/q, so the energy transferred per coulomb is W = εq = 1.5 V × 1 C = 1.5 J (1 V = 1 J C⁻¹).
2 A chemical cell and a solar cell are each used as the energy source in a circuit. Which describes the main energy transfer in each?
Answer and reasoning
Chemical cell: chemical energy to electrical. Solar cell: energy of light to electrical. — In a chemical cell, reactions transfer the chemical energy of the reactants to electrical energy. In a photovoltaic solar cell, the energy of the incident light (electromagnetic radiation) is transferred directly to electrical energy.
Chemical cell: stored electrical energy is let out. Solar cell: light to electrical. — A student who thinks a battery is a store of 'electricity' picks this. A chemical cell stores chemical energy in its reactants; electrical energy is produced only while the reactions run with a current in the circuit.
Chemical cell: chemical energy to electrical. Solar cell: the Sun's heat to electrical. — A student who confuses solar cells with solar water-heating panels picks this. A photovoltaic cell responds to light, not heat: the energy of the light frees charge carriers and produces an emf.
Chemical cell: chemical energy to electrical. Solar cell: its stored light to electrical. — A student who thinks a solar cell stores the light it absorbs, like a battery, picks this. A solar cell stores nothing; it transfers light energy to electrical energy only while light falls on it.
3 A copper wire forms part of a circuit carrying a steady direct current. Which statement about the charge carriers in the wire is correct?
Answer and reasoning
Free electrons drift slowly, in the opposite direction to the conventional current. — In a metal the charge carriers are free (delocalized) electrons. Being negative, they drift from the negative towards the positive terminal, opposite to the conventional current, and their drift speed is small (typically well under 1 mm s⁻¹).
Electrons travel at close to the speed of light, so a lamp lights at the instant the switch closes. — A student who thinks the electrons race round the circuit picks this. The electrons drift slowly; the lamp lights at once because the electric field is set up throughout the circuit almost instantly, so carriers everywhere start moving together.
Protons move along the wire, in the same direction as the conventional current. — A student who thinks conventional current is a flow of positive charges picks this. In a metal the positive charges are in the nuclei of the lattice ions, which stay in place; only electrons move along the wire.
Electrons leave the cell and some are used up in the lamp, so fewer return to the cell. — A student who thinks current is used up in components picks this. Charge is conserved: the current is the same all round a series loop. The lamp transfers energy from the carriers, not the carriers themselves.
4 A steady current of 0.50 A flows through a small electric motor for 2.0 minutes. In this time, 360 J of energy is transferred from electrical energy to other forms in the motor. What is the potential difference across the motor?
Answer and reasoning
360 V — A student who leaves the time in minutes gets q = 0.50 × 2.0 = 1.0 C and V = 360 V. The time must be in seconds, 120 s, before finding the charge.
720 V — A student who uses the current as if it were the charge gets 360/0.50 = 720 V. The charge is the current multiplied by the time: q = 0.50 A × 120 s = 60 C.
3.0 V — A student who thinks pd is energy per second gets 360/120 = 3.0. That is the power, 3.0 W. Potential difference is energy per unit charge, V = W/q.
6.0 V — The charge that passes is q = It = 0.50 × 120 = 60 C. V = W/q = 360/60 = 6.0 V: 6.0 J is transferred for each coulomb passing through the motor.
Working q = IΔt = (0.50 A)(2.0 × 60 s) = 60 C. V = W/q = 360 J / 60 C = 6.0 V.
5 What is the origin of the electrical resistance of a metal wire?
Answer and reasoning
Free electrons collide with the vibrating lattice ions, transferring energy to them. — As the free electrons drift under the electric field, they collide with the positive ions of the lattice, which are vibrating about fixed positions. Each collision transfers energy to the lattice, opposing the drift and heating the metal.
Some of the free electrons are absorbed by the metal atoms, so fewer leave than enter. — A student who thinks current is used up picks this. Charge is conserved, so as many electrons leave a section of wire each second as enter it. Resistance transfers energy from the electrons; it does not remove them.
The electrons rub against the surface of the wire, like water flowing along a pipe. — A student who extends the water-in-pipes analogy too far picks this. The current flows through the whole cross-section of the metal, and resistance arises from collisions with lattice ions throughout its volume, which is why R ∝ L/A.
The electrons bump into one another as they crowd through the wire. — A student who uses a crowd or traffic-jam picture picks this. In the IB model, the energy is transferred to the metal through collisions of electrons with the massive lattice ions, not with each other.
6 The I–V characteristic of a filament lamp is plotted with current I on the vertical axis and pd V on the horizontal axis. Up to 0.10 V the graph is a straight line through the origin, and at 0.10 V the current is 0.050 A. At 6.0 V the current is 0.50 A, and the tangent to the curve at this point has a gradient of 0.040 A V⁻¹. What is the resistance of the lamp at 6.0 V, and how is it found?
Answer and reasoning
25 Ω, from the reciprocal of the tangent's gradient — A student who thinks resistance is the inverse gradient of the I–V graph gets 1/0.040 = 25 Ω. That method works only for a straight line through the origin; for a curve, R = V/I at the point.
2.0 Ω, from V/I on the straight part near the origin — A student who thinks a component has one fixed resistance gets 0.10/0.050 = 2.0 Ω. That is the resistance of the cold filament. The filament heats up at higher pd, and its resistance at 6.0 V is much larger.
0.040 Ω, from the gradient of the tangent at 6.0 V — A student who reads the gradient of an I–V graph directly as a resistance gets 0.040. With I on the vertical axis the gradient has units A V⁻¹, not Ω; and in any case resistance is V/I at the point.
12 Ω, from V/I using the pd and current at 6.0 V — Resistance is defined as R = V/I at the operating point, for any component: R = 6.0 V / 0.50 A = 12 Ω. The tangent's gradient describes how the current changes with pd, not the resistance.
Working R = V/I at the point = 6.0 V / 0.50 A = 12 Ω.
The current in a conductor is proportional to the pd across it, provided its temperature is constant. — Ohm's law states that the current in a conductor is directly proportional to the potential difference across it, provided physical conditions such as temperature stay the same. A metal conductor at constant temperature obeys it.
Every conductor obeys V = IR, so the current in any conductor is proportional to the pd. — A student who thinks V = IR is Ohm's law picks this. V = IR only defines resistance and applies to any component; the current is proportional to the pd only if R stays constant, which is not true of, for example, a filament lamp.
The resistance of a conductor is proportional to the pd across it, at constant temperature. — A student who reads R = V/I as saying that R depends on V picks this. For a conductor obeying Ohm's law, doubling V doubles I, so R = V/I stays the same.
