Current and circuits HL Papers 1& 2

B.5 — Current and Circuits
HL Practice: Paper 2

15 extended-response questions • 1 lab investigation • Expandable full solutions

This original IB-style practice bank covers current, potential difference, emf, resistance, resistivity, electrical power, series and parallel networks, internal resistance, current–voltage characteristics, potential dividers, and practical measurements. The later questions develop circuit modelling, instrument loading, uncertainty, and experimental evaluation.

Instructions and conventions

Show full working, correct units, and explanations. Attempt each question before opening its solution.
Use the magnitude of the electron charge, $e=1.60\times10^{-19}\,\mathrm{C}$ .
Assume steady direct current and negligible lead resistance unless stated otherwise. Ideal sources have zero internal resistance.
Assume ideal meters unless a finite meter resistance is specified: an ideal ammeter has zero resistance; an ideal voltmeter has infinite resistance.
For a discharging cell use $V=\varepsilon-Ir$ , where $I$ is the current supplied by the cell.
Convert millimetres to metres for wire geometry. Keep resistance units consistent in potential-divider ratios.
Give calculated results to three significant figures unless uncertainty dictates otherwise. Bracketed marks are suggested practice marks.

Key relationships

$\displaystyle I=\frac{\Delta q}{\Delta t},\qquad V=\frac{E}{q},\qquad R=\frac{V}{I}$ $\displaystyle R=\rho\frac{L}{A},\qquad A=\pi\left(\frac{d}{2}\right)^2$ $\displaystyle P=VI=I^2R=\frac{V^2}{R},\qquad E=Pt$ For series and parallel resistor networks: $\displaystyle R_{\mathrm{series}}=\sum R_i,\qquad\frac1{R_{\parallel}}=\sum\frac1{R_i}$ For a source of emf $\varepsilon$ and internal resistance $r$ supplying an external resistance $R$ : $\displaystyle I=\frac{\varepsilon}{R+r},\qquad V=\varepsilon-Ir$ For an unloaded divider, with output across its lower resistor: $\displaystyle V_{\mathrm{out}}=V_s\frac{R_{\mathrm{lower}}}{R_{\mathrm{upper}}+R_{\mathrm{lower}}}$ Charge conservation gives the junction rule; energy conservation gives the loop rule.

Extended-response questions

Question 1 — Charge flow and conventional current [7 marks]

(a) A steady current of $0.240\,\mathrm{A}$ flows through a metal wire for $180\,\mathrm{s}$ . Define electric current and calculate the charge passing a cross-section. [3 marks]

(b) Calculate the number of electrons passing that cross-section. [2 marks]

(c) State the electron drift direction relative to conventional current. Explain whether charge is consumed by a resistor. [2 marks]

Show full solution and marking guidance

(a) [3 marks] Current is the rate of flow of electric charge. $\displaystyle I=\frac{\Delta q}{\Delta t},\qquad q=It=(0.240)(180)=43.2\,\mathrm{C}$

(b) [2 marks] $\displaystyle N=\frac{q}{e}=\frac{43.2}{1.60\times10^{-19}}=2.70\times10^{20}$ This is a number of electrons, not a charge.

(c) [2 marks] Electrons drift opposite to conventional current in a metal. Charge is conserved: the steady current entering the resistor equals the current leaving it. The resistor transfers electrical energy to thermal energy; it does not consume charge.

Question 2 — Potential difference and energy transfer [7 marks]

(a) A heater has a potential difference of $12.0\,\mathrm{V}$ across it and carries $2.50\,\mathrm{A}$ . Define potential difference and calculate the energy transferred when $30.0\,\mathrm{C}$ passes through it. [3 marks]

(b) Calculate its power and the energy transferred in $120\,\mathrm{s}$ . [2 marks]

(c) Distinguish the emf of a source from the potential difference across a component. [2 marks]

Show full solution and marking guidance

(a) [3 marks] Potential difference is energy transferred per unit charge between two points. $\displaystyle V=\frac{E}{q},\qquad E=qV=(30.0)(12.0)=360\,\mathrm{J}$

(b) [2 marks] $\displaystyle P=VI=(12.0)(2.50)=30.0\,\mathrm{W}$ $\displaystyle E=Pt=(30.0)(120)=3.60\times10^3\,\mathrm{J}$

(c) [2 marks] Emf is the energy supplied by the source per unit charge, for example by converting chemical energy. A component’s potential difference describes energy transferred per unit charge across that component. Both are measured in volts; emf is not a force.

