B.5 Current and circuits flashcards: IB Physics SL and HL

All 71 B.5 Current and circuits cards

1. DefinitionSL & HL

   Define electric current and state its SI unit and whether it is a scalar or a vector.

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   Electric current is the rate of flow of electric charge past a point in a conductor: I = Δq/Δt. SI unit the ampere (A), where 1 A = 1 C s⁻¹. Current is treated as a scalar in circuit work (it has a sense along the conductor, but currents add algebraically at a junction, not by vector addition). Exam tip: the mark is for "rate of flow of charge", not "flow of electrons" or "amount of charge" — omitting "per unit time" scores zero. In metals the carriers are free (delocalised) electrons; in electrolytes and gases both positive and negative ions carry the current.

2. DefinitionSL & HL

   Define potential difference between two points in a circuit.

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   The potential difference (pd) between two points is the work done per unit charge in moving charge between those points: V = W/q. SI unit the volt (V), where 1 V = 1 J C⁻¹. It is a scalar. Exam tip: the mark requires "work done (or energy transferred) per unit charge" — "energy used by the component" without "per unit charge" scores zero. Pd refers to energy converted FROM electrical form in a component (e.g. to thermal or light energy), whereas emf refers to energy converted TO electrical form; the mark scheme looks for this distinction whenever both terms appear.

3. DefinitionSL & HL

   Define electromotive force (emf) of a cell.

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   The emf ε of a cell is the total work done per unit charge by the source in driving charge around a complete circuit, i.e. the electrical energy supplied per unit charge by the source: ε = W/q. SI unit the volt (V); scalar. Exam tip: emf is NOT a force despite its name — writing units of newtons is an automatic zero. The key discriminating phrase is "energy converted from chemical (or other) form to electrical form per unit charge". Numerically ε equals the terminal pd when the cell delivers zero current (open circuit), which is how it is measured with a high-resistance (ideal) voltmeter.

4. DefinitionSL & HL

   Define internal resistance of a cell and explain its physical origin.

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   Internal resistance r is the resistance of the source itself — of the electrolyte and electrodes — to the current it drives, measured in ohms (Ω). Because of r the cell dissipates power I²r internally, so not all the emf is available to the external circuit. Exam tip: students often say internal resistance "reduces the emf" — it does not; ε is fixed by the chemistry, and it is the TERMINAL pd V = ε − Ir that falls as I rises. State that r is why a cell warms in use, why terminal pd drops when extra lamps are switched on in parallel, and why a short-circuited cell delivers a maximum current I = ε/r.

5. DefinitionSL & HL

   Define terminal potential difference and state how it relates to emf and internal resistance.

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   The terminal pd is the potential difference across the external terminals of a cell, equal to the energy per unit charge delivered to the external circuit: V = ε − Ir. SI unit volt; scalar. Exam tip: state explicitly that V < ε whenever current flows, V = ε only when I = 0 (open circuit or an ideal voltmeter alone), and V can be driven below normal values by a large I. A frequent error is treating the "lost volts" Ir as energy that disappears — it is energy per unit charge dissipated as thermal energy inside the cell, so energy is still conserved.

6. DefinitionSL & HL

   Define electrical resistance and state the unit.

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   Resistance is the ratio of the potential difference across a component to the current through it: R = V/I. SI unit the ohm (Ω), where 1 Ω = 1 V A⁻¹; scalar. Exam tip: R = V/I is the DEFINITION of resistance and applies to every component, ohmic or not; Ohm's law is the separate claim that R is constant. Answers stating "resistance is the opposition to current" alone do not gain the defining mark — the ratio must be given. Microscopically, resistance arises from conduction electrons colliding with the vibrating lattice ions, transferring energy to the lattice.

7. DefinitionSL & HL

   State Ohm's law precisely and define an ohmic conductor.

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   Ohm's law: the current through a conductor is directly proportional to the potential difference across it, provided the temperature (and other physical conditions) remain constant. An ohmic conductor is one that obeys this — R = V/I is constant, so its I–V graph is a straight line through the origin. Exam tip: the mark is lost if "at constant temperature" is omitted; that condition is the whole point of the law. Do not write "V = IR is Ohm's law" — V = IR is a definition and is always true instantaneously. A metallic wire at fixed temperature is ohmic; a filament lamp, diode, thermistor and LDR are not.

8. DefinitionSL & HL

   Define a non-ohmic conductor and describe the behaviour of a filament lamp, a diode, an NTC thermistor and an LDR.

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   A non-ohmic component is one whose resistance changes with the conditions, so its I–V graph is not a straight line through the origin. Filament lamp: as I rises the filament heats, lattice ions vibrate more, electrons collide more often, so R increases and the I–V curve bends towards the V axis. Semiconductor diode: conducts negligibly below ≈ 0.7 V forward bias, then current rises steeply; in reverse bias R is very large. NTC thermistor: R falls sharply as temperature rises (more charge carriers released). LDR: R falls as light intensity rises. Exam tip: always explain thermistor/LDR changes by carrier NUMBER, and the lamp by lattice vibration, not the reverse.

9. DefinitionSL & HL

   Define resistivity and state its unit and what it depends on.

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   Resistivity ρ of a material is defined by ρ = RA/L, where R is the resistance of a uniform sample of cross-sectional area A and length L. SI unit ohm metre (Ω m); scalar. It is a property of the MATERIAL and its temperature, independent of the sample's dimensions. Exam tip: the classic error is calling resistivity "the resistance of a material" — it is the resistance of a sample of unit length and unit cross-sectional area. Typical values: copper ≈ 1.7 × 10⁻⁸ Ω m, nichrome ≈ 1.1 × 10⁻⁶ Ω m, so any answer giving a metal resistivity of order 1 Ω m has slipped a power of ten badly.

10. DefinitionSL & HL

    Define drift speed of charge carriers and comment on its typical magnitude.

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    Drift speed v is the average speed with which the free charge carriers move along the conductor in the direction of the current, superimposed on their much faster random thermal motion. From I = nAvq, v = I/(nAq); unit m s⁻¹. Exam tip: typical drift speeds in a copper wire are of order 10⁻⁴ m s⁻¹ (a fraction of a millimetre per second), yet a lamp lights instantly because the electric field is established through the circuit at close to the speed of light and ALL the electrons in the wire start drifting almost simultaneously. Confusing drift speed with signal speed is one of the most heavily penalised errors in this sub-topic.

11. DefinitionSL & HL

    Define electric power in a circuit and state the three booklet forms.

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    Electric power is the rate at which electrical energy is transferred (converted) in a component: P = E/t = IV. Using V = IR gives P = I²R and P = V²/R. SI unit the watt (W), 1 W = 1 J s⁻¹; scalar. Exam tip: P = I²R and P = V²/R are only valid for a resistive (ohmic) component where V = IR holds; P = IV is general and is the form to use for a cell (power delivered by emf is P = Iε) or a motor. A common trap: doubling the pd across a fixed resistor quadruples the power, not doubles it.

12. DefinitionSL & HL

    State Kirchhoff's first law (junction rule) and the conservation principle behind it.

