IB Physics SL B.5 Current and circuits Questions

SLSL HLHL

Practise IB Physics SL B.5 by solving resistor networks, calculating charge and power, analysing real-cell voltage and predicting thermistor or LDR divider outputs.

143 questions in this syllabus scope, with question parts counted together.

Exam points & syllabus

Exam points

Define current as charge flow per time; convert consistently between current, charge, elapsed time and the number of charge carriers.
Explain current through mobile carriers and distinguish their motion from current direction; opposite-moving opposite-sign ions can add to the same current.
Treat emf as source energy per charge; distinguish chemical energy transformed in a cell from energy dissipated in the load and inside the cell.
Read resistance as V/I at the operating point; do not substitute a tangent slope for the static resistance of a non-ohmic component.
State Ohm’s law with constant temperature or physical conditions; justify graph classifications through proportionality or constant V/I.
Use R=ρL/A with length along charge flow and cross-section perpendicular to it; distinguish resistivity from geometry-dependent resistance.
Compare wire dimensions through L/A, square radius or diameter factors and conserve the stated metal volume when a wire is reshaped.
Use resistance graphs through the model: R–L gradient is ρ/A, while resistance at fixed length varies as 1/d².
Explain heating effects through carrier collisions or resistive power, and connect temperature drift to the measurement or operating condition.
Trace circuit nodes before applying series and parallel rules; identify whether a meter reads a branch current, total current or a voltage difference.
After switching or breaking a branch, recalculate current and voltage sharing; series lamps and surviving parallel lamps can change in opposite directions.
Use the actual divider endpoints and operating curves to find outputs or ranges; check both extreme slider positions.
Use common voltage for parallel nonlinear components and common current for series components; show graph readings and voltage or current sums.
Connect measuring instruments to a functional circuit; for loading questions, compare ammeter and voltmeter resistance with the measured load.
Choose the power equation for the specified component and distinguish source, internal and load power using the correct voltage and current.
Explain power comparisons with the fixed quantity: series components share current, parallel components share voltage, and P depends on the matching resistance.
Use working ratings and design limits consistently; round maximum lamp counts down and identify the area used in a surface heating limit.
Track electrical energy with E=VIt and recharge time with E/P; apply overall efficiency once to a consistent energy or power boundary.
Use ε=I(R+r) and Vterminal=ε−Ir; adding parallel load raises total current and internal voltage loss, so terminal voltage can fall.
Derive and interpret cell graphs: a V–I plot has intercept ε and slope −r; a 1/I–R plot has slope 1/ε and R-intercept −r.
Combine cell emfs and internal resistances by their actual connections; for a charging cell in a single loop, terminal voltage is ε+Ir.
Read or derive the internal-resistance condition supplied: compare internal and load heating powers or apply an explicitly given maximum-power criterion.
For thermistor and LDR dividers, connect the resistance change to the correct voltage share; use the graph’s gradient magnitude to judge sensitivity.
Compare energy sources using the stated operating context and supplied data; distinguish photovoltaic electrical output from solar heating.
State approximations and consequences explicitly when using rated filament resistance, battery condition or current-limiting components.

Covered syllabus points

B.5.1—Cells provide a source of emf B.5.2—Electrical energy sources B.5.3—Resistance B.5.4—Ohmic behaviour B.5.5—Electrical power B.5.6—Potential difference and electrical energy transfer B.5.7—Electric cells B.5.8—Circuit diagrams and conventional current B.5.9—Direct current and charge flow B.5.10—Direct and alternating current boundary B.5.11—Conservation in series and parallel circuits B.5.12—Resistance and resistivity B.5.13—Series and parallel circuits B.5.14—Emf and internal resistance B.5.15—Variable resistance

Exam points

Define emf as energy or work supplied by the source per unit charge; state the source or complete-circuit context when the definition carries two marks.
Find emf from supplied energy per charge, ε = E/q; for energy per electron use the magnitude e and show the numerical ratio.
Identify emf from an ideal open-circuit voltmeter reading; a loaded terminal reading can be lower because current flows through internal resistance.
Infer cell emf by counting series voltage-adding stages in the actual battery diagram; identical parallel cells do not add their emfs.

[Maximum number: 1]

This question is in two parts. Part 1 is about a thermistor circuit. Part 2 is about vibrations and waves.
Part 1 Thermistor circuit
The circuit shows a negative temperature coefficient (NTC) thermistor X and a $100 \mathrm{k} \Omega$ fixed resistor R connected across a battery.

