IB Physics HL B.5 Current and circuits Questions

SLSL HLHL

Practise IB Physics HL B.5 by solving resistor networks, reading V–I curves, analysing real-cell voltage under load and tracing power through AC circuits.

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

Exam points & syllabus

Exam points

Define emf as energy or work supplied by the cell per unit charge around the complete circuit, adding the source context when required.
Distinguish emf from terminal potential difference: emf includes all energy supplied per charge, while some is dissipated inside a real cell.
Calculate emf or load voltage from energy per electron using the charge magnitude e, and identify which energy transfer the question specifies.
Identify emf from an ideal open-circuit voltmeter reading; at a potentiometer null point, no current means no drop across internal resistance.
Distinguish photovoltaic radiation-to-electrical conversion from solar-panel radiation-to-thermal conversion, naming both devices and outputs.
Explain photovoltaic electrical output’s versatility through suitable appliances, storage or grid export; recognise that a PV cell supplies dc.
Compare solar devices quantitatively for the given demand and efficiency, rather than claiming one output form is universally more efficient.
Justify solar-cell research through a stated efficiency or fossil-fuel benefit, linking the reason to useful electrical output.
Explain conduction through mobile charge carriers: fields accelerate free electrons, which are abundant in copper but scarce in an insulator.
Explain why sustained current requires a complete charge-transfer path: accumulating charge can oppose and stop further flow.
Use I = Δq/Δt or q = It, converting current and time units; divide current by e to obtain the number of electrons crossing a section each second.
Define resistance using potential difference across the component and current through it; for a non-ohmic device use the operating-point V/I.
Use the voltage and current of the same circuit section when applying R = V/I, and convert mΩ, mA and graph axes before substituting.
State Ohm’s law with constant temperature or physical conditions, then justify ohmic behaviour by proportional V and I or constant V/I.
Describe resistance changes from the actual non-ohmic characteristic without mistaking a decreasing trend for inverse proportionality.
Explain rising metal resistance with temperature through increased lattice vibration and carrier collisions, lowering current at fixed voltage.
Use R = ρL/A with the path length along current and area perpendicular to it, converting rectangular dimensions or squared units correctly.
For a cylindrical wire use A = πd²/4; convert diameter to radius before finding wire length or radius from resistivity.
Scale same-material resistance with L/A, square radius or diameter factors, and account for both dimensions when material is reshaped at fixed volume.
Trace all segments of a single zigzag heater before calculating its radius; use length fractions for segments of a uniform wire.
If current direction through a film changes, identify both the new path length and new perpendicular area before deducing the resistance change.
Use resistivity as a material property, and derive units or graph axes from ρ = RA/L rather than confusing it with sample resistance.
Reduce parallel groups before adding series resistances, then distinguish source current from the current in a selected branch.
Put an ideal ammeter in series and an ideal voltmeter across the relevant nodes; include any explicitly stated finite meter resistance.
For nonlinear components in parallel, find one shared voltage at which the graph currents add to the required total current.
For a resistor beside a non-ohmic device, read its operating voltage and use series voltage subtraction or parallel branch-current subtraction.
Distinguish a slider tap from an adjustable series resistor; use the actual measured segment and device graph to predict voltage and current limits.
At a null potentiometer balance, use the uniform wire’s voltage fraction because the test-cell branch carries no current.
Explain parallel lamp behaviour through shared voltage and added branch currents; round fuse-limited lamp counts down to the largest safe integer.
Use P = IV, I²R or V²/R for the selected circuit section; identify the quantity held fixed before predicting how a resistance change affects power.
Find operating resistance from rated voltage and power, showing substitution and sufficient precision when the question asks to show a value.
Separate whole-circuit power from component power, using supply current for the whole circuit and local voltage or resistance for a chosen load.
Explain power changes after switching or rearranging a network through the new equivalent resistance and the unchanged ideal-source voltage.
For fixed transmitted power, higher voltage gives lower current and less I²R heating; a voltage factor k reduces line power loss by k².
Keep transmitted power, useful output and cable heating separate when combining P = IV, I²R and efficiency = output/input.
Read the correct power curve before using I = √(P/R), and compare calculated or graph power quantitatively with a stated cell limit.
For identical stored cell energy, relate discharge time to total circuit power, distinguishing each resistor’s dissipation from the total.
Use E = IVt with actual operating duration, including the stated fraction of a year or conversion from hours to seconds.
For solar recharging, divide required electrical energy by panel intensity × area × efficiency, applying efficiency once.
For identical cells, add emf across series stages and combine internal resistances by their actual series-parallel topology.
Check both cell orientations when selecting a recharger circuit: the positive charger terminal must connect to each secondary cell’s positive terminal.
Use ε = I(R + r) and V = IR together; find lost volts from ε − V before calculating r, rather than treating load voltage as emf.
Determine r from two loaded states by eliminating their common emf, then recover emf from either state using I(R + r).
For a terminal V–I graph, emf is the zero-current intercept and r is the negative gradient’s magnitude; use widely separated points and correct units.
Sketch the cell terminal V–I line with intercept ε and gradient −r, even when the load is nonlinear; derive the line from the real-cell model.
Linearise the real-cell relation as 1/I = R/ε + r/ε; identify gradient 1/ε and resistance-axis intercept −r.
Use internal heating I²r or equality of internal and external powers to determine r, and distinguish εI from useful external power.
If Rexternal = r is explicitly supplied as a maximum-transfer condition, apply it to the entire external network before finding branch power.
For a real cell, adding a parallel load raises source current and lost volts Ir, reducing terminal voltage; explain the trend when calculations are forbidden.
Link a nearly discharged battery’s lower emf or greater internal resistance to reduced heater power and a longer time to transfer the required energy.
In a thermistor divider, combine the shared current and voltage division with the actual R–T graph; for an NTC thermistor cooling raises its resistance.
Identify a potentiometer’s adjustable-reference role in the stated temperature monitor without importing amplifier gain or a switching-law calculation.