The current in a conductor is proportional to the power supplied to it, at constant temperature. — A student who mixes up pd with power (energy per second) picks this. Ohm's law relates current to pd, the energy per unit charge. For an ohmic conductor P = I²R, so the power is proportional to the square of the current, not to the current.
8 As the pd across a filament lamp is increased, the ratio V/I for the lamp increases. Which explanation is correct?
Answer and reasoning
More of the current is used up in the filament at higher pd, so less passes through it. — A student who thinks current is used up picks this. The current into the filament equals the current out; charge is conserved. The ratio V/I rises because the resistance of the hot filament increases.
The filament heats up; its ions vibrate more, so electrons collide with them more often. — The current heats the filament (the heating effect of a current). At a higher temperature the lattice ions vibrate with larger amplitude, electron–ion collisions become more frequent, and the resistance V/I increases.
The filament expands as it heats up, and its extra length increases its resistance. — A student who explains the rise with R = ρL/A and expansion picks this. Expansion increases the area as well as the length and is only about 1%; the resistance rises several-fold because the resistivity increases with temperature.
At higher current the electrons crowd together and collide with one another more. — A student who uses a crowd picture of resistance picks this. Resistance arises from electron collisions with the lattice ions, and it rises here because those ions vibrate more at the higher temperature.
9 A 12 V dc supply of negligible internal resistance is connected to a 4.0 Ω resistor in series with a parallel combination of a 6.0 Ω resistor and a 12 Ω resistor. What is the current in the 12 Ω resistor?
Answer and reasoning
0.50 A — The parallel pair has 1/R_p = 1/6.0 + 1/12, so R_p = 4.0 Ω; the total resistance is 4.0 + 4.0 = 8.0 Ω and the supply current is 12/8.0 = 1.5 A. The pd across the parallel pair is 1.5 × 4.0 = 6.0 V, so the current in the 12 Ω resistor is 6.0/12 = 0.50 A.
0.75 A — A student who splits the 1.5 A supply current equally between the branches gets 0.75 A. The branches have the same pd, 6.0 V, so the current divides in inverse proportion to resistance: 1.0 A in the 6.0 Ω resistor and 0.50 A in the 12 Ω resistor.
1.00 A — A student who puts the full supply pd across the 12 Ω resistor gets 12/12 = 1.0 A. The 4.0 Ω series resistor takes 6.0 V of the 12 V, leaving 6.0 V across the parallel pair.
1.50 A — A student who thinks the current is the same everywhere in a circuit gives the supply current, 1.5 A. At the junction the current divides, so each branch carries only part of it.
Working R_p = (1/6.0 + 1/12)⁻¹ = 4.0 Ω. R_total = 4.0 + 4.0 = 8.0 Ω. I = 12 V / 8.0 Ω = 1.5 A. V_p = IR_p = 1.5 × 4.0 = 6.0 V. I₁₂ = 6.0 V / 12 Ω = 0.50 A.
10 A battery of emf 9.0 V and internal resistance 0.50 Ω is connected to a 4.0 Ω resistor. What is the terminal potential difference of the battery?
Answer and reasoning
9.0 V — A student who thinks the terminal pd always equals the emf gives 9.0 V. There is a current in the internal resistance, so a pd Ir = 1.0 V is dropped inside the battery.
7.9 V — A student who finds the current from the external resistance alone gets I = 9.0/4.0 = 2.25 A and V = 9.0 − 2.25 × 0.50 = 7.9 V. The internal resistance is in series in the loop, so I = ε/(R + r) = 2.0 A.
1.0 V — A student who gives the pd across the internal resistance, Ir = 2.0 × 0.50 = 1.0 V, as the terminal pd picks this. That pd is dissipated inside the battery; the terminal pd is what is left, 8.0 V.
8.0 V — ε = I(R + r), so I = 9.0/(4.0 + 0.50) = 2.0 A. The terminal pd is V = IR = 2.0 × 4.0 = 8.0 V (equivalently, ε − Ir = 9.0 − 1.0 = 8.0 V).
Working I = ε/(R + r) = 9.0 V / (4.0 Ω + 0.50 Ω) = 2.0 A. V = IR = (2.0 A)(4.0 Ω) = 8.0 V; check: ε − Ir = 9.0 − (2.0)(0.50) = 8.0 V.
Read the ones marked not yet in Learn, then Verify.
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31 more questions. Every wrong answer here is a real misconception, and you see why it is wrong straight away.
1 A student writes: 'For a garden lamp that must shine all night, a solar cell is a better energy source than a chemical cell in every way.' Which evaluation of this claim is correct?
Answer and reasoning
Right: the solar cell stores the energy of the light it absorbs by day and releases it to the lamp at night. — A student who thinks a solar cell stores energy picks this. A solar cell transfers light energy to electrical energy only while light falls on it. In real solar lamps, the storage is done by a hidden rechargeable chemical cell.
Right: the Sun is a renewable source, so a solar cell gives a steady output at all times, by day or night. — A student who equates 'renewable' with 'always available' picks this. Renewable means the source is not used up, not that it is constant: a solar cell's output falls to zero when there is no light.
Wrong: a solar cell holds no store of charge, so it cannot drive a current even while light falls on it. — A student who thinks a source must contain a store of charge picks this. No source stores charge: the charge carriers are already in the circuit. A solar cell produces an emf, and so drives a current, whenever light falls on it.
Wrong: sunlight is renewable, but a solar cell gives no output at night, so a storage cell is needed. — A solar cell's advantages are real — a renewable source, no fuel to replace, no emissions in use — but its output depends on light intensity and is zero at night. A solar garden lamp therefore also needs a (rechargeable) chemical cell, charged by the solar cell during the day, to supply energy at night. So it is not better 'in every way'.
2 A student wants to measure the current in a lamp and the pd across it, but connects the meters the wrong way round. The circuit is: a cell, a voltmeter and the lamp connected in series in a single loop, with an ammeter connected in parallel with the lamp. Both meters are ideal. What does the student observe?
Answer and reasoning
The lamp is off and both meters read zero, as there is no current in the circuit. — A student who thinks that a pd needs a current picks this. The ammeter does read zero, but a pd exists across the voltmeter even though no charge flows through it: by the loop rule, the emf must appear somewhere, and the only place is across the voltmeter's (infinite) resistance.
The lamp is off, the ammeter reads zero and the voltmeter reads the emf of the cell. — An ideal voltmeter has infinite resistance, so with it in series there is no current in the loop: the lamp is off and the ammeter reads zero. With no current, there is no pd across the lamp (or the ammeter) and no pd across any internal resistance, so the whole emf appears across the voltmeter.
The ammeter short-circuits the lamp, so a very large current flows through the ammeter. — A student who looks only at the ammeter–lamp pair picks this. The ammeter does bypass the lamp, but the voltmeter's infinite resistance is in series with the whole loop, so the current is zero. A short circuit gives a large current only if the rest of the loop has low resistance.