The battery has an electromotive force (emf) of 12.0 V and negligible internal resistance.

Part 2 Vibrations and waves

The cone and dust cap D of a loudspeaker L vibrates with a frequency of 1.25 kHz with simple harmonic motion (SHM).

Define electromotive force (emf).

Exam points

Distinguish useful outputs: photovoltaic cells convert solar radiation to electrical energy, while solar heating panels transfer it to thermal energy.
When operation is requested, add that a thermal panel heats liquid for storage or circulation, whereas a photovoltaic cell generates emf or pd.
Choose an electrical domestic use for a photovoltaic cell, such as lighting, a radio or a suitable low-power appliance.
Explain a stated PV benefit through efficiency improvement, renewable input or reduced fossil use and operational emissions; respect the task’s restrictions.
For a specified replacement power, use Pout = ηIA to find photovoltaic area; take √A only when the requested array is square.

[Maximum number: 1]

Outline why research into solar cell technology is important to society.

Exam points

Define component resistance as potential difference across that component divided by current through it; define symbols if using R = V/I.
At a nonlinear I–V graph point, use the coordinate ratio V/I for resistance; a tangent gradient or its reciprocal is not the requested static resistance.
At the same current, compare graph voltages to compare resistances, and state why R = V/I makes the larger voltage imply larger resistance.
Find the selected resistor’s own voltage and current before taking V/I; subtract series voltages or parallel branch currents when required.
Explain conductor–insulator differences through mobile charge carriers and their response to an electric field, not only a good/bad conduction label.
For the supplied carbon-fibre temperature graph, link lower resistivity to lower resistance and greater current at the fixed source pd.
For a nonlinear component, a circuit change moves its operating point; use its characteristic to justify the change in V/I resistance.
Use infinite resistance for an ideal voltmeter and zero for an ideal ammeter; an ideal voltmeter draws no current.
Inspect meter connections before calculating: an ideal voltmeter placed in series opens the loop, so current is zero and the meter reads the source emf.
Recognise W A⁻² as a resistance unit, and keep milliampere readings and any Ω/kΩ prefix consistent with V/I.
When a parallel branch is added, distinguish total source current from a single branch current before linking lower equivalent resistance to a larger reading.

[Maximum number: 2]

The graph shows how current I varies with potential difference V for a resistor R and a non-ohmic component T.

Deduce, without a numerical calculation, whether R or T has the greater resistance at $I=0.40 \mathrm{~A}$ .

$\text { at } 0.4 \mathrm{~A}: V_{\mathrm{R}}>V_{\mathrm{T}} \text { or } V_{R}=5.6 \mathrm{~V} \text { and } V_{\mathrm{T}}=5.3 \mathrm{~V}$

so $\mathrm{R}_{\mathrm{R}}>\mathrm{R}_{\mathrm{T}}$ because $V=I R / V \propto \mathrm{R}$ «and I same for both»

Marking guidance:

Award [0] for a bald correct answer without deduction or with incorrect reasoning.

Ignore any references to graph gradients.

Both elements must be present for MP2 to be awarded.

Exam points

State Ohm’s law with its condition: for the ohmic conductor at constant temperature or physical conditions, V is proportional to I and resistance is constant.
Classify each characteristic with evidence: a straight line through the origin gives constant V/I; a curved graph or unequal V/I values is non-ohmic.
For the supplied Q curve, state that resistance increases as current increases; do not invent a later constant-resistance region.

[Maximum number: 1]

Which of the following is a statement of Ohm's law?

The resistance of a conductor is constant.

The current in a conductor is inversely proportional to the potential difference across the conductor provided the temperature is constant.

The resistance of a conductor is constant provided that the temperature is constant.

The current in a conductor is proportional to the potential difference across it.

Exam points

Choose P=VI, I²R or V²/R for the specified component; convert kW and mA before substitution, and treat energy per second as power.
Distinguish total source power εI from load power VI or I²R; include internal resistance when finding current and total dissipation.
For circuit power, first identify series and parallel paths and use equivalent resistance with the correct total or branch current.
Explain power comparisons using the invariant quantity: series components share current, while parallel components share voltage.
When resistance changes at fixed branch voltage, compare P=V²/R; do not assume current stays fixed while also changing resistance.
Use rated working voltage and power to find resistance or lamp current; for a maximum lamp count, round down to respect the supply limit.
For non-ohmic circuit power, read the operating point from the actual curves; enforce common series current and voltage sums first.
Relate electrical input and useful output through efficiency before using P=VI to calculate motor current.
Use the stated surface power limit and exposed film area to find allowable power, then calculate maximum current from I=√(P/R).
Explain heating effects with a causal chain: current causes resistive dissipation; limiting current reduces heating and temperature drift.