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 a cell’s emf as energy or work supplied per unit charge in driving charge around the complete circuit.
Identify emf from the terminal voltage when no current flows; distinguish the open-switch reading from the lower voltage under load.
Calculate emf from source energy per charge, using the magnitude e for one electron and showing working for a stated approximate value.

[Maximum number: 1]

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 .

State the emf of the cell.

$12 \mathrm{~V}$

Exam points

Distinguish solar radiation converted to electrical energy in a photovoltaic cell from radiation converted to thermal energy in a solar heating panel.
Compare operation as well as output: solar heating circulates or stores hot liquid, while a photovoltaic cell generates an emf.
Explain the versatility of photovoltaic electrical output through suitable domestic appliances, storage or grid export.
Recognise photovoltaic output as direct current when identifying an incorrect claim that the cell generates alternating current.
Compare collector areas using the stated power outputs and efficiencies; for the given water-heating load, solar thermal needs less area than the PV array.
Justify solar-cell research through improved conversion efficiency or reduced dependence on fossil fuels, linking the benefit to its electrical output.

[Maximum number: 3]

This question is in two parts. Part 1 is about solar power and climate models. Part 2 is about digital storage of data.

(a)

Distinguish, in terms of the energy changes involved, between a solar heating panel and a photovoltaic cell.

[ 2 ]

(b)

State an appropriate domestic use for a

[ 1 ]

(i)

photovoltaic cell.

[ 1 ]

Exam points

Define resistance as potential difference across the component divided by current through it, identifying both quantities when using R = V/I.
For a non-ohmic device, calculate resistance from V/I at the stated operating point rather than the tangent gradient of its characteristic.
At the same current, compare the component voltages to compare resistance, and state why the larger voltage means the larger R.
In a parallel branch, use the shared branch voltage and subtract currents at the junction before applying R = V/I to the chosen resistor.
For a series resistor beside a non-ohmic device, read the device voltage at the given current and subtract it from the supply voltage before calculating R.
When a circuit change moves a non-ohmic device’s operating point, use its characteristic to justify the resulting change in V/I.
Find a lamp’s resistance at its rated brightness from its voltage and power, using I = P/V then R = V/I or the equivalent V²/P.
Use zero resistance for an ideal ammeter and infinite resistance for an ideal voltmeter when interpreting measurement circuits.
Recognise equivalent resistance units through Ω = V/A = W/A², keeping power and current factors in the correct positions.

[Maximum number: 1]

This question is in two parts

The plates in (c) are replaced by a cell that has an emf of 12.0 V and internal resistance $5.00 \Omega$ . A resistor of resistance R is connected in series with the cell. The energy transferred by the cell to an electron as it moves through the resistor is $1.44 \times 10^{-18} \mathrm{~J}$ .

Define resistance of a resistor.

ratio potential difference/voltage (across resistor) to current (in resistor) / $\frac{V}{I}$
with symbols defined;

Exam points

State Ohm’s law as V proportional to I, or constant resistance, with temperature and other physical conditions held constant.
Classify a straight characteristic through the origin as ohmic; justify a curved characteristic as non-ohmic using non-constant V/I.
Describe the direction of resistance change from the actual characteristic; a decrease as current rises does not establish inverse proportionality.