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    At any junction in a circuit the sum of the currents entering equals the sum of the currents leaving: ΣI_in = ΣI_out, equivalently ΣI = 0 taking signs into account. It is a statement of conservation of electric charge — charge does not accumulate at a junction. Exam tip: the mark for the "why" is conservation of CHARGE, not energy; that is the single most common lost mark. State that the rule implies the current is the same at every point in a single series loop, and that in a parallel combination the branch currents sum to the main current, with the larger current in the smaller-resistance branch.

13. DefinitionSL & HL

    State Kirchhoff's second law (loop rule) and the conservation principle behind it.

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    Around any closed loop in a circuit the sum of the emfs equals the sum of the potential drops: Σε = ΣIR. Equivalently the algebraic sum of all potential differences around a closed loop is zero. It follows from conservation of ENERGY — a unit charge returning to its starting point must have gained and lost equal amounts of energy. Exam tip: the mark for the reason is energy, not charge. In practice: choose a loop direction, count an emf as positive when traversed from − to +, count IR as a drop when moving with the current, and remember to include the internal resistance term Ir for every real cell in the loop.

14. DefinitionSL & HL

    Define a potential divider and state what it is used for.

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    A potential divider is two (or more) resistors in series across a supply, arranged so that a chosen fraction of the supply pd is tapped off across one of them: V_out = V_in × R₂/(R₁ + R₂). It provides a continuously variable or sensor-controlled output voltage. Exam tip: describe the divider as sharing pd in the RATIO of the resistances, since the current through both is the same. Replacing R₂ with a thermistor or LDR makes a temperature- or light-sensing circuit; state clearly which way V_out moves when the sensor's resistance falls. The formula only holds if the output draws negligible current, i.e. the load has a very high resistance.

15. DefinitionSL & HL

    Define an ideal ammeter and an ideal voltmeter and explain how a non-ideal meter affects a reading.

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    An ideal ammeter has zero resistance and is connected in SERIES so that it does not change the current it measures. An ideal voltmeter has infinite resistance and is connected in PARALLEL so that it draws no current from the component. Exam tip: a real ammeter's small resistance adds to the circuit, so the measured current is slightly LOW; a real voltmeter's finite resistance draws current in parallel with the component, reducing the combined resistance, so the measured pd is slightly LOW. Both are systematic errors. Connecting an ammeter in parallel with a cell effectively short-circuits it — a standard "suggest what happens" answer.

16. EquationSL & HLData booklet: Yes

    State and explain the defining equation for electric current.

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    I = Δq/Δt (data booklet). I = current in amperes (A); Δq = charge that passes a cross-section in coulombs (C); Δt = time interval in seconds (s). Valid for a steady current; for varying current it gives the average over Δt. Rearranged: Δq = IΔt, and the number of carriers N = Δq/q where q = 1.60 × 10⁻¹⁹ C for electrons. Common misuse: using the total charge stored somewhere rather than the charge that has PASSED the point. Sanity check: a 0.50 A current for 2.0 minutes transfers Δq = 0.50 × 120 = 60 C, which is 60/1.60 × 10⁻¹⁹ = 3.8 × 10²⁰ electrons.

17. EquationSL & HLData booklet: Yes

    State and explain the equation defining potential difference.

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    V = W/q (data booklet). V = potential difference in volts (V); W = work done or energy transferred in joules (J); q = charge moved in coulombs (C). Applies between any two points; the same relation W = qΔV gives the energy gained by a charge accelerated through a pd, which is the basis of the electronvolt: 1 eV = 1.60 × 10⁻¹⁹ J. Common misuse: substituting the total circuit energy rather than the energy transferred in the specific component. Sanity check: 15 C passing through a 6.0 V lamp transfers W = 6.0 × 15 = 90 J.

18. EquationSL & HLData booklet: Yes

    State the defining equation for resistance and its rearrangements.

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    R = V/I (data booklet). R = resistance in ohms (Ω); V = pd across the component in volts (V); I = current through it in amperes (A). Valid for every component at the instant of measurement — for a non-ohmic component R is simply the value at that operating point, given by the RECIPROCAL of the gradient of the chord from the origin to the point on the I–V graph, NOT by the tangent gradient. Rearranged: V = IR and I = V/R. Common misuse: reading the resistance of a filament lamp as 1/(tangent gradient). Sanity check: 3.0 V across a component carrying 0.25 A gives R = 12 Ω.

19. EquationSL & HLData booklet: Yes

    State the resistivity equation and all its rearrangements.

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    ρ = RA/L (data booklet). ρ = resistivity in ohm metres (Ω m); R = resistance in ohms (Ω); A = cross-sectional area in m² (for a wire of diameter d, A = πd²/4); L = length in metres (m). Valid for a uniform sample at constant temperature. Rearranged: R = ρL/A, so R ∝ L and R ∝ 1/A ∝ 1/d². Common misuse: substituting the diameter for the radius, which makes A four times too large. Sanity check: a copper wire (ρ = 1.7 × 10⁻⁸ Ω m) of length 2.0 m and diameter 0.50 mm has A = 1.96 × 10⁻⁷ m², so R = 1.7 × 10⁻⁸ × 2.0/1.96 × 10⁻⁷ = 0.17 Ω.

20. EquationSL & HLData booklet: Yes

    State the drift speed equation and identify every symbol.

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    I = nAvq (data booklet). I = current in amperes (A); n = number of charge carriers per unit volume in m⁻³ (for copper ≈ 8.5 × 10²⁸ m⁻³); A = cross-sectional area in m²; v = drift speed in m s⁻¹; q = charge on each carrier in coulombs (C). Valid for a uniform conductor with one type of carrier. Rearranged: v = I/(nAq). Common misuse: using the number of carriers rather than the number DENSITY, or forgetting that if the same current flows into a thinner section then v must increase since nAv q is fixed. Sanity check: I = 1.0 A in a wire of A = 1.0 × 10⁻⁶ m² gives v = 1.0/(8.5 × 10²⁸ × 1.0 × 10⁻⁶ × 1.6 × 10⁻¹⁹) ≈ 7 × 10⁻⁵ m s⁻¹.

21. EquationSL & HLData booklet: Yes

    State the three expressions for electric power and the conditions on each.

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    P = IV, P = I²R, P = V²/R (all data booklet). P = power in watts (W); I = current in amperes (A); V = pd in volts (V); R = resistance in ohms (Ω). P = IV is completely general. P = I²R and P = V²/R follow from V = IR and so apply only to a resistor of resistance R. Energy transferred E = Pt = IVt in joules. Common misuse: using P = V²/R with the SUPPLY voltage rather than the pd across that particular resistor in a series circuit. Sanity check: a 1200 W heater on 240 V draws I = 1200/240 = 5.0 A and has R = 240²/1200 = 48 Ω.

22. EquationSL & HLData booklet: Yes

    State the rules for combining resistors in series and give the consequences.

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    R_series = R₁ + R₂ + R₃ + … (data booklet). Each R in ohms (Ω). Valid when the same current passes through every resistor in turn. Consequences: the total is always LARGER than the largest individual resistor; the current is identical in all of them; the pd divides in direct proportion to resistance, V_n = I R_n; the total pd is the sum of the individual pds. Common misuse: adding resistances of components that are actually in parallel because the circuit diagram is drawn with awkward geometry — always trace the junctions. Sanity check: 4.0 Ω and 6.0 Ω in series give 10.0 Ω, and across 20 V the current is 2.0 A with 8.0 V and 12 V across them.