[Maximum number: 2]

A cable consisting of many copper wires is used to transfer electrical energy from a generator to an electrical load. The copper wires are protected by an insulator.

The cable consists of 32 copper wires each of length 35 km . Each wire has a resistance of $64 \Omega$ . The resistivity of copper is $1.7 \times 10^{-8} \Omega \mathrm{~m}$ .

There is a current of 730 A in the cable. Show that the power loss in 1 m of the cable is about 30 W .

$R=2\,\Omega$ $P=I^2R=(730\,\mathrm{A})^2\times2\,\Omega=1.07\times10^6\,\mathrm{W}=1.07\,\mathrm{MW}$ $\text{power loss per metre}=\frac{1.07\times10^6\,\mathrm{W}}{35\times10^3\,\mathrm{m}}=30.6\,\mathrm{W\,m^{-1}}$

Allow [2] for a solution using the resistivity and the answer to (b)(i), giving a resistance per unit length of $5.7\times10^{-5}\,\Omega\,\mathrm{m^{-1}}$ .

Exam points

Calculate discharged electrical energy using E=VIt; convert hours to seconds when voltage and current give power in watts.
For solar recharge time, use panel area and incident intensity, apply overall efficiency once, then divide required energy by useful power.

[Maximum number: 3]

A fully charged cell of emf 6.0 V delivers a constant current of 5.0 A for a time of 0.25 hour until it is completely discharged.

The cell is then re-charged by a rectangular solar panel of dimensions $0.40 \mathrm{~m} \times 0.15 \mathrm{~m}$ at a place where the maximum intensity of sunlight is $380 \mathrm{Wm}^{-2}$ .

The overall efficiency of the re-charging process is 18 %.
Calculate the minimum time required to re-charge the cell fully.

ALTERNATIVE 1

«energy output of the panel = » VIt OR $6 \times 5 \times 0.25 \times 3600$ OR 27000 «J»

«available power = » $380 \times 0.4 \times 0.15 \times 0.18$ OR 4.1 «W»

$t=\frac{27000}{4.1}\approx6600\,\mathrm{s}$

ALTERNATIVE 2

«energy needed from Sun = » $\frac{V I t}{\text { eff }}$ OR $\frac{6 \times 5 \times 0.25 \times 3600}{0.18}$ OR 150000 «J»

« incident power=» $380 \times 0.4 \times 0.15$ OR 22.8 «W»

$t=\frac{150000}{22.8}\approx6600\,\mathrm{s}$

Marking guidance:

Allow ECF for MP3

Accept final answer in minutes (110) or hours (1.8).

Exam points

A discharging cell converts chemical energy to electrical energy; a primary cell is not rechargeable.
Use emf to identify chemical energy transformed per coulomb in the cell; with internal resistance, this is not all delivered to the load.

[Maximum number: 1]

With reference to internal energy conversion and ability to be recharged, what are the characteristics of a primary cell?

Internal energy conversion

Ability to be recharged

chemical to electrical

rechargeable

chemical to electrical

not rechargeable

electrical to chemical

rechargeable

electrical to chemical

not rechargeable

Exam points

Define current as the rate of flow of charge; use q=It and convert hours and microamps consistently.
Convert between electron count and charge using q=Ne: an electron rate gives current Ne/t, and a current over time gives N=It/e.
For opposite-moving positive and negative ions, track charge sign and motion direction; add contributions in the same current direction.
For charge removed from a moving surface, calculate area passing per second, then multiply by surface charge density to obtain current.
Explain field-driven current using mobile charge carriers: copper has free electrons that respond to the field; an insulator has few mobile carriers.

[Maximum number: 1]

In an experiment a student constructs the circuit shown in the diagram. The ammeter and the voltmeter are assumed to be ideal.

A connecting wire in the circuit has a radius of 1.2 mm and the current in it is 3.5 A . The number of electrons per unit volume of the wire is $2.4 \times 10^{28} \mathrm{~m}^{-3}$ . Show that the drift speed of the electrons in the wire is $2.0 \times 10^{-4} \mathrm{~ms}^{-1}$ .