[Maximum number: 1]

The graph shows how current I varies with potential difference V across a component X .

Outline why component X is considered non-ohmic.

Exam points

Use P = IV for the chosen circuit or component, converting kW and mA before substituting the matching voltage and current.
Calculate operating resistance from rated voltage and power using R = V²/P, showing substitution and sufficient precision when requested.
For component power, use the voltage across that component or its resistance with its current; supply voltage may include other drops.
Find the shared current and individual voltages from a nonlinear series operating point before calculating the selected device’s power.
Reduce the actual series and parallel paths before calculating total power; open switches remove branches from the circuit.
At fixed supply voltage, adding an identical parallel heater branch doubles total power; an unchanged separate branch keeps its own power.
Distinguish device current from total supply current: an added parallel path can raise total power while the device current stays unchanged.
Compare series resistor powers using their common current and P = I²R; the larger resistance dissipates more power.
Compare parallel resistor powers using their common voltage and P = V²/R; the smaller resistance dissipates more power.
For fixed-voltage resistance changes, use P = V²/R: doubling resistance halves power, and reducing 4R to R quadruples it.
Separate total cell power εI from load power VI and internal heating I²r when the cell has internal resistance.
Read the correct power curve at the stated resistance, then obtain current from I = √(P/R).
Check a cell’s power limit quantitatively at the relevant external resistance, including a short circuit with internal resistance still present.
Explain transmission efficiency by keeping transmitted power fixed: higher voltage lowers current, reducing I²R heating in the wires.
For fixed transmitted power and line resistance, a voltage factor k reduces current by k and line power loss by k².
Calculate transmission current from transmitted P/V and then line heating from I²R, keeping transmitted and lost powers distinct.
Use efficiency = output power/input power to obtain source input power, pairing each power with the voltage and current at that stage.
For an identical cell energy store, compare discharge time using total circuit power; each resistor’s power may differ from the total.
When equal internal and external resistance is given, use half the emf across the external network, then find the chosen branch power.

[Maximum number: 1]

This question is in two parts

Calculate the total power supplied by the cell.

Exam points

Calculate electrical energy from E = IVt, converting the operating time to seconds and applying the stated fraction of time in use.
For solar charging, find useful input power from intensity × panel area × efficiency, then divide the required electrical energy by this power.
Apply efficiency once: divide the useful energy by efficiency or multiply the incident power by efficiency before finding charging time.

[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»
$t=\frac{27000}{4.1}=6600\ \mathrm{s}$

ALTERNATIVE 2

«energy needed from Sun = » $\frac{V I t}{e f f}$ 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»
$t=\frac{150000}{22.8}=6600\ \mathrm{s}$

Marking guidance:

Allow ECF for MP3

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

Exam points

To recharge both secondary cells, connect the recharger’s positive terminal to each cell’s positive terminal and check both cell orientations.

[Maximum number: 1]

Two secondary cells require recharging. Which circuit will recharge both cells?

Figure for Question 1 — IB Physics HL — Option A

Figure for Question 1 — IB Physics HL — Option B

Figure for Question 1 — IB Physics HL — Option C

Figure for Question 1 — IB Physics HL — Option D

[Maximum number: 1]

A1.0 mF capacitor is connected in series to a $2.3 \mathrm{k} \Omega$ resistor and a cell of negligible internal resistance.

The capacitor is initially uncharged. At time t=0 the switch is moved to position A. The initial current in the resistor is 3.9 mA .

Draw, on the circuit diagram above, an arrow showing the direction of the conventional current in the resistor for $t>5.0 \mathrm{~s}$ .

Exam points

Explain conduction using mobile charge carriers: an electric field accelerates free electrons, which are abundant in copper but scarce in an insulator.
For a sustained current, explain the need for charge transfer around a complete circuit; charge build-up can oppose and stop further flow.
Calculate average current as charge transferred divided by the full time interval, converting milliseconds to seconds.
Use q = It for a steady current, converting microamps to amperes and hours to seconds before calculating charge.
Obtain the number of electrons crossing a section each second from I/e; a supplied electron speed is unnecessary for this charge-counting rate.

[Maximum number: 3]

A cable consisting of many copper wires is used to transfer electrical energy from an alternating current (ac) generator to an electrical load. The copper wires are protected by an insulator.

The copper wires and insulator are both exposed to an electric field. Discuss, with reference to charge carriers, why there is a significant electric current only in the copper wires.