23. EquationSL & HLData booklet: Yes

    State the rule for combining resistors in parallel and give the consequences.

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    1/R_parallel = 1/R₁ + 1/R₂ + 1/R₃ + … (data booklet); each R in ohms (Ω). Valid when the same pd is across every resistor. Consequences: the total is always SMALLER than the smallest individual resistance; adding another parallel branch always DECREASES total resistance and increases the total current; the branch currents divide in inverse proportion to resistance. For exactly two resistors R = R₁R₂/(R₁ + R₂); for N identical resistors R = R₁/N. Common misuse: forgetting to invert at the end — the single most frequent arithmetic slip in Paper 1. Sanity check: 4.0 Ω and 6.0 Ω in parallel give 24/10 = 2.4 Ω, which is less than 4.0 Ω.

24. EquationSL & HLData booklet: Yes

    State the emf equation for a cell with internal resistance and its rearrangements.

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    ε = I(R + r) (data booklet). ε = emf in volts (V); I = current in amperes (A); R = total external (load) resistance in ohms (Ω); r = internal resistance in ohms (Ω). Valid for a single-loop circuit; it is Kirchhoff's loop rule written out. Rearranged: I = ε/(R + r), and the terminal pd V = IR = ε − Ir. Maximum (short-circuit) current is ε/r when R = 0. Common misuse: using ε in place of the terminal pd when calculating power dissipated in the external resistor. Sanity check: ε = 9.0 V, R = 7.0 Ω, r = 2.0 Ω gives I = 1.0 A, terminal pd 7.0 V and lost volts 2.0 V.

25. EquationSL & HLData booklet: Yes

    State the terminal potential difference equation and describe how it is used to find ε and r experimentally.

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    V = ε − Ir (data booklet form ε = I(R + r) rearranged). V = terminal pd in volts (V); ε = emf in volts (V); I = current drawn in amperes (A); r = internal resistance in ohms (Ω). Valid for a cell of constant ε and r. Written as V = −rI + ε it is y = mx + c, so plotting V (y-axis) against I (x-axis) gives a straight line of gradient −r and V-intercept ε; the I-intercept is the short-circuit current ε/r. Common misuse: quoting r as a negative number because the gradient is negative — r is the MAGNITUDE of the gradient. Sanity check: intercept 1.5 V, gradient −0.50 V A⁻¹ gives ε = 1.5 V, r = 0.50 Ω.

26. EquationSL & HLData booklet: No – derive

    State the potential divider equation and its conditions of validity.

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    V_out = V_in × R₂/(R₁ + R₂), where V_out is taken across R₂. Not printed in the data booklet — derive it from V = IR with I = V_in/(R₁ + R₂), so it counts as "No – derive". V_in, V_out in volts (V); R₁, R₂ in ohms (Ω). Valid only if negligible current is drawn from the output (load resistance ≫ R₂); otherwise R₂ must first be combined in parallel with the load. Common misuse: applying it after connecting a low-resistance load across the output, which lowers V_out substantially. Sanity check: 12 V across 3.0 kΩ and 1.0 kΩ gives 3.0 V across the 1.0 kΩ resistor.

27. EquationSL & HLData booklet: No – memorise

    State the equations for electrical energy transfer including the kilowatt-hour.

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    E = Pt = IVt = I²Rt = V²t/R, all following from P = IV (booklet). E = energy in joules (J); P = power in watts (W); t = time in seconds (s). For domestic use E (in kW h) = P (in kW) × t (in hours), with 1 kW h = 3.6 × 10⁶ J — this conversion is "No – memorise". Valid for constant power; otherwise E is the area under a power–time graph. Common misuse: mixing units, e.g. using watts with hours, or seconds with kilowatts. Sanity check: a 60 W lamp left on for 24 h uses 1.44 kW h = 5.2 × 10⁶ J.

28. EquationSL & HLData booklet: No – derive

    State the condition for maximum power transfer to an external load and how it is obtained.

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    The power delivered to the external resistance is P = I²R = ε²R/(R + r)², which is a maximum when R = r. Not in the data booklet — treat as "No – derive" (differentiate P with respect to R, or read the maximum from a P–R graph). Symbols: P in watts (W); ε in volts (V); R external resistance in ohms (Ω); r internal resistance in ohms (Ω). At that point the maximum power is P_max = ε²/(4r) and the efficiency is only 50%, since equal power is wasted inside the cell. Common misuse: assuming maximum power means maximum efficiency. Sanity check: ε = 6.0 V, r = 3.0 Ω gives P_max = 36/12 = 3.0 W at R = 3.0 Ω.

29. Graph/diagramSL & HL

    Describe the I–V graph of an ohmic conductor (metal wire at constant temperature) and what can be extracted from it.

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    Axes: current I/A on the y-axis against potential difference V/V on the x-axis (IB convention; the reverse is also accepted if labelled). Shape: a straight line through the origin, extending into the third quadrant with the same gradient when the pd is reversed. The gradient is 1/R, so R = 1/gradient; the reciprocal of the chord from the origin gives the same value at every point, which is the graphical statement of Ohm's law. Area under the graph has no useful meaning here. Changing a parameter: increasing the wire's length or reducing its area increases R, so the line becomes LESS steep; raising the temperature also reduces the gradient. Non-zero intercepts indicate a systematic (zero) error in a meter.

30. Graph/diagramSL & HL

    Describe the I–V characteristic of a filament lamp and explain its shape.

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    Axes: I/A against V/V. Shape: a curve through the origin that is initially straight then bends TOWARDS the V axis (gradient decreasing) as V increases, and is symmetric about the origin for reversed pd. Interpretation: R = V/I is the reciprocal of the chord gradient from the origin, and it increases with V because the filament's temperature rises, lattice ions vibrate with greater amplitude, electron–lattice collisions become more frequent and the drift speed for a given field falls. Extraction: pick a point, read V and I, compute R = V/I — never use the tangent. Change of parameter: a lamp of higher power rating carries more current at each V, so its curve lies above.

31. Graph/diagramSL & HL

    Describe the I–V characteristic of a semiconductor diode and how to read the threshold voltage.

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    Axes: I/A (often mA) against V/V, showing both forward and reverse bias. Shape: in forward bias the current is essentially zero until about 0.6–0.7 V, then rises very steeply and almost vertically; in reverse bias the current is negligible (a few μA) until breakdown, so the curve is strongly ASYMMETRIC about the origin. Extraction: the threshold (turn-on) voltage is where the forward curve leaves the axis — extrapolate the steep section back to the V axis. Resistance is very low forward-biased, very high reverse-biased, and R = V/I differs at every point. Change of parameter: raising the temperature lowers the threshold voltage slightly and increases the current at a given V.

32. Graph/diagramSL & HL

    Describe the resistance–temperature graph of an NTC thermistor and the resistance–intensity graph of an LDR.

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    Thermistor: axes R/Ω (y) against temperature θ/°C or T/K (x). Shape: a smooth curve falling steeply at low temperature and flattening at high temperature (approximately exponential decay, never linear, never reaching zero). Reason: rising temperature releases many more charge carriers into the conduction band, and this outweighs the increased lattice vibration. LDR: axes R/Ω against light intensity/lux, again a falling curve; a log–log plot linearises it. Extraction: read R at a stated temperature or intensity, then feed it into a potential divider calculation. Change of parameter: a thermistor in the LOWER arm of a divider gives a V_out that FALLS as temperature rises — state this direction explicitly.