$3.5=2.4 \times 10^{28} \times \pi\left(1.2 \times 10^{-3}\right)^{2} \times 1.6 \times 10^{-19} \times v\left\|\Rightarrow v=2.0 \times 10^{-4} \mathrm{~ms}^{-1}\right\| v$

Exam points

Apply R=ρL/A with the area perpendicular to charge flow; convert mm², cm, km and micrometres consistently before substitution.
At fixed material and temperature, compare R in proportion to L/A; square radius or diameter ratios before comparing resistances.
For reshaped metal, use conserved volume or stated mass to find the new area before applying R=ρL/A and combining the resistors.
Find wire length or radius by rearranging R=ρL/A; use the resistance and resistivity appropriate to the stated working condition.
For a changed current direction through a film, identify the new path length and perpendicular area; explain both effects on resistance.
Distinguish resistivity as a material property from resistance, which also depends on geometry; use resistivity to compare or select materials.
Use an R–L graph gradient as ρ/A; multiply by cross-sectional area for ρ, and give a resistance-per-length constant in Ω m⁻¹.
At fixed length, sketch resistance decreasing as 1/d² through the given point; for R against L, a thicker wire gives a smaller straight-line slope.
Combine geometry-based resistance ratios with circuit rules: add series resistances and use the shared series current for power comparisons.
Interpret supplied resistivity data at the relevant temperature, then check whether the calculated resistance matches the design requirement.
Eliminate geometry or electrical variables explicitly: use A=volume/L for ρ, and R=V²/P before inferring wire length from a power rating.

[Maximum number: 2]

A cable consisting of many copper wires is used to transfer electrical energy from a generator to an electrical load. The copper wires are protected by an insulator.

The cable consists of 32 copper wires each of length 35 km . Each wire has a resistance of $64 \Omega$ . The resistivity of copper is $1.7 \times 10^{-8} \Omega \mathrm{~m}$ .

Calculate the radius of each wire.

A=\frac{(1.7\times10^{-3}\,\Omega\,\mathrm{m})(35\times10^3\,\mathrm{m})}{64\,\Omega}=9.3\times10^{-6}\,\mathrm{m^2}

Exam points

Redraw circuits by connected nodes before combining resistors; ideal wires can make components parallel despite their apparent arrangement.
Use the same voltage across parallel branches and the same current through series components; distinguish branch current from supply current.
For a potential divider, identify the measured endpoints and use Vout=Vsupply Rout/Rtotal; evaluate both extremes when a range is requested.
For parallel non-ohmic components, use one common voltage and add currents from their curves; give the graph reasoning as well as the result.
For series and parallel graph calculations, enforce the supply voltage sum and show the arithmetic connecting the chosen component voltages.
After opening or adding a branch, recalculate voltage sharing and total current; different lamps can brighten and dim in the same change.
For ideal meter circuits, place an ammeter in series and a voltmeter across the required component; check the complete functional current path.
When meter loading is assessed, compare resistances to the measured load: the ammeter must be much smaller and the voltmeter much larger.
Draw a genuine potential-divider arrangement when requested; connect the test resistor to the slider output and measure its current and voltage.
Explain power changes with equivalent resistance, fixed source voltage and the power relation; a bare recalculated power may earn only partial credit.
For parallel lamp limits, add branch currents and round the allowed lamp count down; state any fixed-resistance approximation in estimates.
Explain parallel-lamp or cell-module benefits using component voltage, independent operation or distinct series and parallel effects.

[Maximum number: 4]

The battery has an electromotive force (emf) of 12.0 V and negligible internal resistance.

Part 2 Vibrations and waves

The cone and dust cap D of a loudspeaker L vibrates with a frequency of 1.25 kHz with simple harmonic motion (SHM).

The graph below shows the variation with temperature T of the resistance $R_{\mathrm{x}}$ of the thermistor.

Determine the temperature of X when the potential difference across R is 4.5 V .

$V_X=7.5\,\mathrm{V}$ $I=\frac{4.5\,\mathrm{V}}{100\times10^3\,\Omega}=4.5\times10^{-5}\,\mathrm{A}$ $R_X=\frac{7.5\,\mathrm{V}}{4.5\times10^{-5}\,\mathrm{A}}=1.67\times10^5\,\Omega$ $T=-37\text{ to }-38\,^{\circ}\mathrm{C}$

Alternatively, use the potential-divider relation V_X/V_R=R_X/R_R or the equivalent resistance calculation.