Question 3 — Ohm’s law and microscopic resistance [7 marks]

(a) A resistor at constant temperature carries $0.150\,\mathrm{A}$ at $3.00\,\mathrm{V}$ and $0.300\,\mathrm{A}$ at $6.00\,\mathrm{V}$ . Determine its resistance and state the evidence for ohmic behaviour over this range. [3 marks]

(b) Calculate its current and power at $9.00\,\mathrm{V}$ if its temperature remains unchanged. [2 marks]

(c) Explain why the resistance of an ordinary metal wire generally increases as its temperature rises. [2 marks]

Show full solution and marking guidance

(a) [3 marks] $\displaystyle R=\frac{V}{I}=\frac{3.00}{0.150}=20.0\,\Omega$ The second point also gives $20.0\,\Omega$ . Current is proportional to voltage over the measured range at constant temperature.

(b) [2 marks] $\displaystyle I=\frac{9.00}{20.0}=0.450\,\mathrm{A},\qquad P=VI=(9.00)(0.450)=4.05\,\mathrm{W}$

(c) [2 marks] Greater temperature produces stronger lattice vibrations. Conduction electrons scatter more strongly from the vibrating lattice, reducing their drift response to a given electric field and increasing resistivity.

Question 4 — Resistivity and wire geometry [7 marks]

(a) A uniform wire has length $2.00\,\mathrm{m}$ , diameter $0.500\,\mathrm{mm}$ , and resistance $5.00\,\Omega$ . Calculate its cross-sectional area and resistivity. [3 marks]

(b) A second wire of the same material and temperature has twice the length and twice the diameter. Determine its resistance. [2 marks]

(c) The first wire’s resistance, length, and diameter have fractional uncertainties of $2.0\%$ , $1.0\%$ , and $3.0\%$ . Estimate the maximum fractional uncertainty in its calculated resistivity. [2 marks]

Show full solution and marking guidance

(a) [3 marks] $\displaystyle A=\pi\left(\frac{d}{2}\right)^2=\pi(2.50\times10^{-4})^2=1.96\times10^{-7}\,\mathrm{m^2}$ $\displaystyle \rho=\frac{RA}{L}=\frac{(5.00)(1.9635\times10^{-7})}{2.00}=4.91\times10^{-7}\,\Omega\,\mathrm{m}$

(b) [2 marks] Area increases by a factor of four. $\displaystyle \frac{R_2}{R_1}=\frac{L_2/L_1}{A_2/A_1}=\frac24=\frac12$ Hence $R_2=2.50\,\Omega$ .

(c) [2 marks] Since $\rho=R\pi d^2/(4L)$ , add the fractional contributions for a maximum estimate: $\displaystyle \frac{\delta\rho}{\rho}\approx\frac{\delta R}{R}+2\frac{\delta d}{d}+\frac{\delta L}{L}=2.0\%+6.0\%+1.0\%=9.0\%$

Question 5 — Series circuit and energy conservation [7 marks]

(a) Resistors of $4.00\,\Omega$ and $8.00\,\Omega$ are connected in series across an ideal $12.0\,\mathrm{V}$ supply. Calculate the total resistance and circuit current. [2 marks]

(b) Calculate the potential difference and power for each resistor. [3 marks]

(c) Explain why the current is the same in both resistors and why their potential differences add to the supply voltage. [2 marks]

Show full solution and marking guidance

(a) [2 marks] $\displaystyle R_{\mathrm{total}}=4.00+8.00=12.0\,\Omega,\qquad I=\frac{12.0}{12.0}=1.00\,\mathrm{A}$