33. Graph/diagramSL & HL

    Describe the graph of terminal potential difference against current for a real cell and how ε and r are found from it.

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    Axes: terminal pd V/V on the y-axis against current I/A on the x-axis, obtained by varying a rheostat across the cell. Shape: a straight line of NEGATIVE gradient, since V = −rI + ε. Extraction: the y-intercept is the emf ε (the terminal pd at zero current); the magnitude of the gradient is the internal resistance r; the x-intercept is the maximum short-circuit current ε/r. Uncertainty work: draw maximum and minimum gradient lines through the error bars and take r = (r_max − r_min)/2 as the absolute uncertainty. Change of parameter: an older cell has larger r, so the line is steeper while the intercept ε is almost unchanged — a very common "explain the difference" question.

34. Graph/diagramSL & HL

    Describe the linearised graphs used in the practical to determine the resistivity of a wire.

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    Primary graph: R/Ω (y) against length L/m (x), with the wire's cross-section constant. Shape: a straight line through the origin, since R = (ρ/A)L. Gradient = ρ/A, so ρ = gradient × A, where A = πd²/4 from a micrometer measurement of the diameter (take several readings at different points and orientations). A non-zero intercept reveals contact/lead resistance — a systematic error. Alternative linearisation: plot R against 1/A for fixed L, giving gradient ρL; or R against L/A, giving gradient ρ directly. Change of parameter: using a wire of higher resistivity (nichrome rather than copper) increases the gradient; allowing the wire to heat up raises all R values, curving the line upward at large currents.

35. Graph/diagramSL & HL

    Describe the graph of power dissipated in the external load against external resistance for a cell with internal resistance.

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    Axes: power in the load P/W (y) against external resistance R/Ω (x), with ε and r fixed. Shape: starts at P = 0 when R = 0 (all power lost internally), rises to a single maximum at R = r, then falls asymptotically towards zero for large R. Extraction: the R value at the peak equals the internal resistance r, and the peak height is P_max = ε²/(4r), which gives ε once r is known. Note the curve is markedly asymmetric — it falls away slowly on the high-R side. Change of parameter: doubling ε multiplies every P value by four without moving the peak position; doubling r moves the peak to larger R and halves its height.

36. Graph/diagramSL & HL

    Describe how the output voltage of a potential divider varies as the sliding contact of a potentiometer is moved, and contrast it with a series rheostat.

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    Axes: V_out/V (y) against fraction x of the track between the slider and the lower end (x from 0 to 1). Shape with negligible load: a straight line from 0 V to the full supply V_in, gradient V_in, because V_out = xV_in. With a finite load resistance the line SAGS below the ideal straight line, the droop being greatest around the middle of the track and vanishing at both ends (V_out is 0 at x = 0 and V_in at x = 1 whatever the load), since the load lowers the effective resistance of the lower arm. Extraction: the intercepts show the full available range — a potentiometer can reach 0 V, whereas a rheostat in series with the load cannot: its output only falls from V_in to V_in R_load/(R_load + R_max). This limited range is the standard reason for preferring the potential-divider arrangement.

37. Concept/explainSL & HL

    Explain, in terms of the free-electron model, what electrical resistance is and why the resistance of a metal wire increases as its temperature rises.

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    • A metal contains a lattice of positive ions and a sea of delocalised free electrons in random thermal motion
    • An applied pd sets up an electric field in the wire that superimposes a slow drift on this random motion, giving a current
    • Resistance arises because drifting electrons collide with (are scattered by) the vibrating lattice ions, transferring kinetic energy to the lattice as internal energy
    • Raising the temperature increases the amplitude and frequency of lattice vibrations, so collisions are more frequent, the mean time between collisions falls and the drift speed for a given field decreases
    • Hence I falls for the same V and R = V/I increases
    • Real-world: this is why a filament lamp is non-ohmic and why platinum resistance thermometers work. Exam tip: incomplete answers say "the electrons hit the atoms harder" or "the atoms get bigger"; the marking point is more frequent scattering reducing drift speed, not larger ions.
38. Concept/explainSL & HL

    Distinguish conventional current from electron flow and explain why the historical convention is retained.

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    • Conventional current is defined as the direction of flow of positive charge, i.e. from the positive terminal round the external circuit to the negative terminal
    • In a metallic conductor the actual mobile carriers are electrons, which drift in the opposite direction, from negative to positive
    • The convention was fixed by Franklin before the electron was discovered, and all circuit rules, the left-hand/right-hand rules and F = BIL are consistent with it, so it is kept
    • Current I = Δq/Δt is a scalar with an assigned direction, unit ampere (A = C s⁻¹)
    • In electrolytes and plasmas both positive and negative carriers move, and the conventional current is the net rate of positive charge transfer. Exam tip: students lose the mark by writing "current flows the wrong way"; state that conventional current direction is a definition, not an error.
39. Concept/explainSL & HL

    Explain why a lamp lights essentially instantly when the switch is closed even though the drift speed of the electrons is less than a millimetre per second.

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    • Drift speed from I = nAvq is typically ~10⁻⁴ m s⁻¹ because the free-electron number density n in a metal is enormous (~10²⁸–10²⁹ m⁻³)
    • Closing the switch establishes an electric field along the whole circuit at a speed approaching c, so all the free electrons everywhere in the circuit begin to drift almost simultaneously
    • The conductor already contains mobile electrons throughout its length, so electrons in the filament start moving at once; no electron has to travel from the cell to the lamp
    • Energy is transferred by the field, not carried bodily by individual electrons
    • Hence the delay is set by the field propagation time, not the drift time. Exam tip: the common wrong answer is "electrons travel very fast"; drift speed is small and must be distinguished from field propagation speed and from random thermal speed (~10⁵ m s⁻¹).
40. Concept/explainSL & HL

    Explain the shape of the I–V characteristic of a filament lamp and why it is described as non-ohmic.

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    • At low pd the graph is almost linear through the origin: the filament is cool and R is roughly constant
    • As V increases the power dissipated P = IV heats the filament to over 2000 K
    • Increased lattice vibration scatters electrons more, so resistivity and therefore resistance increase
    • The gradient of the I–V curve therefore decreases: the curve bends towards the V axis
    • R = V/I at any point is still defined and is given by the reciprocal of the gradient of the chord from the origin, not by the tangent
    • The component is non-ohmic because I is not proportional to V at constant physical conditions — here the temperature is not constant
    • The curve is symmetric about the origin because the filament behaves the same for either current direction. Exam tip: many answers say "Ohm's law is broken"; say instead that Ohm's law applies only at constant temperature, which is not satisfied here.
41. Concept/explainSL & HL

    Explain how the resistance of an NTC thermistor and of an LDR change with the external conditions, and outline one sensing use of each.