Exam points

Use ε=I(R+r) and Vterminal=ε−Ir; calculate current using the whole series resistance, and use terminal voltage across the external load.
For parallel loads on a real cell, add branch currents before calculating Ir; adding a parallel branch can reduce terminal voltage.
Identify a short-circuit path before using the load resistance; internal resistance can limit current even when the external load is bypassed.
Define internal resistance in context: materials within the cell dissipate energy as current flows, with internal heating power I²r.
Derive terminal voltage from ε=I(R+r) and V=IR, then identify ε as the V–I intercept and −r as its slope.
For a 1/I versus R graph, rearrange to 1/I=R/ε+r/ε: the gradient is 1/ε and the R-axis intercept is −r.
When load resistance rises, current and internal heating fall while terminal voltage rises; retain r when comparing current ratios.
For cell combinations, combine internal resistances by their actual paths and add series emfs; identical parallel strings retain one string’s emf.
In a single loop with opposing cells, use the net emf and total internal resistance; a charging cell has terminal voltage ε+Ir.
Use total cell energy per charge for emf; include both internal and external energy losses, and distinguish emf from a component’s voltage.
Use the correct power curve to find internal resistance: r=Pinternal/I², or r=R where internal and load heating powers are equal.
When a maximum-power condition is explicitly supplied, apply it to the complete external equivalent resistance before solving for one variable resistor.
Explain measurement bias or battery condition through a complete chain of cause, current or power change, and its effect on the inferred result.
Read the cell-model boundary in diagrams: a labelled ideal emf plus a separate internal resistor must be distinguished from the terminal voltage.

[Maximum number: 5]

(a)

A cell is connected to an ideal voltmeter, a switch S and a resistor R. The resistance of R is $4.0 \Omega$ .

When S is open the reading on the voltmeter is 12 V . When S is closed the voltmeter reads 8.0 V .

[ 2 ]

(i)

Deduce the internal resistance of the cell.

[ 2 ]

I=2.0 A OR 12=I(r+4) OR 4=I r OR 8=4 I

«Correct working to get » r=2.0 « $\Omega$ »

Marking guidance:

Allow ECF from (b)(i)

(b)

The voltmeter is used in another circuit that contains two secondary cells.

Cell A has an emf of 10 V and an internal resistance of $1.0 \Omega$ . Cell B has an emf of 4.0 V and an internal resistance of $2.0 \Omega$ .

Calculate the reading on the voltmeter.

[ 3 ]

Exam points

For a potentiometer, evaluate both slider endpoints and use the component’s characteristic to convert the resulting voltage range into a current range.
Read thermistor sensitivity from the magnitude of the R–T gradient; the steepest region gives the largest resistance change per degree.
For a series NTC thermistor and fixed resistor, cooling raises thermistor resistance and increases Vthermistor/Vfixed because the current is shared.
In an LDR divider, increasing light lowers LDR resistance and its voltage share, even though the total circuit current increases.

[Maximum number: 3]

The battery has an electromotive force (emf) of 12.0 V and negligible internal resistance.

Part 2 Vibrations and waves

The cone and dust cap D of a loudspeaker L vibrates with a frequency of 1.25 kHz with simple harmonic motion (SHM).

(a)

The graph below shows the variation with temperature T of the resistance $R_{\mathrm{x}}$ of the thermistor.

[ 3 ]

(i)

State the range of temperatures for which the change in the resistance of the thermistor is most sensitive to changes in temperature.

[ 1 ]

50 to (up to) $-30^{\circ} \mathrm{C} /$ at low temperatures;

(ii)

State and explain the effect of a decrease in temperature on the ratio

\frac{\text { voltage across } \mathrm{X}}{\text { voltage across } \mathrm{R}}

[ 2 ]

as the temperature decreases $R_{X}$ increases;
same current through R and X so the ratio increases or $V_{X}$ increases and $V_{R}$ decreases so the ratio increases;

More syllabus points

B.5.8—Circuit diagrams and conventional currentNo questions yet

No public sample questions are available for this syllabus point yet.

B.5.10—Direct and alternating current boundaryNo questions yet

No public sample questions are available for this syllabus point yet.

B.5.11—Conservation in series and parallel circuitsNo questions yet

No public sample questions are available for this syllabus point yet.

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