(b) [3 marks] $\displaystyle V_4=(1.00)(4.00)=4.00\,\mathrm{V},\qquad V_8=(1.00)(8.00)=8.00\,\mathrm{V}$ $\displaystyle P_4=I^2R_4=4.00\,\mathrm{W},\qquad P_8=I^2R_8=8.00\,\mathrm{W}$

(c) [2 marks] There is one path, so charge conservation requires the same steady current throughout. Conservation of energy per unit charge around the loop gives a supply rise equal to the sum of the potential drops: 12.0 V = 4.00 V + 8.00 V.

Question 6 — Parallel branches and an opened switch [7 marks]

(a) A $6.00\,\Omega$ resistor and a $3.00\,\Omega$ resistor form parallel branches across an ideal $12.0\,\mathrm{V}$ supply. Calculate the equivalent resistance and total current. [3 marks]

(b) Determine the branch currents and verify the junction rule. [2 marks]

(c) A switch opens only the $3.00\,\Omega$ branch. Determine the new total current and explain what happens to the current in the other branch. [2 marks]

Show full solution and marking guidance

(a) [3 marks] $\displaystyle \frac1{R_{\mathrm{eq}}}=\frac16+\frac13=\frac12,\qquad R_{\mathrm{eq}}=2.00\,\Omega$ $\displaystyle I_{\mathrm{total}}=\frac{12.0}{2.00}=6.00\,\mathrm{A}$

(b) [2 marks] Each branch has $12.0\,\mathrm{V}$ across it: $\displaystyle I_6=2.00\,\mathrm{A},\qquad I_3=4.00\,\mathrm{A}$ Their sum is $6.00\,\mathrm{A}$ , equal to the current entering the junction.

(c) [2 marks] The remaining branch still has $12.0\,\mathrm{V}$ across it because the supply is ideal. Its current stays at $2.00\,\mathrm{A}$ ; this is now also the total current. An ideal supply maintains its voltage as the load changes.

Question 7 — A mixed circuit and component ratings [8 marks]

(a) A $2.00\,\Omega$ resistor is in series with a parallel combination of $6.00\,\Omega$ and $3.00\,\Omega$ . The network is connected across an ideal $12.0\,\mathrm{V}$ supply. Determine the supply current. [3 marks]

(b) Calculate the voltage across the parallel combination and the current in each branch. [3 marks]

(c) The series resistor is rated at $10.0\,\mathrm{W}$ . Calculate its power and decide whether this rating is exceeded. [2 marks]

Show full solution and marking guidance

(a) [3 marks] $\displaystyle R_{\parallel}=\frac{(6.00)(3.00)}{6.00+3.00}=2.00\,\Omega$ $\displaystyle R_{\mathrm{total}}=2.00+2.00=4.00\,\Omega,\qquad I=\frac{12.0}{4.00}=3.00\,\mathrm{A}$

(b) [3 marks] The series resistor drops $(3.00)(2.00)=6.00\,\mathrm{V}$ , leaving $6.00\,\mathrm{V}$ across each parallel branch. $\displaystyle I_6=\frac{6.00}{6.00}=1.00\,\mathrm{A},\qquad I_3=\frac{6.00}{3.00}=2.00\,\mathrm{A}$

(c) [2 marks] $\displaystyle P_{\mathrm{series}}=I^2R=(3.00)^2(2.00)=18.0\,\mathrm{W}$ This exceeds the $10.0\,\mathrm{W}$ rating. A resistor with a suitable higher rating or a redesigned circuit is required.