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    • Both are semiconducting devices in which the number density n of charge carriers is not fixed
    • In an NTC thermistor, raising the temperature gives more electrons enough energy to be released into the conduction band, so n rises sharply and resistance falls, typically from kilohms to tens of ohms
    • This dominates over the increased lattice scattering that would raise R in a metal
    • In an LDR, increasing light intensity delivers more photons, releasing more charge carriers, so n rises and resistance falls (from ~MΩ in darkness to ~hundreds of Ω in bright light)
    • Both are non-ohmic: I is not proportional to V
    • Use: thermistor in a potential-divider thermostat or car coolant sensor; LDR in an automatic street-lamp or camera exposure meter. Exam tip: state the carrier-density mechanism, not just "resistance goes down".
42. Concept/explainSL & HL

    Explain the I–V characteristic of a semiconductor diode and how a diode is used to protect a circuit.

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    • A diode conducts appreciably in one direction only
    • In forward bias almost no current flows until the pd exceeds the threshold (about 0.6–0.7 V for silicon), after which the current rises very steeply and the resistance becomes very small
    • In reverse bias the current is negligible (microamps) and the resistance is extremely high, until breakdown at large reverse pd
    • The characteristic is therefore strongly asymmetric about the origin and the device is non-ohmic
    • A series diode blocks current if the supply is connected the wrong way round, protecting polarity-sensitive components; diodes also rectify AC
    • An LED is a diode that emits light when forward biased and needs a series protective resistor. Exam tip: students often sketch a straight line through the origin for the forward region — the steep curve must not pass linearly through the origin, and the reverse branch must lie along the V axis.
43. Concept/explainSL & HL

    Explain the difference between the emf of a cell and the potential difference across its terminals, and account for the difference using internal resistance.

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    • Emf ε is the total energy transferred from chemical to electrical form per unit charge driven round the complete circuit, unit volt (J C⁻¹)
    • Potential difference is the energy transferred from electrical to other forms per unit charge between two points
    • A real cell has internal resistance r, so some energy per unit charge, Ir, is dissipated inside the cell as internal energy
    • Hence terminal pd V = ε − Ir, which is always less than ε when current flows
    • V = ε only when I = 0, i.e. on open circuit or measured with an ideal voltmeter
    • Under short circuit R → 0 and I → ε/r, the maximum current, and the cell heats up
    • Energy conservation: εI = I²R + I²r. Exam tip: writing "emf is the voltage of the battery" gains nothing; emf must be defined as energy per unit charge supplied by the source.
44. Concept/explainSL & HL

    State Kirchhoff's two circuit laws and explain the conservation principle underlying each.

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    • Junction (first) law: the sum of the currents entering any junction equals the sum leaving, ΣI_in = ΣI_out
    • This follows from conservation of electric charge — charge does not accumulate at a point in a steady current, so charge per second in equals charge per second out
    • Loop (second) law: around any closed loop the sum of the emfs equals the sum of the pd drops, Σε = ΣIR
    • This follows from conservation of energy — a unit charge taken once round a loop returns to the same potential, so energy gained from sources equals energy dissipated
    • Signs matter: assign a loop direction and treat pd as negative when traversing a resistor with the current
    • Internal resistance appears in the loop equation as a drop Ir. Exam tip: quoting the laws without naming charge conservation and energy conservation loses the explain mark.
45. Concept/explainSL & HL

    Explain why components connected in series carry the same current while components in parallel have the same potential difference across them.

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    • In series there is only one conducting path, so by charge conservation (Kirchhoff's first law) the same rate of charge flow must pass through every component: I is identical
    • The pds add: ε = V₁ + V₂ + …, giving R_total = R₁ + R₂ + … so total resistance is always larger than any single resistor
    • In parallel each branch is connected between the same two junctions, so each has the same pair of potentials and therefore the same pd
    • Currents divide in inverse proportion to branch resistance, I_total = I₁ + I₂ + …, giving 1/R = 1/R₁ + 1/R₂ + …
    • Adding a parallel branch provides an extra path, lowering total resistance below the smallest branch and increasing total current drawn. Exam tip: the frequent error is "current is used up in the first lamp" — current is not consumed, energy is.
46. Concept/explainSL & HL

    Explain how a potential divider works and describe the effect of connecting a load across the output.

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    • Two resistors in series across a supply share the pd in proportion to their resistances
    • The output taken across R₂ is V_out = V_in R₂/(R₁ + R₂), because the same current flows through both
    • Replacing one resistor with a thermistor, LDR or variable resistor makes V_out vary with temperature, light or a slider position, so the divider acts as a sensing or control circuit
    • Connecting a load of resistance R_L across R₂ places R_L in parallel with R₂, reducing that combined resistance, so V_out falls below the unloaded value
    • The loading effect is negligible only if R_L is much larger than R₂
    • A potentiometer (continuously variable divider) can give any output from 0 to V_in, unlike a series rheostat. Exam tip: many students forget the parallel combination and quote the unloaded formula when a voltmeter or lamp is attached.
47. Concept/explainSL & HL

    Explain why an ideal ammeter has zero resistance and an ideal voltmeter infinite resistance, and describe the systematic errors caused by real meters.

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    • An ammeter is connected in series, so any resistance it has adds to the circuit resistance, reducing the current it is meant to measure; ideally R_A = 0 so the pd across it is zero
    • A voltmeter is connected in parallel with a component, so it provides an alternative path; unless its resistance is very large it draws current, reduces the combined resistance of that section and lowers the pd being measured
    • Both effects are systematic: the readings are consistently too low, and repeating readings does not remove them
    • The errors are worst when the ammeter resistance is comparable with the circuit resistance, or when the voltmeter resistance is comparable with the resistor it is across
    • Digital meters approach ideal behaviour (R_V ~ 10 MΩ). Exam tip: state "systematic, reading too low", not merely "the meter affects the circuit".
48. Concept/explainSL & HL

    Explain why resistivity, not resistance, is quoted as a property of a material, and deduce what happens to the resistance of a wire that is stretched.

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    • Resistance depends on the sample: R = ρL/A, so it grows with length and falls with cross-sectional area
    • Resistivity ρ = RA/L is independent of the shape and size of the specimen and depends only on the material and its temperature, unit Ω m
    • It therefore allows fair comparison of conductors (copper ~1.7 × 10⁻⁸ Ω m) with alloys (nichrome ~1.1 × 10⁻⁶ Ω m) and insulators (~10¹⁵ Ω m)
    • Stretching a wire conserves its volume V = AL, so if L is multiplied by k then A is divided by k
    • Hence R = ρL/A is multiplied by k², e.g. doubling the length quadruples the resistance
    • Resistivity itself is unchanged by stretching. Exam tip: the common error is to multiply only by k, forgetting that A also changes.
49. Concept/explainSL & HL

    Explain why the power delivered to an external resistor by a real cell is a maximum when the external resistance equals the internal resistance.

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    • For a cell of emf ε and internal resistance r driving external resistance R, I = ε/(R + r)
    • Power in the load P = I²R = ε²R/(R + r)²
    • When R ≪ r the current is large but almost all the energy is dissipated inside the cell, so P is small
    • When R ≫ r the terminal pd approaches ε but the current is tiny, so P is again small
    • Between these limits P passes through a maximum at R = r, where P_max = ε²/4r and the efficiency is only 50%
    • The P against R curve rises steeply, peaks at R = r and falls slowly
    • Real-world: loudspeaker and aerial impedance matching uses this result, while power stations deliberately work far from it to keep efficiency high. Exam tip: do not confuse maximum power transfer with maximum efficiency.
50. Concept/explainSL & HL

    Explain why electrical energy for domestic use is measured in kilowatt-hours and relate the kilowatt-hour to the joule.