The lamp lights normally, and the meters show its current and pd as when correctly connected. — A student who treats meters as passive readers that work wherever they are placed picks this. An ideal voltmeter in series acts as a break in the circuit, and an ideal ammeter in parallel acts as a zero-resistance bypass, so the circuit is changed completely.
3 A steady current of 0.40 A flows in a copper wire for 3.0 minutes. The magnitude of the charge on an electron is e = 1.60 × 10⁻¹⁹ C. How many electrons pass a point in the wire in this time?
Answer and reasoning
7.5 × 10¹⁸ — A student who leaves the time in minutes gets Δq = 0.40 × 3.0 = 1.2 C and N = 7.5 × 10¹⁸. The ampere is a coulomb per second, so the time must be 180 s.
2.5 × 10¹⁸ — A student who treats the current as an amount of charge takes Δq = 0.40 C and gets N = 0.40/(1.60 × 10⁻¹⁹) = 2.5 × 10¹⁸. That is the number passing in one second. Current is a rate, so the charge in 180 s is Δq = IΔt = 72 C.
4.5 × 10²⁰ — Δq = IΔt = 0.40 A × 180 s = 72 C. Each electron carries 1.60 × 10⁻¹⁹ C, so N = Δq/e = 72/(1.60 × 10⁻¹⁹) = 4.5 × 10²⁰.
1.4 × 10¹⁶ — A student who rearranges I = Δq/Δt as Δq = I/Δt gets Δq = 0.40/180 = 2.2 × 10⁻³ C and N = 1.4 × 10¹⁶. Multiplying both sides by Δt gives Δq = IΔt = 72 C. Check the units: A × s = C.
Working Δt = 3.0 min = 180 s. Δq = IΔt = (0.40 A)(180 s) = 72 C. N = Δq/e = 72 C / (1.60 × 10⁻¹⁹ C) = 4.5 × 10²⁰ electrons.
4 A battery of emf 9.0 V, a switch and a lamp are connected in series. The switch is open. An ideal voltmeter is connected first across the switch and then across the lamp. What are the two readings?
Answer and reasoning
Across both the switch and the lamp 0 V — A student who thinks there can be no pd without a current picks this. The lamp's pd is indeed zero, but the pd across the open switch is the full emf: the pds around the loop must still add up to 9.0 V.
Across the switch 9.0 V; across the lamp 0 V — With the switch open there is no current, so the pd across the lamp is IR = 0 and there is no pd across any internal resistance. By the loop rule the pds must add up to the emf, so the full 9.0 V appears across the gap in the switch: work would be done on a charge moved across it, even though none flows.
Across the switch 4.5 V; across the lamp 4.5 V — A student who thinks the pd is always shared equally between series components picks this. The pd divides in proportion to resistance: the open switch has (effectively) infinite resistance and the lamp, carrying no current, has no pd at all.
Across each of the switch and lamp 9.0 V — A student who thinks the full supply pd appears across every component picks this. The pds around the loop must add up to the emf, 9.0 V, so they cannot both be 9.0 V. With no current, the lamp's pd is zero.
Working No current flows, so V_lamp = IR = 0 and the pd across any internal resistance is 0. Loop rule: ε = V_switch + V_lamp, so V_switch = 9.0 V − 0 = 9.0 V.
5 Copper is a good electrical conductor and polythene is an insulator. Which statement explains the difference?
Answer and reasoning
Copper contains charged particles, but polythene contains no charged particles at all. — A student who thinks insulators have no charges picks this. Polythene, like all matter, contains protons and electrons; they are simply not free to move.
Copper has protons that are free to move; polythene's protons are all held in place. — A student who thinks conventional current is a flow of positive charges picks this. The protons in copper are in the nuclei of the lattice ions and do not move along the wire; the mobile carriers are electrons.
Copper has electrons free to move through it; polythene's electrons are bound to its molecules. — Conduction needs mobile charge carriers. In copper, some electrons are delocalized and free to move through the lattice; in polythene every electron is held in a bond or an atom, so almost no carriers can move.
Copper lets the charge stored in a cell pass through; polythene blocks this charge. — A student who thinks a cell supplies charge from a store picks this. The charge carriers are part of the conductor itself; a cell provides energy, not charge. Polythene does not conduct because its own electrons cannot move.
6 A copper cable has length 5.0 m and diameter 1.0 mm. The resistivity of copper is 1.7 × 10⁻⁸ Ω m. What is the resistance of the cable?
Answer and reasoning
2.7 × 10⁻² Ω — A student who uses the diameter as the radius gets A = π × (1.0 × 10⁻³)² = 3.14 × 10⁻⁶ m², four times too large, and R = 2.7 × 10⁻² Ω. The radius is half the diameter, 5.0 × 10⁻⁴ m.
1.1 × 10⁻¹ Ω — r = 0.50 mm = 5.0 × 10⁻⁴ m, so A = πr² = π × (5.0 × 10⁻⁴)² = 7.85 × 10⁻⁷ m². R = ρL/A = (1.7 × 10⁻⁸ × 5.0)/(7.85 × 10⁻⁷) = 0.11 Ω.
1.1 × 10⁻⁴ Ω — A student who works out A = π × 0.50² = 0.785 mm² and converts with 10⁻³ gets A = 7.85 × 10⁻⁴ m² and R = 1.1 × 10⁻⁴ Ω. Since 1 mm = 10⁻³ m, 1 mm² = 10⁻⁶ m², so A = 7.85 × 10⁻⁷ m².
5.4 × 10⁻⁵ Ω — A student who forgets to square the radius gets A = π × 5.0 × 10⁻⁴ = 1.57 × 10⁻³ and R = 5.4 × 10⁻⁵ Ω. The cross-sectional area is πr², so the radius must be squared.
Working r = d/2 = 0.50 mm = 5.0 × 10⁻⁴ m. A = πr² = π(5.0 × 10⁻⁴ m)² = 7.85 × 10⁻⁷ m². R = ρL/A = (1.7 × 10⁻⁸ Ω m)(5.0 m)/(7.85 × 10⁻⁷ m²) = 0.108 Ω ≈ 1.1 × 10⁻¹ Ω.
7 Wire X has resistance R and is made of a material of resistivity ρ. Wire Y is made of the same material at the same temperature, but it is twice as long as X and has twice the diameter of X. What are the resistance and the resistivity of wire Y?