Question 8 — Emf, internal resistance, and terminal voltage [8 marks]

(a) A cell with emf $12.0\,\mathrm{V}$ and internal resistance $1.00\,\Omega$ supplies a $5.00\,\Omega$ load. Calculate the current and terminal voltage. [3 marks]

(b) Calculate the power supplied by the source, the power delivered to the load, and the power dissipated internally. [3 marks]

(c) Determine the fraction of the source power delivered to the load. Explain the terminal voltage when the circuit is open. [2 marks]

Show full solution and marking guidance

(a) [3 marks] $\displaystyle I=\frac{\varepsilon}{R+r}=\frac{12.0}{5.00+1.00}=2.00\,\mathrm{A}$ $\displaystyle V=\varepsilon-Ir=12.0-(2.00)(1.00)=10.0\,\mathrm{V}$

(b) [3 marks] $\displaystyle P_{\mathrm{source}}=\varepsilon I=24.0\,\mathrm{W}$ $\displaystyle P_{\mathrm{load}}=I^2R=20.0\,\mathrm{W},\qquad P_{\mathrm{internal}}=I^2r=4.00\,\mathrm{W}$ The load and internal powers sum to the source power.

(c) [2 marks] $\displaystyle \eta_{\mathrm{delivery}}=\frac{20.0}{24.0}=0.833=83.3\%$ In an open circuit $I=0$ , so there is no internal voltage drop and $V=\varepsilon=12.0\,\mathrm{V}$ , assuming negligible voltmeter current.

Question 9 — Inferring cell properties from measurements [8 marks]

(a) A cell’s terminal voltage is $5.40\,\mathrm{V}$ when its current is $0.400\,\mathrm{A}$ and $4.80\,\mathrm{V}$ when its current is $0.800\,\mathrm{A}$ . Determine its internal resistance and emf. [3 marks]

(b) Predict the terminal voltage when a $9.00\,\Omega$ load is connected. [3 marks]

(c) Explain why the terminal voltage decreases when load resistance is reduced, assuming the cell properties remain constant. [2 marks]

Show full solution and marking guidance

(a) [3 marks] Use $V=\varepsilon-Ir$ . $\displaystyle r=-\frac{\Delta V}{\Delta I}=-\frac{4.80-5.40}{0.800-0.400}=1.50\,\Omega$ $\displaystyle \varepsilon=V+Ir=5.40+(0.400)(1.50)=6.00\,\mathrm{V}$

(b) [3 marks] $\displaystyle I=\frac{6.00}{9.00+1.50}=0.571\,\mathrm{A}$ $\displaystyle V=IR=(0.571429)(9.00)=5.14\,\mathrm{V}$

(c) [2 marks] A smaller external resistance increases current. The internal voltage drop Ir then increases, reducing the terminal voltage. The source emf itself is held constant in this model.

Question 10 — A filament lamp’s current–voltage characteristic [7 marks]

(a) A filament lamp carries $0.200\,\mathrm{A}$ at $2.00\,\mathrm{V}$ and $0.350\,\mathrm{A}$ at $6.00\,\mathrm{V}$ . Calculate its resistance at each operating point. [2 marks]

(b) Explain whether the lamp is ohmic and give a physical reason for the change. [2 marks]

(c) Describe an experimental method to obtain its current–voltage characteristic, including meter positions and measurements for both polarities. [3 marks]

Show full solution and marking guidance

(a) [2 marks] $\displaystyle R_1=\frac{2.00}{0.200}=10.0\,\Omega,\qquad R_2=\frac{6.00}{0.350}=17.1\,\Omega$ Use $V/I$ at each point; the slope between the two points is not the lamp’s resistance at either point.

(b) [2 marks] The resistance is not constant, so the lamp is non-ohmic over this operating range. Increasing current heats the filament, increasing its resistivity through stronger scattering of electrons by the lattice.

(c) [3 marks] Use an adjustable low-voltage supply or variable resistor, with an ammeter in series with the lamp and a high-resistance voltmeter in parallel across it. Change the voltage in steps and allow readings to settle at each operating point. Record corresponding current and voltage values, then reverse the supply connections and repeat within the lamp’s rating. Plot current vertically against voltage horizontally; the increasing-temperature lamp gives a curve whose gradient decreases in magnitude away from the origin.