Exam points

Use R = ρL/A with the current-path length and the cross-sectional area perpendicular to current; convert all dimensions consistently.
For a cylindrical wire, use A = πr² = πd²/4; a diameter mistaken for a radius changes the calculated area by a factor of four.
At fixed material and temperature, scale resistance with length and inverse area, squaring any radius or diameter scale factor.
When resistance and material stay fixed, wire area is proportional to length, so radius changes with the square root of the length factor.
Scale each wire’s resistance before combining resistors: a wire of length 4L and area 2A has 2R and gives 3R in series with R.
For material reshaped at constant volume, account for both longer path and smaller area; quadrupling length raises resistance sixteenfold.
Use resistance proportional to length for segments of one uniform wire, and show the length fraction when asked to show a resistance.
For a maximum permitted wire resistance, solve R = ρL/(πr²) at the limit to find the minimum radius, with the required significant figures.
Trace the full length of a single zigzag heater, including connecting segments, before finding its radius from resistivity and rated resistance.
When current direction through a film changes, identify the new path length and perpendicular area; explain both effects if calculation is forbidden.
Treat resistivity as a material property for comparing and choosing materials, while resistance also depends on the sample’s dimensions.
For fixed wire length, recognise that plotting 1/R against diameter squared gives a straight line, since 1/R is proportional to d².
At fixed resistance, rearrange ρ = Rπr²/L to obtain resistivity proportional to r²/L.
Derive resistivity units from ρ = RA/L: Ωm is kg m³ s⁻³ A⁻² in fundamental SI units.
If R = kL for a uniform wire, identify k = ρ/A and give its unit as Ω/m, distinguishing it from resistivity in Ωm.
Explain rising metal resistance with temperature through greater lattice vibration and electron collisions, reducing current at fixed voltage.

[Maximum number: 2]

A cable consisting of many copper wires is used to transfer electrical energy from an alternating current (ac) 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 cable is connected to the ac generator which has an output power of 110 MW when the peak potential difference is 150 kV . The resistivity of copper is $1.7 \times 10^{-8} \Omega \mathrm{~m}$ .
output power =110 MW

Calculate the radius of each wire.

$\begin{aligned} \text{area} &= \frac{1.7\times10^{-8}\times35\times10^{3}}{64}=9.3\times10^{-6}\ \mathrm{m^2}\\ \text{radius} &= \sqrt{\frac{9.3\times10^{-6}}{\pi}}=0.00172\ \mathrm{m} \end{aligned}$

Exam points

Reduce each parallel group before adding series resistances, then use the supply voltage to find the total current.
Trace which branch contains the ammeter or selected resistor; its current can differ from the total cell current.
Place an ideal ammeter in series and an ideal voltmeter across the relevant two nodes, checking both instruments when both are requested.
Include any stated meter resistance: a finite voltmeter loads its parallel branch, while a resistive ammeter adds resistance in series.
For parallel nonlinear devices, find the common voltage at which the graph currents add to the specified total current.
Use a nonlinear device’s graph to find current at the slider’s voltage limits; zero potential difference gives the zero-current endpoint.
Distinguish a potentiometer tap from a variable series resistor: identify which resistance or measured segment changes before predicting voltage.
At a potentiometer null point, use the uniform wire’s voltage fraction: no current in the test-cell branch means it does not load the divider.
Explain that adding a parallel lamp branch lowers total resistance and adds source current while existing branches keep the same supply voltage.
For a fuse-limited parallel lamp circuit, add lamp currents and round the allowable lamp count down to the largest safe whole number.
For identical cells in mixed arrangements, add emf across series stages and keep the same emf across cells connected in parallel.
Combine cell internal resistances as resistors: two identical cells in parallel have r/2, and five such pairs in series have 5r/2.
For parallel cell strings, explain the lower internal resistance or greater current capability; surviving strings can still supply power if one fails.
With an ideal voltmeter drawing no current between divider junctions, find each junction potential from the same reference before comparing them.
At equal currents in two parallel nonlinear components, read their common graph intersection, then add the currents to obtain equivalent resistance.
Explain increased power after a network change using lower equivalent resistance, the unchanged supply voltage and greater total current.
For a lamp in series with parallel lamps, use total power to find source current, then distinguish series-lamp and parallel-branch voltages.
In a thermistor divider, subtract the fixed-resistor voltage from the supply, use the shared current to find thermistor resistance, then read temperature.
Explain thermistor voltage ratios using the common series current: Vthermistor/Vfixed equals Rthermistor/Rfixed and rises as an NTC thermistor cools.

[Maximum number: 3]

A girl rides a bicycle that is powered by an electric motor. A battery transfers energy to the electric motor. The emf of the battery is 16 V and it can deliver a charge of 43 kC when discharging completely from a full charge.