Answer and reasoning
resistance R, resistivity ρ — A student who thinks doubling the diameter only doubles the area gets R × 2/2 = R. The area is πr², so doubling the diameter makes it four times as large.
resistance R/2, resistivity ρ/2 — A student who treats resistivity as if it were resistance lets it change with the dimensions too. The resistance does halve, but resistivity depends only on the material and temperature, so it stays ρ.
resistance 8R, resistivity ρ — A student who thinks a thicker wire has more resistance multiplies by 2 for the extra length and by 4 for the extra cross-section ('more material in the way'). A larger area gives more paths for the charge carriers, so it reduces the resistance: R ∝ 1/A.
resistance R/2, resistivity ρ — R = ρL/A. Doubling the length doubles R; doubling the diameter quadruples the area, which divides R by 4. So R_Y = R × 2/4 = R/2. Resistivity is a property of the material at a given temperature, so it is unchanged: ρ.
Working R = ρL/A with A = πd²/4. For Y: L_Y = 2L and A_Y = 4A, so R_Y = ρ(2L)/(4A) = ½ ρL/A = R/2. ρ is a property of the material, so ρ_Y = ρ.
8 Three components X, Y and Z are tested. The currents at pds of 1.0 V, 2.0 V and 4.0 V are: X — 0.10 A, 0.20 A, 0.40 A; Y — 0.05 A, 0.15 A, 0.35 A; Z — 0.20 A, 0.30 A, 0.40 A. Which of the components are ohmic, and why?
Answer and reasoning
X and Y: each gives a straight-line graph of I against V. — A student who thinks any straight-line graph means ohmic picks this. Y's line (I = 0.10V − 0.05) does not pass through the origin, so V/I changes with V and Y is non-ohmic.
None of them: V/I should double as V doubles, and does not. — A student who reads R = V/I as 'resistance is proportional to pd', and takes that to be Ohm's law, expects V/I to double when V doubles, and no component does this. Ohm's law says I ∝ V, so V/I stays constant. X has V/I = 10 Ω at every reading, so X is ohmic; Y (20 Ω, 13 Ω, 11 Ω) and Z (5.0 Ω, 6.7 Ω, 10 Ω) are not.
X only: V/I has the same value at every reading. — For X, V/I = 10 Ω at every reading, so I ∝ V: X is ohmic. For Y, V/I is 20 Ω, 13 Ω and 11 Ω; its graph is a straight line that does not pass through the origin. For Z, V/I is 5.0 Ω, 6.7 Ω and 10 Ω. Neither Y nor Z has I proportional to V.
X, Y and Z: V = IR applies to every one of the readings. — A student who thinks Ohm's law is V = IR picks this. Resistance can be calculated for any component at any pd; Ohm's law requires that this V/I stays the same as V changes, which is true only for X.
9 A steady current of 0.25 A flows in a 24 Ω resistor for 2.0 minutes. How much energy is dissipated in the resistor?
Answer and reasoning
720 J — A student who forgets to square the current uses P = IR = 0.25 × 24 = 6.0 and gets 6.0 × 120 = 720 J. IR is the pd (6.0 V), not the power; P = I²R.
3.0 J — A student who leaves the time in minutes gets 1.5 × 2.0 = 3.0 J. The watt is a joule per second, so t = 120 s.
180 J — P = I²R = 0.25² × 24 = 1.5 W. E = Pt = 1.5 × 120 = 180 J.
1.5 J — A student who takes power to be the energy gives the power, 1.5 W, as 1.5 J. Power is energy per second; the energy is E = Pt = 1.5 W × 120 s.
Working P = I²R = (0.25 A)²(24 Ω) = 1.5 W. t = 2.0 min = 120 s. E = Pt = (1.5 W)(120 s) = 180 J.
10 A 4.0 Ω resistor and an 8.0 Ω resistor are connected in parallel across a dc supply. Which resistor dissipates more power, and why?
Answer and reasoning
The 8.0 Ω resistor dissipates twice as much, since P = I²R and its resistance is larger. — A student who thinks the larger resistance always dissipates more power picks this. P = I²R compares powers only when the current is the same, as in series. Here the pd is common, and the 8.0 Ω resistor carries half the current of the 4.0 Ω resistor.
The 4.0 Ω resistor dissipates twice as much, since both have the same pd and P = V²/R. — Parallel components have the same pd. With V common, P = V²/R, so halving R doubles P: the 4.0 Ω resistor dissipates twice the power of the 8.0 Ω resistor (it also carries twice the current).
Both dissipate the same power, since each has the same pd and the current splits equally. — A student who thinks current divides equally at a junction picks this. The pds are indeed equal, but then I = V/R in each branch, so the 4.0 Ω resistor carries twice the current of the 8.0 Ω resistor and, with P = IV, twice the power.
The 8.0 Ω resistor dissipates twice as much, since it takes the larger share of the pd. — A student who thinks parallel components share the supply pd, as series ones do, picks this. Each parallel branch has the full pd between the two junctions, so both resistors have the same pd.
Working V is the same for both. P = V²/R, so P₄/P₈ = (V²/4.0)/(V²/8.0) = 2.
11 Two identical filament lamps P and Q are connected in series with a cell of negligible internal resistance, in the order: positive terminal, lamp P, lamp Q, negative terminal. How do the brightnesses of the lamps compare?
Answer and reasoning
They are equally bright, and each is as bright as a single lamp on the cell. — A student who thinks the cell supplies a fixed current picks this. The current depends on the total resistance: with two lamps in series the resistance is larger and the current smaller, so each lamp is dimmer than a single lamp.
P is brighter, since some of the current is used up in P before it reaches Q. — A student who thinks current is used up in components picks this. Charge is conserved: the current is the same at every point in a series loop, so P and Q carry equal currents.
Q is brighter, as the electrons leave the negative terminal and so reach Q first. — A student who reasons sequentially, following the electrons round the loop, picks this. The current in a series loop is set by the whole circuit and is the same everywhere; the order of the lamps makes no difference.
They are equally bright, and each is dimmer than a single lamp on the cell. — In series the current is the same in both lamps (I = I₁ = I₂), and identical lamps carrying the same current have the same pd and the same power, so they are equally bright. The two lamps share the supply pd, so each has less pd and less current than one lamp alone, and each is dimmer.
12 A filament lamp A is connected across a cell of negligible internal resistance. An identical lamp B is then connected in parallel with A. What happens to the brightness of A and to the current from the cell?
Answer and reasoning
A becomes dimmer, since the cell's current is now shared between two lamps. — A student who thinks a cell supplies a fixed current picks this. The cell provides a fixed emf; adding a parallel branch lowers the total resistance, so the cell supplies more current, and A's share is unchanged.
A becomes dimmer, since the pd across A is now shared with lamp B. — A student who thinks parallel components share the supply pd, as series components do, picks this. Each parallel branch is connected across the same two points, so A still has the full emf across it.