Question 11 — A thermistor potential divider [7 marks]

(a) An ideal $6.00\,\mathrm{V}$ supply connects through a $2.00\,\mathrm{k\Omega}$ fixed resistor to an output junction. An NTC thermistor connects the junction to the negative terminal. The output is measured between the junction and negative terminal with negligible loading. Calculate the output when the thermistor is $4.00\,\mathrm{k\Omega}$ . [2 marks]

(b) Calculate the output when warming reduces its resistance to $1.00\,\mathrm{k\Omega}$ . Explain the direction of change. [3 marks]

(c) A separate controller changes state at an output of $3.00\,\mathrm{V}$ . Determine the thermistor resistance at this threshold. [2 marks]

Show full solution and marking guidance

(a) [2 marks] The output is across the thermistor: $\displaystyle V_{\mathrm{out}}=V_s\frac{R_T}{R_f+R_T}=6.00\frac{4.00}{2.00+4.00}=4.00\,\mathrm{V}$

(b) [3 marks] $\displaystyle V_{\mathrm{out}}=6.00\frac{1.00}{2.00+1.00}=2.00\,\mathrm{V}$ An NTC thermistor’s resistance decreases on warming. Its fraction of the total series resistance falls, so the voltage across it falls.

(c) [2 marks] $\displaystyle 3.00=6.00\frac{R_T}{2.00+R_T}\quad\Rightarrow\quad R_T=2.00\,\mathrm{k\Omega}$ The resistance units must be consistent throughout the ratio.

Question 12 — Loading a potential divider [7 marks]

(a) Two $10.0\,\mathrm{k\Omega}$ resistors form a divider across an ideal $12.0\,\mathrm{V}$ source. Determine the unloaded output across the lower resistor. [2 marks]

(b) A meter with input resistance $10.0\,\mathrm{k\Omega}$ is connected across the lower resistor. Calculate the loaded output and the percentage reduction from the unloaded value. [3 marks]

(c) Explain how a meter with much greater input resistance improves the measurement. [2 marks]

Show full solution and marking guidance

(a) [2 marks] $\displaystyle V_{\mathrm{out}}=12.0\frac{10.0}{10.0+10.0}=6.00\,\mathrm{V}$

(b) [3 marks] The meter and lower resistor are in parallel: $\displaystyle R_{\mathrm{lower,eff}}=\frac{(10.0)(10.0)}{10.0+10.0}=5.00\,\mathrm{k\Omega}$ $\displaystyle V_{\mathrm{loaded}}=12.0\frac{5.00}{10.0+5.00}=4.00\,\mathrm{V}$ $\displaystyle \text{reduction}=\frac{6.00-4.00}{6.00}\times100\%=33.3\%$

(c) [2 marks] A high input resistance draws little current. Its parallel combination with the lower resistor is then close to the lower resistor’s original resistance, so the divider output is disturbed less. An ideal voltmeter has infinite resistance.

Question 13 — A light-dependent resistor and circuit response [7 marks]

(a) An LDR connects an ideal $9.00\,\mathrm{V}$ supply’s positive terminal to an output junction. A $3.00\,\mathrm{k\Omega}$ fixed resistor connects that junction to the negative terminal. Output is measured across the fixed resistor. The LDR is $1.00\,\mathrm{k\Omega}$ in bright light and $12.0\,\mathrm{k\Omega}$ in darkness. Calculate both outputs. [3 marks]

(b) Explain why the output changes in that direction as illumination increases. [2 marks]

(c) Describe how to rearrange the same two components so the output increases in darkness. Calculate the dark output for this arrangement. [2 marks]

Show full solution and marking guidance

(a) [3 marks] $\displaystyle V_{\mathrm{bright}}=9.00\frac{3.00}{1.00+3.00}=6.75\,\mathrm{V}$ $\displaystyle V_{\mathrm{dark}}=9.00\frac{3.00}{12.0+3.00}=1.80\,\mathrm{V}$

(b) [2 marks] More illumination lowers LDR resistance, reducing the total series resistance and increasing current. The fixed resistor’s voltage IR therefore increases.