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    • Power is the rate of energy transfer, P = IV, so energy transferred is E = Pt
    • One kilowatt-hour is the energy transferred by a device of power 1 kW operating for 1 hour
    • 1 kW h = 1000 W × 3600 s = 3.6 × 10⁶ J
    • The joule is inconveniently small for household use — a 2 kW heater running for an evening transfers tens of megajoules — so the kW h is used as a practical commercial unit
    • Cost = number of kW h × price per kW h
    • The kW h is a unit of energy, not power, despite containing "kilowatt"
    • Real-world link: energy meters record kW h, and efficiency labelling uses kW h per year. Exam tip: students frequently call the kW h a unit of power, or divide by 3600 instead of multiplying; check the answer is larger in joules than in kW h.
51. Worked problemSL & HLData booklet: Yes

    A current of 0.25 A passes through a lamp for 2.0 minutes. Determine the charge that flows and the number of electrons that pass a point in the filament.

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    Principle: I = Δq/Δt so Δq = IΔt. Substitution: Δt = 2.0 × 60 = 120 s; Δq = 0.25 A × 120 s = 30 C. Number of electrons: N = Δq/e = 30 C / (1.60 × 10⁻¹⁹ C) = 1.875 × 10²⁰. Final answers: charge = 30 C (2 s.f.); N = 1.9 × 10²⁰ electrons (2 s.f., no unit — it is a pure number). Check/Trap: convert minutes to seconds before substituting — using t = 2.0 gives 0.50 C, a factor of 60 too small. A coulomb is an ampere-second, so the units confirm the working. Do not attach the unit C to N, and do not write the electron charge as negative here; only the magnitude is needed for counting.

52. Worked problemSL & HLData booklet: Yes

    A copper wire of cross-sectional area 1.0 mm² carries a current of 3.0 A. The free-electron number density of copper is 8.5 × 10²⁸ m⁻³. Calculate the drift speed of the electrons.

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    Principle: I = nAvq, so v = I/(nAq). Substitution: A = 1.0 mm² = 1.0 × 10⁻⁶ m²; v = 3.0 / (8.5 × 10²⁸ m⁻³ × 1.0 × 10⁻⁶ m² × 1.60 × 10⁻¹⁹ C). Intermediate: denominator = 8.5 × 10²⁸ × 1.6 × 10⁻²⁵ = 1.36 × 10⁴ C m⁻¹. v = 3.0/1.36 × 10⁴ = 2.2 × 10⁻⁴ m s⁻¹ (2 s.f.), about 0.22 mm s⁻¹. Check/Trap: the area conversion is the usual killer — 1 mm² = 10⁻⁶ m², not 10⁻³ m². The tiny answer is physically correct; it does not contradict the lamp lighting immediately, because the electric field is established through the circuit at close to the speed of light and electrons are already present throughout the wire.

53. Worked problemSL & HLData booklet: Yes

    A nichrome wire of resistivity 1.1 × 10⁻⁶ Ω m has length 0.80 m and diameter 0.40 mm. Determine its resistance.

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    Principle: ρ = RA/L, so R = ρL/A, with A = πd²/4 for a circular cross-section. Substitution: d = 0.40 mm = 4.0 × 10⁻⁴ m; A = π(4.0 × 10⁻⁴)²/4 = π × 1.6 × 10⁻⁷/4 = 1.26 × 10⁻⁷ m². R = (1.1 × 10⁻⁶ Ω m × 0.80 m)/(1.26 × 10⁻⁷ m²) = 8.8 × 10⁻⁷ / 1.26 × 10⁻⁷. Final answer: R = 7.0 Ω (2 s.f.). Check/Trap: the commonest error is substituting the diameter as if it were the radius, which makes A four times too large and R four times too small (1.8 Ω). Convert millimetres to metres before squaring. Units cancel correctly: (Ω m × m)/m² = Ω.

54. Worked problemSL & HLData booklet: Yes

    A 12 Ω resistor and a 6.0 Ω resistor are connected in parallel, and this combination is in series with a 2.0 Ω resistor across a supply of negligible internal resistance and emf 12 V. Determine the total resistance, the total current and the current in each parallel branch.

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    Principle: 1/R_p = 1/R₁ + 1/R₂, then R_total = R_p + R_series, then I = V/R. Parallel: 1/R_p = 1/12 + 1/6.0 = 1/12 + 2/12 = 3/12, so R_p = 4.0 Ω. Total: R = 4.0 + 2.0 = 6.0 Ω. Total current: I = 12 V / 6.0 Ω = 2.0 A. Pd across the 2.0 Ω resistor = 2.0 × 2.0 = 4.0 V, so pd across the parallel section = 12 − 4.0 = 8.0 V. Branch currents: I₁ = 8.0/12 = 0.67 A; I₂ = 8.0/6.0 = 1.3 A. Check/Trap: 0.67 + 1.3 = 2.0 A, agreeing with Kirchhoff's first law. Never add parallel resistances directly (12 + 6 = 18 Ω is wrong) and remember to invert 3/12 at the end.

55. Worked problemSL & HLData booklet: Yes

    A cell delivers 1.2 A when connected to a 4.0 Ω resistor and 0.60 A when connected to a 9.0 Ω resistor. Determine the emf and the internal resistance of the cell, and the terminal pd in the first case.

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    Principle: ε = I(R + r) for both cases, then solve simultaneously. Equations: ε = 1.2(4.0 + r) and ε = 0.60(9.0 + r). Setting equal: 4.8 + 1.2r = 5.4 + 0.60r, so 0.60r = 0.60 and r = 1.0 Ω. Substituting back: ε = 1.2 × (4.0 + 1.0) = 6.0 V. Terminal pd in the first case: V = ε − Ir = 6.0 − (1.2 × 1.0) = 4.8 V, which equals IR = 1.2 × 4.0 = 4.8 V as a check. Final answers: ε = 6.0 V, r = 1.0 Ω, V = 4.8 V. Check/Trap: the terminal pd is always less than the emf when current flows; a bigger current gives a bigger lost volts Ir. Do not equate ε to the voltmeter reading across the external resistor.

56. Worked problemSL & HLData booklet: No – derive

    In a potential divider a 8.0 kΩ resistor and a 4.0 kΩ resistor are in series across a 12 V supply, with the output taken across the 4.0 kΩ resistor. Determine the output pd, and then the output when a 4.0 kΩ load is connected across the output terminals.

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    Principle: V_out = V_in R₂/(R₁ + R₂). Unloaded: V_out = 12 × 4.0/(8.0 + 4.0) = 12 × 4.0/12 = 4.0 V. Loaded: the 4.0 kΩ load is in parallel with the 4.0 kΩ output resistor, giving 1/R = 1/4.0 + 1/4.0 so R = 2.0 kΩ. New V_out = 12 × 2.0/(8.0 + 2.0) = 12 × 2.0/10 = 2.4 V. Final answers: 4.0 V unloaded, 2.4 V loaded. Check/Trap: the load always reduces the output — quoting 4.0 V for the loaded case is the standard lost mark. Units of kΩ cancel in the ratio, so there is no need to convert to ohms. If the load were 400 kΩ the drop would be negligible, which is why voltmeters must have high resistance.