(a)

The battery is made from an arrangement of 10 identical cells as shown.

Calculate

[ 3 ]

(i)

the emf of one cell.

[ 1 ]

$\frac{16}{5}=3.2$ « V »

(ii)

the internal resistance of one cell.

[ 2 ]

ALTERNATIVE 1:

2.5 r=0.62

r=0.25 «Ω»

ALTERNATIVE 2:

$\frac{0.62}{5}=0.124$ « $\Omega$ »

r=2(0.124)=0.248 « $\Omega$ »

Marking guidance:

Allow ECF from (d).

Exam points

Distinguish emf from terminal voltage: emf is total source energy per charge, including energy dissipated internally as well as in the load.
Explain internal resistance through resistance of the cell’s contents and energy transferred to those contents, leaving less energy for the external circuit.
Use ε = I(R + r) with the total resistance, and distinguish external terminal voltage IR from internal lost volts Ir.
At zero current an internal resistance produces no voltage drop: an ideal open-circuit voltmeter reads emf and a null-potentiometer balance is unchanged.
Use two measured load states of the same cell to eliminate the emf and solve for r; do not work backwards from a given approximate answer.
After finding internal resistance, calculate emf from either measured state using ε = I(R + r), allowing consistent earlier-result substitution.
Convert energy transferred per electron to the load voltage using V = E/e, then use the remaining emf across r to find the common current or resistance.
For a real battery, compute internal heating with I²r after including both R and r in the current calculation.
Use internal heating Pinternal = I²r to find r, or identify r where internal and external powers are equal; justify the graph reading.
To collect cell V–I data, connect a voltmeter across the cell and an ammeter in series with an adjustable load, without adding another source.
Derive V = ε − Ir by explicitly combining ε = I(R + r) with V = IR rather than quoting only the final relation.
For terminal voltage plotted against current, emf is the extrapolated zero-current intercept and internal resistance is the negative gradient’s magnitude.
Use widely separated points on the fitted line for internal resistance; convert mA to A and show the gradient working needed by the mark scheme.
In a show-that estimate from a graph, give full substitution or enough significant figures to substantiate the stated approximate internal resistance.
Sketch the cell’s terminal V–I characteristic as a straight line with intercept ε and negative gradient −r, even when the external load is non-ohmic.
Linearise ε = I(R + r) as 1/I = R/ε + r/ε: the gradient is 1/ε and the intercept on the resistance axis is −r.
For identical parallel cells, keep the emf unchanged and combine internal resistances in parallel before finding load current.
For a mixed cell module, add emf and internal resistance within each series string, then combine equal strings in parallel.
When maximum power transfer is explicitly stated to require Rexternal = r, apply that condition to the complete external resistor network.
Explain how a load change alters terminal voltage through the changed current and internal drop Ir; an Ohm’s-law statement alone is insufficient.
Adding an external parallel branch lowers equivalent load resistance and raises cell current, increasing Ir and reducing the terminal voltage.
For an almost discharged battery, link reduced emf or increased internal resistance to lower heater power and a longer melting time.

[Maximum number: 4]

This question is in two parts

Show that the value of R is $15.0 \Omega$ .

pd across $R=\frac{1.44 \times 10^{-18}}{1.6 \times 10^{-19}}=9.00 \mathrm{~V}$ ;
pd across internal resistance $=12.0-9.00=3.00 \mathrm{~V}$ ;
current in circuit $=\frac{3.00}{5.00}=0.600 \mathrm{~A}$ ;
$R=\frac{9.00}{0.600}=15.0 \Omega$ ;

Exam points

For an NTC thermistor above a fixed resistor in a divider, explain that cooling raises the thermistor’s voltage share and lowers the junction potential.
Identify a potentiometer’s role in a stated temperature monitor as setting the temperature at which an action or signal occurs.

[Maximum number: 3]

This question is about an operational amplifier (op-amp).

An op-amp is used as a comparator in a circuit, as shown.

The circuit is used to monitor the temperature in a greenhouse. The resistance of a thermistor decreases as its temperature increases.

(a)

Explain how the potential at the inverting input changes as the temperature decreases.

[ 2 ]

(b)

Describe the purpose of component X in the circuit.
When the potential of the non-inverting input is set at 0 V , the heater switches on at a temperature of around $10^{\circ} \mathrm{C}$ . The ideal temperature to grow plants in the greenhouse is $18^{\circ} \mathrm{C}$ or greater. At $18^{\circ} \mathrm{C}$ the resistance of the thermistor is $193 \Omega$ .

[ 1 ]

More syllabus points

B.5.10—Direct and alternating current boundaryNo questions yet

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

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