(c) [2 marks] Place the fixed resistor above the junction and the LDR below it; measure output across the LDR to the negative terminal. $\displaystyle V_{\mathrm{dark}}=9.00\frac{12.0}{3.00+12.0}=7.20\,\mathrm{V}$ The bright output in this arrangement would be $2.25\,\mathrm{V}$ .

Question 14 — Changing a load: power and delivery efficiency [8 marks]

(a) A source has emf $6.00\,\mathrm{V}$ and internal resistance $2.00\,\Omega$ . Calculate the current and load power for an external resistance of $2.00\,\Omega$ . [3 marks]

(b) Repeat for an external resistance of $8.00\,\Omega$ , and calculate the power-delivery efficiency in each case. [3 marks]

(c) Compare the two operating conditions and explain why greater delivery efficiency need not imply greater load power. [2 marks]

Show full solution and marking guidance

(a) [3 marks] $\displaystyle I=\frac{6.00}{2.00+2.00}=1.50\,\mathrm{A},\qquad P_R=I^2R=(1.50)^2(2.00)=4.50\,\mathrm{W}$

(b) [3 marks] With $R=8.00\,\Omega$ : $\displaystyle I=\frac{6.00}{8.00+2.00}=0.600\,\mathrm{A},\qquad P_R=(0.600)^2(8.00)=2.88\,\mathrm{W}$ $\displaystyle \eta_{\mathrm{delivery}}=\frac{I^2R}{\varepsilon I}=\frac{R}{R+r}$ Efficiencies are $50.0\%$ for the $2.00\,\Omega$ load and $80.0\%$ for the $8.00\,\Omega$ load.

(c) [2 marks] The higher resistance delivers a larger fraction of source power externally, but reduces the current and total source power. Thus its load power is lower despite higher delivery efficiency. Efficiency is a ratio; power is an energy-transfer rate.

Question 15 — Meter resistance and measurement uncertainty [8 marks]

(a) A $20.0\,\Omega$ resistor is connected across an ideal $6.00\,\mathrm{V}$ supply through an ammeter of resistance $1.00\,\Omega$ . A voltmeter of effectively infinite resistance is connected across the resistor. Calculate both meter readings. [3 marks]

(b) Determine the resistance obtained from the voltmeter reading divided by the ammeter reading. Explain the error if the supply voltage is used instead. [2 marks]

(c) In a separate measurement, $V=(4.00\pm0.04)\,\mathrm{V}$ and $I=(0.200\pm0.005)\,\mathrm{A}$ . Calculate resistance and estimate its maximum absolute uncertainty. [3 marks]

Show full solution and marking guidance

(a) [3 marks] $\displaystyle I=\frac{6.00}{20.0+1.00}=0.286\,\mathrm{A}$ The voltmeter measures only the resistor’s voltage: $\displaystyle V_R=I R=(0.285714)(20.0)=5.71\,\mathrm{V}$

(b) [2 marks] $\displaystyle R_{\mathrm{measured}}=\frac{5.714286}{0.285714}=20.0\,\Omega$ Using supply voltage gives $6.00/0.285714=21.0\,\Omega$ , including the ammeter resistance. Meter placement matters even though dividing the correctly measured resistor voltage by current gives the right value here.

(c) [3 marks] $\displaystyle R=\frac{4.00}{0.200}=20.0\,\Omega$ $\displaystyle \frac{\delta R}{R}\approx\frac{0.04}{4.00}+\frac{0.005}{0.200}=0.035$ $\displaystyle \delta R=(20.0)(0.035)=0.7\,\Omega$ Report $R=(20.0\pm0.7)\,\Omega$ using the stated maximum-uncertainty estimate.

Lab investigation — Measuring emf and internal resistance

Investigation 1 — Terminal voltage and load current [15 marks]

A student varies the external resistance connected to a cell and measures terminal voltage and current. Treat the following data as idealised central values and initially assume negligible voltmeter current. Use the quoted uncertainties when answering part (c).