57. Worked problemSL & HLData booklet: No – derive

    A 9.0 V supply is connected in series with a 2.0 kΩ fixed resistor and a thermistor whose resistance is 4.0 kΩ at 20 °C and 1.0 kΩ at 60 °C. The output is taken across the fixed resistor. Determine the output pd at each temperature.

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    Principle: series circuit, V_out = ε R_fixed/(R_fixed + R_thermistor), assuming negligible internal resistance. At 20 °C: V_out = 9.0 × 2.0/(2.0 + 4.0) = 9.0 × 2.0/6.0 = 3.0 V. At 60 °C: V_out = 9.0 × 2.0/(2.0 + 1.0) = 9.0 × 2.0/3.0 = 6.0 V. Final answers: 3.0 V at 20 °C and 6.0 V at 60 °C, so the output rises as the temperature rises. Check/Trap: identify which component the output is across — if it were taken across the NTC thermistor instead, the output would fall from 6.0 V to 3.0 V as the temperature rose. Check the two pds in each case add to 9.0 V. An NTC thermistor's resistance decreases with rising temperature.

58. Worked problemSL & HLData booklet: Yes

    A lamp is rated "12 V, 36 W". Determine the current in it and its resistance at normal operating conditions, and comment on the power dissipated if it is instead connected to a 6.0 V supply.

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    Principle: P = IV and P = V²/R. Current: I = P/V = 36 W / 12 V = 3.0 A. Resistance: R = V²/P = (12)²/36 = 144/36 = 4.0 Ω (or R = V/I = 12/3.0 = 4.0 Ω). At 6.0 V, if R stayed at 4.0 Ω then P = V²/R = 36/4.0 = 9.0 W, one quarter of the rated power. In practice the filament runs cooler, so its resistance is less than 4.0 Ω and the actual power is somewhat greater than 9.0 W. Final answers: 3.0 A, 4.0 Ω, and P > 9.0 W at 6.0 V. Check/Trap: 4.0 Ω is the hot resistance only; a cold filament may measure well under 1 Ω, which is why lamps often fail at switch-on. Halving V quarters P only for an ohmic conductor.

59. Worked problemSL & HLData booklet: Yes

    Two lamps rated "12 V, 24 W" and "12 V, 6.0 W" are connected in series across a 12 V supply of negligible internal resistance. Assuming their resistances stay at their rated values, deduce which lamp is brighter.

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    Principle: find each rated resistance from R = V²/P, then use the fact that the current is common in series and P = I²R. Resistances: R₁ = 144/24 = 6.0 Ω; R₂ = 144/6.0 = 24 Ω. Total R = 30 Ω, so I = 12/30 = 0.40 A. Powers: P₁ = I²R₁ = (0.40)² × 6.0 = 0.96 W; P₂ = I²R₂ = 0.16 × 24 = 3.8 W. Deduction: the lamp rated 6.0 W dissipates the greater power and is the brighter one. Check/Trap: the powers sum to 4.8 W = IV = 0.40 × 12, confirming the arithmetic. The trap is to assume the higher-rated lamp is always brighter — in series the same current flows, so P = I²R means the larger resistance (lower power rating) gets the larger share of the pd and the power.

60. Worked problemSL & HLData booklet: No – memorise

    An electric heater rated 2.5 kW is used for 3.0 hours each day for 30 days. Determine the energy transferred in kW h and in joules, and the cost at $0.18 per kW h.

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    Principle: E = Pt with P in kW and t in hours gives energy in kW h. Time: 3.0 h/day × 30 days = 90 h. Energy: E = 2.5 kW × 90 h = 225 kW h ≈ 2.3 × 10² kW h (2 s.f.). In joules: 1 kW h = 1000 W × 3600 s = 3.6 × 10⁶ J, so E = 225 × 3.6 × 10⁶ = 8.1 × 10⁸ J. Cost: 225 × $0.18 = $40.50 ≈ $41. Check/Trap: multiply, do not divide, by 3.6 × 10⁶ when converting kW h to J — the joule is the smaller unit so the number must get bigger. Keep power in kilowatts and time in hours for the kW h calculation; mixing watts with hours gives 2.25 × 10⁵ "units", which is meaningless.

61. Worked problemSL & HLData booklet: Yes

    A cell of emf 9.0 V and internal resistance 0.50 Ω is connected to a 4.0 Ω resistor in series with a parallel combination of 6.0 Ω and 3.0 Ω. Determine the current from the cell, the terminal pd, the current in the 3.0 Ω resistor and the power dissipated inside the cell.

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    Principle: reduce the network, then ε = I(R + r). Parallel: 1/R_p = 1/6.0 + 1/3.0 = 3/6.0, so R_p = 2.0 Ω. External R = 4.0 + 2.0 = 6.0 Ω; total = 6.5 Ω. Current: I = 9.0/6.5 = 1.3846 ≈ 1.4 A. Terminal pd: V = ε − Ir = 9.0 − (1.3846 × 0.50) = 9.0 − 0.69 = 8.3 V. Pd across the parallel section: V_p = 1.3846 × 2.0 = 2.77 V, so I₃ = 2.77/3.0 = 0.92 A (and I₆ = 0.46 A). Power in the cell: P = I²r = (1.3846)² × 0.50 = 0.96 W. Check/Trap: 0.92 + 0.46 = 1.38 A, matching the cell current. Carry unrounded values through and round only at the end; rounding I to 1.4 A too early shifts the final power by about 2%.

62. Worked problemSL & HLData booklet: Yes

    For a cell, a graph of terminal pd V against current I is a straight line with intercept 1.55 V on the V axis and passing through the point (2.0 A, 1.25 V). Determine the emf and internal resistance, and the power dissipated inside the cell at 2.0 A.

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    Principle: V = ε − Ir, so a V–I graph is a straight line of intercept ε and gradient −r. Intercept: ε = 1.55 V (the pd when I = 0, i.e. open circuit). Gradient: (1.25 − 1.55)/(2.0 − 0) = −0.30/2.0 = −0.15 V A⁻¹, so r = 0.15 Ω. Internal power at I = 2.0 A: P = I²r = (2.0)² × 0.15 = 0.60 W. Final answers: ε = 1.55 V, r = 0.15 Ω, P = 0.60 W. Check/Trap: r is the magnitude of the gradient — quoting a negative resistance loses the mark, and the negative sign only shows that V falls as I rises. Take gradient values from a large triangle on the drawn line, not from two raw data points, and state the unit V A⁻¹ = Ω.

63. Worked problemSL & HLData booklet: Yes

    A uniform wire has resistance 2.0 Ω. It is stretched, without change of volume or resistivity, until its length is three times the original. Determine the new resistance.

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    Principle: R = ρL/A with volume V = AL constant. If L → 3L then, since AL is constant, A → A/3. New resistance: R' = ρ(3L)/(A/3) = 9ρL/A = 9R. Substitution: R' = 9 × 2.0 Ω = 18 Ω. Final answer: 18 Ω (2 s.f.). Check/Trap: the standard error is to multiply only by 3, giving 6.0 Ω, because the change in cross-sectional area is forgotten. In general stretching by a factor k multiplies R by k². Resistivity ρ is a material property and does not change on stretching; only the geometry changes. A quick sanity check: the wire is longer and thinner, so both effects raise the resistance, and 18 Ω > 6.0 Ω > 2.0 Ω is consistent.