A is unchanged in brightness, and the current from the cell doubles. — A and B are both connected directly across the cell, so each has the full emf across it (V = V₁ = V₂). A's pd, current and power are unchanged. The cell now supplies both branches, so the total current is I = I₁ + I₂, twice the original.
A is unchanged, and so is the cell current, as B's branch is separate from A's. — A student who looks only at A's branch picks this. A is indeed unchanged, but the cell's current passes through both branches: it is I₁ + I₂, so connecting B doubles it.
13 An ideal voltmeter is connected across the terminals of a battery. With nothing else connected it reads 6.0 V. Identical 10 Ω resistors are then connected across the terminals one at a time, each in parallel with those already there. With one, two and three resistors connected, the voltmeter reads 5.7 V, 5.5 V and 5.2 V. When the resistors are removed, the reading returns at once to 6.0 V. Which explanation of the falling readings is correct?
Answer and reasoning
The battery supplies more current, so the pd across its internal resistance increases. — Each added parallel resistor lowers the external resistance, so the current I = ε/(R + r) increases. The pd Ir across the internal resistance rises, so the terminal pd ε − Ir falls. The return to 6.0 V on removing the load shows that the emf itself has not changed. (The data fit ε = 6.0 V and r ≈ 0.5 Ω.)
The battery's emf falls, since each extra resistor draws on more of its emf. — A student who thinks the emf is used up by the load picks this. If the emf had fallen, the reading would not return at once to 6.0 V when the resistors are removed. The emf is constant; the terminal pd falls because of the pd across the internal resistance.
The 6.0 V is shared among more resistors, so the voltmeter gets a smaller share. — A student who thinks parallel components share the supply pd picks this. Each resistor, and the voltmeter, is connected across the same two terminals and has the full terminal pd across it. The terminal pd falls for a different reason: the pd across the internal resistance rises.
The battery's fixed current is shared among more resistors, so each one's pd falls. — A student who thinks a battery supplies a fixed current picks this. The current from the battery increases as resistors are added in parallel, because the total resistance falls; it is the larger current through the internal resistance that lowers the terminal pd.
14 An NTC thermistor is warmed, and separately a light-dependent resistor (LDR) is moved into brighter light. How does the resistance of each change?
Answer and reasoning
The thermistor's rises, as in a hot metal wire; the LDR's falls. — A student who generalizes the behaviour of metals to every resistor picks this. A metal's resistance does rise with temperature, but an NTC thermistor is a semiconductor: warming it releases more charge carriers, so its resistance falls.
Both fall, since more charge carriers are released in each device. — Both are semiconductor devices. Raising the temperature of an NTC thermistor, or the light intensity on an LDR, releases more charge carriers, so the resistance of each decreases.
The thermistor's falls; the LDR's rises, as it resists brighter light. — A student who reads 'light-dependent resistor' as 'resists light' picks this. Light releases charge carriers in the LDR, so the brighter the light, the lower its resistance.
Neither changes, since a component's resistance is a fixed property. — A student who thinks every component has one fixed resistance picks this. Thermistors and LDRs are variable resistors: their resistance changes with temperature and with light intensity respectively.
15 A potentiometer has a uniform resistance track connected across a 9.0 V dc supply of negligible internal resistance. An ideal voltmeter is connected between one end of the track and the sliding contact, which is one third of the way along the track from that end. What does the voltmeter read?
Answer and reasoning
3.0 V — The two parts of the track are in series and carry the same current, so the pd divides in proportion to resistance. The part between the end and the slider has one third of the track's resistance, so its pd is 9.0 × 1/3 = 3.0 V.
9.0 V — A student who thinks the full supply pd appears across any part of a circuit gives 9.0 V. The pds across the two parts of the track add up to 9.0 V, so each part has only a share.
4.5 V — A student who thinks the pd is always shared equally between series parts gives 4.5 V. The pd divides in proportion to resistance, and the parts are in the ratio 1 : 2.
6.0 V — A student who thinks the smaller resistance takes the larger share of the pd gives 9.0 × 2/3 = 6.0 V. In series, the same current flows through both parts, so V = IR is larger across the larger resistance.
Working The section of track between the end and the slider has 1/3 of the total resistance. Series parts share the pd in proportion to resistance: V = 9.0 V × 1/3 = 3.0 V.
16 A cell of emf 6.0 V drives a steady current of 0.25 A through a lamp for 2.0 minutes. How much energy does the cell transfer from its chemical store to the charge that passes through it in this time?
Answer and reasoning
180 J — The emf is the energy the cell transfers per coulomb: W = εq. The charge that passes is q = IΔt = 0.25 × 120 = 30 C, so W = 6.0 × 30 = 180 J.
3.0 J — A student who puts the time into q = IΔt in minutes gets q = 0.25 × 2.0 = 0.50 C and W = 6.0 × 0.50 = 3.0 J. Convert the time to seconds first: 2.0 minutes = 120 s.
1.5 J — A student who treats the current of 0.25 A as if it were a charge of 0.25 C gets W = εq = 6.0 × 0.25 = 1.5 J. The current is the charge per second; the charge that passes is IΔt = 30 C.
720 J — A student who takes the emf to be an energy per second gets W = 6.0 × 120 = 720 J. The volt is a joule per coulomb, not a joule per second: multiply the emf by the charge that passes, not by the time.
Working Charge passing: q = IΔt = 0.25 A × (2.0 × 60 s) = 30 C. Emf is the energy transferred per unit charge, ε = W/q, so W = εq = 6.0 V × 30 C = 180 J.
17 Two resistors, each of resistance 100 kΩ, are connected in series across a 6.0 V supply of negligible internal resistance. A voltmeter that is not ideal, with a constant resistance of 100 kΩ, is connected across one of the resistors. What does the voltmeter read?
Answer and reasoning
3.0 V — A student who thinks a meter simply reads the circuit without affecting it gives the value for the circuit alone, 3.0 V. Because this voltmeter's resistance is comparable to the resistor's, it draws current and lowers the pd across the pair to 2.0 V.
6.0 V — A student who thinks the full supply pd appears across every component gives 6.0 V. The pds across the series parts add up to 6.0 V, so the measured part has only a share of it.
1.5 V — A student who thinks the 3.0 V across the resistor is now shared between the resistor and the voltmeter gives 1.5 V. Parallel components have the same pd; what changes is that the parallel pair, 50 kΩ, now takes a smaller share of the 6.0 V.
2.0 V — The voltmeter's own 100 kΩ is in parallel with the resistor, making 50 kΩ. That 50 kΩ is in series with the other 100 kΩ, so it takes 50/150 of the supply pd: 6.0 × 1/3 = 2.0 V. A non-ideal voltmeter changes the pd it is measuring.