Terminal-voltage measurements
Current / A	Terminal voltage / V
0.200	5.70
0.400	5.40
0.600	5.10
0.800	4.80
1.000	4.50

(a) Describe a circuit and procedure to obtain the data, identifying meter positions and important controls. [3 marks]

(b) State the graph to plot. Use its gradient and intercept to determine internal resistance and emf. [3 marks]

(c) Use the readings at $0.200\,\mathrm{A}$ and $1.000\,\mathrm{A}$ to estimate the maximum uncertainty in the gradient by adding fractional uncertainty contributions. Each current has uncertainty $\pm0.020\,\mathrm{A}$ and each voltage $\pm0.050\,\mathrm{V}$ . [3 marks]

(d) Calculate the external load resistance and internal power dissipation at $0.800\,\mathrm{A}$ . [2 marks]

(e) Identify one limitation of the constant-internal-resistance model and one specific improvement to test it. [2 marks]

(f) Explain how voltmeter current can affect the interpretation of an ammeter placed in the external load branch. [2 marks]

Show investigation solution and evaluation

(a) [3 marks] Connect a switch, ammeter, and variable load resistor in series with a low-voltage cell. Connect a high-resistance voltmeter directly across the cell terminals. Adjust the load to obtain a range of simultaneous current and terminal-voltage readings, keeping within component ratings. Open the switch between readings to limit heating and discharge; keep cell temperature and state of charge as constant as possible. Repeat readings, including in reverse order, to check drift. Use negligible-resistance leads and an ammeter whose resistance is accounted for when interpreting the external load.

(b) [3 marks] Plot terminal voltage $V$ vertically against current $I$ horizontally. $\displaystyle V=\varepsilon-rI$ The idealised points lie on a line with gradient: $\displaystyle m=\frac{4.50-5.70}{1.000-0.200}=-1.50\,\mathrm{V\,A^{-1}}$ Therefore $r=-m=1.50\,\Omega$ . The vertical intercept is $\varepsilon=V+Ir=5.70+(0.200)(1.50)=6.00\,\mathrm{V}$ .

(c) [3 marks] The voltage difference has magnitude $1.20\,\mathrm{V}$ with maximum uncertainty $0.050+0.050=0.100\,\mathrm{V}$ . The current difference is $0.800\,\mathrm{A}$ with maximum uncertainty $0.020+0.020=0.040\,\mathrm{A}$ . $\displaystyle \frac{\delta r}{r}\approx\frac{0.100}{1.20}+\frac{0.040}{0.800}=0.133$ $\displaystyle \delta r\approx(1.50)(0.133)=0.20\,\Omega$ Thus $r=(1.50\pm0.20)\,\Omega$ . This is the requested endpoint estimate, not a regression uncertainty.

(d) [2 marks] $\displaystyle R_{\mathrm{external}}=\frac{V}{I}=\frac{4.80}{0.800}=6.00\,\Omega$ $\displaystyle P_{\mathrm{internal}}=I^2r=(0.800)^2(1.50)=0.960\,\mathrm{W}$ The external resistance includes any series meter and lead resistance; it equals the load resistor alone only if those are negligible.

(e) [2 marks] Internal resistance may change with cell temperature, state of charge, or operating current. Take repeated readings at the same currents before and after the sequence while monitoring temperature; compare the slopes or residuals for drift or curvature. Short measurement intervals and rests between readings help control heating and discharge.

(f) [2 marks] If the ammeter reads only load current $I_A$ , the cell also supplies voltmeter current $V/R_V$ . $\displaystyle I_{\mathrm{cell}}=I_A+\frac{V}{R_V}$ The voltage relation is therefore $V=\varepsilon-r(I_A+V/R_V)$ , rather than simply $V=\varepsilon-rI_A$ . A sufficiently high input-resistance voltmeter makes the extra current negligible; otherwise include it in the analysis.