64. Exam technique/trapSL & HL

    Explain the trap in the statement "Ohm's law says V = IR", and state how to answer a question asking whether a component is ohmic.

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    The trap: V = IR is the definition of resistance (R = V/I) and applies to every component, ohmic or not. Ohm's law is the experimental statement that for a metallic conductor at constant temperature the current is directly proportional to the applied pd, so R is constant. Students fall for it because both are taught together and the symbol R appears in each. Correct approach: to decide whether a component is ohmic, check that the I–V graph is a straight line through the origin over the range in question; a filament lamp, diode, thermistor and LDR all fail this. Command-term guidance: "state" needs only the proportionality plus the constant-temperature condition; "explain" requires the microscopic reason; "deduce whether ohmic" requires you to quote evidence from the graph, e.g. "the line is not straight, so R varies, hence non-ohmic".

65. Exam technique/trapSL & HL

    Identify and correct the most common misconceptions about what happens to current and energy around a simple series circuit.

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    Trap 1: "current is used up by the lamp". It is not — Kirchhoff's first law and charge conservation mean the current is the same at every point in a single loop; what is transferred is energy. Trap 2: "the current comes out of the cell and the first component gets it first". Charge carriers are already present throughout the circuit and all begin to drift together. Trap 3: "the cell supplies constant current". A cell supplies a nearly constant emf; the current depends on the total resistance. Trap 4: "adding a resistor in parallel increases total resistance". It provides another path, so total resistance falls and the cell current rises. Why students fall for it: everyday language about "using electricity". Correct approach: argue explicitly from charge conservation for current and energy conservation for pd, and quote the relevant Kirchhoff law by name.

66. Exam technique/trapSL & HL

    Outline how to interpret the command terms used in circuit questions and what each demands in an answer.

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    "State" — a bare answer, no working, e.g. "the current is unchanged". "Outline" — a brief account of the main points. "Describe" — say what happens, without needing the reason. "Explain" — give the physics reason, usually a named law or mechanism. "Determine" or "calculate" — full working with substitution, unit and appropriate significant figures; a bare answer scores at most one mark. "Deduce" — reach a conclusion and justify it from given data or a graph. "Sketch" — draw a shape with labelled axes, correct intercepts and asymptotes but no accurate plotting. "Suggest" — propose a plausible reason for an unfamiliar situation. "Compare" — treat both cases and use comparative language ("larger than", "the same as"). Common loss: answering "determine" with a value only, or answering "compare" by describing one case then the other with no linking statement.

67. Exam technique/trapSL & HL

    Explain how to handle uncertainties when a resistivity is calculated from measurements of resistance, length and wire diameter.

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    Use ρ = RA/L = πRd²/4L, so fractional uncertainties combine as Δρ/ρ = ΔR/R + 2Δd/d + ΔL/L, with the diameter contribution doubled because d is squared. This doubling is the point most often missed. Absolute uncertainty comes last: Δρ = ρ × (fractional total), quoted to one significant figure with ρ rounded to the same decimal place. Micrometer readings of d are the dominant term because d is small, so measure the diameter at several places and in perpendicular directions and average, to reduce random error and detect a non-circular wire. Add ABSOLUTE uncertainties for quantities that are added or subtracted and add FRACTIONAL uncertainties for products and quotients; uncertainties are never multiplied together. Also state whether a zero error exists on the micrometer — that is systematic and must be subtracted, not treated as an uncertainty.

68. Exam technique/trapSL & HL

    Describe an experiment to determine the resistivity of a metal wire, including variables, graph, limitations and improvements.

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    Apparatus: reel of the wire, metre rule, micrometer screw gauge, ammeter, voltmeter, cell, switch, crocodile-clip flying lead. Method: measure the diameter at several points with the micrometer (check zero error) and calculate the mean area A = πd²/4. Connect a measured length L into the circuit, close the switch briefly, record V and I, and obtain R = V/I. Repeat for at least six lengths. Independent variable L, dependent R, controlled: same wire, same diameter, constant temperature (small current, switch off between readings). Graph: R against L, a straight line through the origin of gradient ρ/A, so ρ = gradient × A. Limitations: heating raises R; contact resistance at the clips adds a systematic positive intercept; the wire may be non-uniform. Improvements: use low current and switch off between readings, clean the contacts, average several diameters, use a longer wire to reduce the percentage uncertainty in L.

69. Exam technique/trapSL & HL

    Describe how to determine the emf and internal resistance of a cell experimentally, and evaluate the main sources of error.

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    Apparatus: cell, variable resistor (rheostat), ammeter in series, voltmeter across the cell terminals, switch. Method: vary the rheostat to obtain at least six pairs of readings of I and terminal pd V over a wide range; close the switch only while reading. Independent variable I, dependent V, controlled: same cell, readings taken quickly to keep the cell temperature and state of charge constant. Analysis: V = ε − Ir, so plot V (y axis) against I (x axis); the intercept on the V axis is ε and the magnitude of the gradient is r. Draw maximum and minimum gradient lines through the error bars to obtain uncertainties in both ε and r. Errors: the cell warms and its emf falls if large currents are drawn for too long (systematic drift); the voltmeter resistance is not infinite, so it draws a small current that the ammeter does not register, making I too small and the gradient (hence r) slightly too large. Improvement: use a digital voltmeter, keep currents modest, and use a fresh cell.

70. Exam technique/trapSL & HL

    Explain the difference between systematic and random errors in circuit measurements, and identify which category typical meter faults fall into.

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    Random errors scatter readings either side of the true value, are shown by the spread of repeats, and are reduced by repeating and averaging or by drawing a best-fit line through many points; examples are judging a needle position on an analogue scale and fluctuating contact resistance. Systematic errors shift every reading in the same direction and are not reduced by averaging; examples are a zero error on an ammeter or micrometer, the loading effect of a non-ideal voltmeter (readings always low), the extra resistance of a real ammeter (current always low), and heating of a cell during a run. On a graph, a systematic error usually shows as an unexpected intercept while the gradient may still be correct. Precision refers to the spread of repeats, accuracy to closeness to the true value; a set of readings can be precise but inaccurate if a zero error is present.

71. Exam technique/trapSL & HL

    Identify the common traps when a second resistor is added in parallel to a load supplied by a cell with internal resistance, and set out the correct approach.

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    The trap: students assume adding a parallel branch simply adds current without changing anything else. In fact the external resistance falls, so the current drawn from the cell rises, the lost volts Ir rise and the terminal pd V = ε − Ir falls, so the original lamp becomes slightly dimmer. They fall for it because the idealised r = 0 case, used in most textbook exercises, hides the effect. Correct approach: always reduce the external network first, calculate I = ε/(R + r), then find the terminal pd, then work back to individual branches. Related traps: forgetting to invert 1/R at the end of a parallel calculation; treating a voltmeter reading across the terminals as the emf; quoting an answer to more significant figures than the least precise datum. Finish by checking that branch currents sum to the cell current.

Practise this topic with exam-style questions: B.5 Current and circuits questions (SL) · B.5 Current and circuits questions (HL) · all flashcards

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