Working The voltmeter in parallel with one 100 kΩ resistor gives 1/R_p = 1/100 + 1/100, R_p = 50 kΩ. This is in series with the other 100 kΩ, total 150 kΩ. The pd divides in proportion to resistance: V = 6.0 V × 50/150 = 2.0 V.
18 A copper wire and a solution of sodium chloride form part of the same series circuit, and a steady current flows in both. What are the charge carriers in the wire and in the solution?
Answer and reasoning
Free electrons in the wire and also in the solution, because a current is a flow of electrons in any conductor. — A student who equates 'current' with 'flow of electrons' picks this. The solution contains no free electrons; its current is carried by ions. Current is the flow of any mobile charge, not only of electrons.
Protons in the wire, moving from the positive terminal to the negative; positive and negative ions in the solution. — A student who takes conventional current to be a flow of positive particles picks this. The protons in copper are fixed in the nuclei of the lattice ions; the mobile carriers in a metal are electrons, which move opposite to the conventional current.
Charge released by the cell, which flows out through the wire, on through the solution and back to the cell. — A student who pictures the cell as a store of charge picks this. The carriers are already present in the wire (electrons) and in the solution (ions); the cell transfers energy to them but supplies no charge.
Free electrons in the wire; in the solution, positive ions and negative ions that move through it in opposite directions. — A current is a flow of whatever charge carriers are mobile. In copper the delocalized electrons move; in the solution the mobile carriers are Na⁺ and Cl⁻ ions, which drift in opposite directions and both contribute to the current.
19 A constant pd is applied across a metal wire, so an electric field acts on the free electrons. Why do the electrons drift along the wire at a constant average velocity rather than accelerating continuously?
Answer and reasoning
Friction between the electrons and the inner surface of the wire balances the force from the field. — A student who models resistance as rubbing against the walls, like water in a pipe, picks this. Electrons move throughout the volume of the metal, and what limits them is collisions with the lattice ions in the bulk of the wire, not its surface.
They gain speed between collisions with the lattice ions but lose it again at each collision. — The field accelerates a free electron only until its next collision with a vibrating lattice ion, where it transfers the energy gained to the lattice. Repeated over many collisions this gives a small constant average drift velocity, and the energy transferred heats the wire: this is the origin of resistance.
Collisions with the other electrons crowding through the wire keep them all moving at one speed. — A student who imagines the electrons jostling one another picks this. In the IB model the electrons transfer energy to the massive lattice ions, which vibrate about fixed positions; electron–electron collisions do not remove energy from the electron gas as a whole.
They already travel at almost the speed of light, so the field cannot make them go any faster. — A student who thinks electrons race round a circuit at the speed of light picks this. The drift velocity in a copper wire is typically well under a millimetre per second; it is collisions, not a speed limit, that stop the electrons accelerating.
20 The pd across and the current in four components are measured. W: 9.0 V and 3.0 A. X: 2.0 V and 4.0 A. Y: 3.0 V and 0.60 A. Z: 0.80 V and 0.20 A. Which component has the greatest resistance?
Answer and reasoning
W — A student who thinks resistance grows with the pd across a component picks the one with the largest pd. W has 9.0 V across it, but it also carries a large current: R = 9.0/3.0 = 3.0 Ω, less than Y's 5.0 Ω.
Y — Resistance is the ratio R = V/I. Component Y has 3.0 V / 0.60 A = 5.0 Ω, larger than W (3.0 Ω), X (0.50 Ω) and Z (4.0 Ω). Neither the largest pd nor the smallest current on its own identifies the largest resistance.
X — A student who thinks a larger current means more collisions and so more resistance picks the component with the largest current. X carries 4.0 A with only 2.0 V across it, so its resistance is the smallest, 0.50 Ω.
Z — A student who judges resistance from the current alone picks the component with the smallest current. Z carries only 0.20 A, but only 0.80 V drives it: R = 0.80/0.20 = 4.0 Ω, less than Y's 5.0 Ω.
Working R = V/I for each: W: 9.0/3.0 = 3.0 Ω; X: 2.0/4.0 = 0.50 Ω; Y: 3.0/0.60 = 5.0 Ω; Z: 0.80/0.20 = 4.0 Ω. The greatest is component Y, 5.0 Ω.
21 A student measures the resistance of a length of metal wire as 5.0 Ω, using a pd of 1.0 V. The student claims that, by Ohm's law, the current in the wire will be 2.4 A when the pd is increased to 12 V. Which statement about this claim is correct?
Answer and reasoning
It holds only if the wire is at the same temperature at 12 V as it was at 1.0 V. — Ohm's law states that I ∝ V for a metal conductor at constant temperature. At 12 V the wire dissipates about 29 W and will get hot unless cooled, raising its resistance above 5.0 Ω and giving less than 2.4 A. The claim is correct only if the temperature is unchanged.
It holds, because 5.0 Ω is a fixed property of this wire whatever pd is applied to it. — A student who thinks a component has one resistance under all conditions picks this. The resistance of a metal depends on its temperature; 5.0 Ω was measured with the wire cool, and at 12 V it will be hotter unless it is deliberately kept cool.
It holds only if the wire is a metal, because every metal conductor obeys Ohm's law. — A student who thinks every metal obeys Ohm's law under all conditions picks this. The wire is a metal, but Ohm's law holds for a metal only at constant temperature; at 12 V the wire will be hotter than at 1.0 V unless it is kept cool, and its resistance will exceed 5.0 Ω.
It fails, because the resistance rises to 60 Ω, in proportion to the pd across the wire. — A student who reads R = V/I as saying that R grows with V picks this. For a wire at constant temperature, doubling V doubles I and R = V/I stays the same; any rise in R at 12 V comes from heating, not from the pd itself.
Working If R stays 5.0 Ω, I = V/R = 12 V / 5.0 Ω = 2.4 A. Ohm's law (I ∝ V) holds for a metal only at constant temperature. At 12 V the wire would dissipate P = V²/R = 144/5.0 = 29 W, far more than the 0.20 W at 1.0 V, so it will heat up and its resistance rise unless it is kept at the same temperature.
22 The circuit diagram shows two lamps, L₁ and L₂, of different resistances connected in parallel to a cell, together with ammeters A₁ and A₂. Which statement about the ammeter readings is correct?
Answer and reasoning
A₁ and A₂ give the same reading, because the current is the same at every point in the circuit. — A student who thinks the current is the same everywhere in any circuit picks this. That is true only in a single loop; here the current divides at the junction, and A₂ measures only the part that goes through L₂.
A₂ reads exactly half of A₁, because the current splits into equal parts at a junction. — A student who thinks a junction always splits the current equally picks this. The lamps have different resistances, so the branch with the smaller resistance takes the larger share; A₂ reads half only if the lamps are identical.