Exam tips and common pitfalls

Current is charge per unit time. Resistors transfer energy; they do not use up charge.
Series components carry the same current. Parallel branches have the same potential difference.
Use the voltage across the component and the current through that component when calculating resistance or power.
Ohm’s law describes proportional current and voltage at constant physical conditions. The definition $R=V/I$ also applies at a particular operating point of a non-ohmic component.
For a non-ohmic characteristic, $V/I$ is not generally the local gradient of a voltage–current graph.
Wire area depends on diameter squared. Resistivity has units of $\Omega\,\mathrm{m}$ .
Include internal resistance when calculating source current. Terminal voltage equals emf at zero current in the simple cell model.
Identify the component across which divider output is measured. Include any load in parallel with that component.
In a terminal-voltage versus current graph, the gradient is $-r$ and the vertical intercept is $\varepsilon$ when the current is the total cell current.

Faith connection — Light, fellowship, and practical love

Hebrews 1:3 presents Jesus as sustaining creation and providing purification for sins. In 1 John 1:3–7, fellowship with God and one another is linked with walking in His light and cleansing through Jesus. Revelation 22:5 points to the hope of God’s light with His people.

As you study how a circuit transfers energy through connected components, reflect on opportunities to serve others through practical care, honest work, and wise use of resources. Let your learning encourage gratitude for creation and trust in Jesus’ love, forgiveness, and promised hope.

Further reading: OpenStax — Emf and internal resistance; OpenStax — Kirchhoff’s rules.

B.5 — Current and Circuits
HL Practice: Paper 2

Instructions and conventions

Key relationships

Extended-response questions

Question 1 — Charge flow and conventional current [7 marks]

Question 2 — Potential difference and energy transfer [7 marks]

Question 3 — Ohm’s law and microscopic resistance [7 marks]

Question 4 — Resistivity and wire geometry [7 marks]

Question 5 — Series circuit and energy conservation [7 marks]

Question 6 — Parallel branches and an opened switch [7 marks]

Question 7 — A mixed circuit and component ratings [8 marks]

Question 8 — Emf, internal resistance, and terminal voltage [8 marks]

Question 9 — Inferring cell properties from measurements [8 marks]

Question 10 — A filament lamp’s current–voltage characteristic [7 marks]

Question 11 — A thermistor potential divider [7 marks]

Question 12 — Loading a potential divider [7 marks]

Question 13 — A light-dependent resistor and circuit response [7 marks]

Question 14 — Changing a load: power and delivery efficiency [8 marks]

Question 15 — Meter resistance and measurement uncertainty [8 marks]

Lab investigation — Measuring emf and internal resistance

Investigation 1 — Terminal voltage and load current [15 marks]

Exam tips and common pitfalls

Faith connection — Light, fellowship, and practical love

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IB Physics HL Book 2025; Papers 1 & 2 Practice 5 Sections65 Lessons54 Weeks

Curriculum

Current and circuits HL Papers 1& 2

B.5 — Current and CircuitsHL Practice: Paper 2

Instructions and conventions

Key relationships

Extended-response questions

Question 1 — Charge flow and conventional current [7 marks]

Question 2 — Potential difference and energy transfer [7 marks]

Question 3 — Ohm’s law and microscopic resistance [7 marks]

Question 4 — Resistivity and wire geometry [7 marks]

Question 5 — Series circuit and energy conservation [7 marks]

Question 6 — Parallel branches and an opened switch [7 marks]

Question 7 — A mixed circuit and component ratings [8 marks]

Question 8 — Emf, internal resistance, and terminal voltage [8 marks]

Question 9 — Inferring cell properties from measurements [8 marks]

Question 10 — A filament lamp’s current–voltage characteristic [7 marks]

Question 11 — A thermistor potential divider [7 marks]

Question 12 — Loading a potential divider [7 marks]

Question 13 — A light-dependent resistor and circuit response [7 marks]

Question 14 — Changing a load: power and delivery efficiency [8 marks]

Question 15 — Meter resistance and measurement uncertainty [8 marks]

Lab investigation — Measuring emf and internal resistance

Investigation 1 — Terminal voltage and load current [15 marks]

Exam tips and common pitfalls

Faith connection — Light, fellowship, and practical love

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B.5 — Current and Circuits
HL Practice: Paper 2