A₁ reads more than A₂, because A₁ measures the total current and A₂ measures only the current in L₂. — A₁ is in the single wire before the junction, so it reads the total current. A₂ is in the branch with L₂ only, so it reads the current in L₂. At the junction I = I₁ + I₂, so A₁ − A₂ is the current in L₁.
A₂ reads less than A₁, because L₂ uses up some of the current passing through it. — A student who thinks current is consumed by a lamp picks this. A₂ does read less, but because the current divided at the junction, not because L₂ used any up: the current leaving L₂ equals the current entering it.
23 In the circuit shown, ammeter A₁ reads 3.0 A. What is the reading on ammeter A₂?
Answer and reasoning
1.0 A — The parallel pair has R_p = 2.0 Ω, so the pd across it is 3.0 × 2.0 = 6.0 V. A₂ is in the 6.0 Ω branch, where the current is 6.0/6.0 = 1.0 A; the 3.0 Ω branch carries the other 2.0 A.
1.5 A — A student who thinks the current splits equally at a junction gives 3.0/2 = 1.5 A. The branches have different resistances: the same pd drives twice the current through 3.0 Ω as through 6.0 Ω.
3.0 A — A student who thinks the current is the same at every point in any circuit gives 3.0 A. The current divides at the junction; A₂ measures only the part that flows through the 6.0 Ω resistor.
4.5 A — A student who adds resistances in parallel takes R_p = 9.0 Ω, giving a pd of 3.0 × 9.0 = 27 V and a branch current of 27/6.0 = 4.5 A, more than the total. In parallel 1/R_p = 1/R₁ + 1/R₂, so R_p = 2.0 Ω.
Working 1/R_p = 1/3.0 + 1/6.0, so R_p = 2.0 Ω. The pd across the parallel pair is V = IR_p = 3.0 A × 2.0 Ω = 6.0 V. A₂ is in the 6.0 Ω branch: I = V/R = 6.0/6.0 = 1.0 A. Check: 6.0/3.0 = 2.0 A in the other branch, and 2.0 + 1.0 = 3.0 A.
24 The circuit shown is a potential divider made from an NTC thermistor and a fixed resistor R connected across a supply of constant pd. The voltmeter is ideal. The temperature of the thermistor rises. What happens to the voltmeter reading?
Answer and reasoning
It decreases, because the resistance of the thermistor rises when it is heated, just as a metal's does. — A student who assumes every resistor behaves like a metal picks this. The resistance of an NTC thermistor falls with temperature, so the thermistor takes a smaller share of the supply pd and the share across R increases.
It stays equal to the supply pd, because the full supply pd appears across each component. — A student who thinks every component has the whole supply pd across it picks this. The thermistor and R are in series, so their pds add up to the supply pd; the voltmeter reads only the share across R.
It stays at half the supply pd, because two components in series share the supply pd equally between them. — A student who thinks series components share the pd equally picks this. The pd divides in proportion to resistance, so as the thermistor's resistance changes, the share across R changes too.
It increases, because the resistance of the thermistor falls, so that R takes a larger share of the supply pd. — The voltmeter is across R. An NTC thermistor's resistance falls as it warms, so R becomes a larger fraction of the total series resistance and, since the pd divides in proportion to resistance, the pd across R rises.
25 The graph shows the I–V characteristics of two components, P and Q. Which statement is correct?
Answer and reasoning
Both P and Q obey Ohm's law, because V = IR can be applied at every point on each graph. — A student who thinks V = IR is Ohm's law picks this. R = V/I merely defines resistance and can be evaluated for any component. Ohm's law is the stronger claim that I ∝ V, which only the straight line through the origin, P, satisfies.
P obeys Ohm's law, and the resistance of Q increases as the pd across it increases. — P is a straight line through the origin, so I ∝ V and its resistance V/I is constant: it is ohmic. For Q, as V increases the current rises less and less, so the ratio V/I at each point grows: its resistance increases, as for a filament lamp that heats up.
The resistance of Q decreases as the pd increases, because the gradient of its graph decreases. — A student who reads the gradient of an I–V graph as the resistance picks this. With I on the vertical axis the gradient is I/V, the reciprocal of the resistance for a line through the origin; a falling gradient means a rising resistance.
Q has a single fixed resistance, which can be found from V/I at any one point on its curve. — A student who thinks every component has one fixed resistance picks this. Because Q's graph is curved, V/I is different at different points: Q's resistance depends on the pd (and hence its temperature).
26 The graph shows how the terminal pd V of a cell varies with the current I that it delivers. What is the internal resistance of the cell?
Answer and reasoning
0.40 Ω — A student who takes the terminal pd to be the pd across the internal resistance divides 1.0 V by 2.5 A to get 0.40 Ω. The terminal pd is the pd across the external circuit; the pd across r is the lost volts, ε − V = 0.50 V.
0.60 Ω — A student who divides the emf by the current gets 1.5/2.5 = 0.60 Ω. That is the total resistance R + r of the whole circuit at this current, not the internal resistance on its own.
0.20 Ω — Rearranging ε = I(R + r) gives V = ε − Ir, so the magnitude of the gradient is r. The line falls from 1.5 V to 1.0 V as the current rises from 0 to 2.5 A: r = 0.50 V / 2.5 A = 0.20 Ω.
0.50 Ω — A student who reads the fall in terminal pd, 1.5 − 1.0 = 0.50 V, as the internal resistance picks this. The lost volts are Ir; dividing by the current, 2.5 A, gives r = 0.20 Ω.
Working V = ε − Ir, so the graph of V against I is a straight line with intercept ε and gradient −r. Intercept: ε = 1.5 V. At I = 2.5 A, V = 1.0 V, so r = (ε − V)/I = (1.5 − 1.0)/2.5 = 0.20 Ω.
27 In the circuit shown, the cell has emf 9.0 V and negligible internal resistance, and the voltmeter is ideal. What does the voltmeter read?
Answer and reasoning
6.0 V — A student who shares the 9.0 V equally between the three resistors gives 3.0 V each, so 6.0 V across two of them. The pd divides in proportion to resistance: the 3.0 Ω resistor alone takes half of the 9.0 V.
4.5 V — The voltmeter is connected across the 1.0 Ω and 2.0 Ω resistors together, 3.0 Ω out of a series total of 6.0 Ω. Series pds divide in proportion to resistance, so V = 9.0 × 3.0/6.0 = 4.5 V (the current is 1.5 A).
9.0 V — A student who thinks the full supply pd appears across any part of the circuit gives 9.0 V. The three series pds add up to 9.0 V, and the 3.0 Ω resistor outside the voltmeter's connections takes 4.5 V of it.
7.4 V — A student who thinks the smaller resistances take the larger shares of the pd divides 9.0 V in the ratio 1 : ½ : ⅓, giving 4.9 V + 2.5 V ≈ 7.4 V across the first two. In series the same current flows in each, so V = IR is larger for the larger resistance.
Working Series: total R = 1.0 + 2.0 + 3.0 = 6.0 Ω, I = 9.0/6.0 = 1.5 A. The voltmeter spans the 1.0 Ω and 2.0 Ω resistors: V = I(R₁ + R₂) = 1.5 × 3.0 = 4.5 V. Equivalently, 9.0 V × 3.0/6.0 = 4.5 V.
28 The circuit shows two lamps connected in parallel to a cell of negligible internal resistance. Lamp L₁ has resistance R and lamp L₂ has resistance 2R. Which lamp is brighter, and why?
Answer and reasoning
L₂, because P = I²R, so the lamp with the larger resistance must dissipate the larger power. — A student who thinks the larger resistance always dissipates more power picks this. P = I²R applies when the current is the same, as in series. In parallel the pd is the same, and L₂ carries only half the current of L₁, so P = V²/R is smaller for L₂.
They are equally bright, because the current from the cell divides equally between the two branches. — A student who thinks a junction always splits the current equally picks this. The branches have the same pd but different resistances, so L₁ takes twice the current of L₂ and dissipates twice the power.
L₁, because the pd across each lamp is the same and P = V²/R is larger for the smaller resistance. — In parallel both lamps have the cell's pd across them. With V fixed, P = V²/R, so the lamp of resistance R dissipates twice the power of the lamp of resistance 2R (it also carries twice the current, since I = V/R).
L₁, because the current reaches it first, and only what is then left over passes along to L₂. — A student who reasons sequentially, with the current 'arriving' at the nearer branch first, picks this. Parallel branches are supplied simultaneously with the same pd; L₁ is brighter because of its smaller resistance, not its position in the diagram.
29 The graph shows how the resistance R of a length L of uniform wire varies with L. The cross-sectional area of the wire is 0.50 mm². What is the resistivity of the material of the wire?
Answer and reasoning
2.0 × 10⁻⁷ Ω m — From ρ = RA/L, the gradient of the R–L graph is ρ/A = 0.80/2.0 = 0.40 Ω m⁻¹. With A = 0.50 mm² = 5.0 × 10⁻⁷ m², ρ = 0.40 × 5.0 × 10⁻⁷ = 2.0 × 10⁻⁷ Ω m, a sensible order of magnitude for a metal.
2.0 × 10⁻⁴ Ω m — A student who converts mm² to m² with the length factor 10⁻³ uses A = 0.50 × 10⁻³ m² and gets 2.0 × 10⁻⁴ Ω m. Because 1 mm = 10⁻³ m, 1 mm² = 10⁻⁶ m².
4.0 × 10⁻¹ Ω m — A student who takes resistivity to mean resistance per metre gives the gradient, 0.40, on its own. Resistivity is RA/L: the gradient must be multiplied by the cross-sectional area, and it is measured in Ω m, not Ω m⁻¹.
2.0 × 10⁻¹ Ω m — A student who substitutes the area in mm² directly gets 0.40 × 0.50 = 0.20 Ω m. Metals have resistivities of order 10⁻⁸ to 10⁻⁶ Ω m; converting 0.50 mm² to 5.0 × 10⁻⁷ m² gives 2.0 × 10⁻⁷ Ω m.
Working R = ρL/A, so the gradient of R against L is ρ/A. Gradient = 0.80 Ω / 2.0 m = 0.40 Ω m⁻¹. A = 0.50 mm² = 0.50 × 10⁻⁶ m². ρ = gradient × A = 0.40 × 5.0 × 10⁻⁷ = 2.0 × 10⁻⁷ Ω m.
30 In the potential divider shown, the supply has emf 8.0 V and negligible internal resistance, and the voltmeter is ideal. At the light level in the room the resistance of the LDR is 400 Ω. What does the voltmeter read?
Answer and reasoning
2.0 V — A student who thinks the smaller resistance takes the larger share of the pd gives the 1200 Ω resistor the share 400/1600, 2.0 V. That is the pd across the LDR; in series the larger resistance has the larger pd, V = IR.
4.0 V — A student who thinks series components always share the pd equally gives 4.0 V. The shares are in the ratio of the resistances, 400 : 1200, so the 1200 Ω resistor has three quarters of the 8.0 V.
6.0 V — The LDR and the 1200 Ω resistor are in series, so the pd divides in proportion to resistance. The voltmeter spans the 1200 Ω resistor, which takes 1200/1600 of the supply: 8.0 × 3/4 = 6.0 V.
8.0 V — A student who thinks the full supply pd appears across every component gives 8.0 V. The pds across the LDR and the resistor add up to 8.0 V; the resistor has only its share, 6.0 V.
Working Series: total R = 400 + 1200 = 1600 Ω, I = 8.0/1600 = 5.0 × 10⁻³ A. The voltmeter is across the 1200 Ω resistor: V = IR = 5.0 × 10⁻³ × 1200 = 6.0 V, i.e. 8.0 V × 1200/1600.
31 The graph shows how the current I in a wire varies with time t. How much charge passes a point in the wire during the 1.6 s shown?
Answer and reasoning
6.4 C — A student who uses Δq = IΔt with the final current gets 4.0 × 1.6 = 6.4 C. The current was smaller than 4.0 A for the whole of the 1.6 s; the charge is the area under the graph, half of that.
2.5 C — A student who rearranges I = Δq/Δt as Δq = I/Δt gets 4.0/1.6 = 2.5 C. Multiplying both sides by Δt gives Δq = IΔt (for a steady current); A/s is not a unit of charge.
3.2 C — Because I = Δq/Δt, the charge that passes is the area under the current–time graph. The area of the triangle is ½ × 1.6 × 4.0 = 3.2 C; the average current over the 1.6 s is 2.0 A.
4.0 C — A student who takes the final current of 4.0 A to be the charge, 4.0 C, picks this. Current is charge per second; over the 1.6 s the charge is the area under the graph, 3.2 C.
Working I = Δq/Δt, so the charge is the area under the I–t graph. The graph is a straight line from the origin to (1.6 s, 4.0 A): area = ½ × 1.6 s × 4.0 A = 3.2 C.
That was your twenty minutes. Real practice on B.5 is past-paper questions marked against the mark scheme.
Paper 1A tests it as multiple choice; Paper 1B through data you have not seen before; Paper 2 with short answers and, at HL, extended responses. Look for the command words — outline, explain, compare, evaluate — and give exactly what each asks for.
Compiled from the IB Physics guide (first assessment 2025, updated November 2023) and our question bank · Specialist review in progress. How these pages are made · Free, no account ·
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