Electricity: A-level Physics AQA revision notes

3.5.1

Current electricity

Current, potential difference and power 3.5.1.1

Definitions

Current, $I$ : the rate of flow of charge, $I = \frac{Δ Q}{Δ t}$ , measured in amperes ( $1 A = 1 C s^{- 1}$ ).
Potential difference, $V$ : the energy transferred per unit charge between two points, $V = \frac{W}{Q}$ , measured in volts ( $1 V = 1 J C^{- 1}$ ).
Resistance, $R$ : $R = \frac{V}{I}$ , measured in ohms ( $Ω$ ).

Key results

Power dissipated: $P = I V = I^{2} R = \frac{V ^{2}}{R}$ .
Energy transferred: $W = Q V = I V t$ .
Microscopic current equation: $I = n A v q$ (charge-carrier number density $n$ , cross-sectional area $A$ , mean drift velocity $v$ , charge per carrier $q$ ).

Notes

An ideal ammeter has zero resistance and is always connected in series, so it never itself changes the current it measures; an ideal voltmeter has infinite resistance and is always connected in parallel, so it draws no current from the branch it measures across.
Conventional current is defined as flowing from + to − around the external circuit — the opposite direction to the actual drift of the negatively charged electrons that carry it in a metal.
The microscopic current equation explains a result that often feels counterintuitive: drift velocity is typically only a fraction of a millimetre per second, even though a circuit appears to respond almost instantly when a switch is closed.
The near-instant response is the near-instant propagation of the electric field through the conductor, not the physical transport speed of individual charge carriers — worth holding onto explicitly rather than treating current as 'electrons rushing along a wire'.
$I = n A v q$ shows a thicker wire (larger $A$ ) carries more current at the same drift velocity — the physical reason thicker cables are used for higher-current applications, independent of any resistance argument.
$P = I V t$ combines with $W = Q V$ as the same energy-transfer idea read two ways: power is energy transferred per second, so $W = Pt$ recovers $W = I V t$ directly.

Current–voltage characteristics 3.5.1.2

Definitions

Ohmic conductor: a component for which current is directly proportional to potential difference, provided physical conditions (notably temperature) stay constant.

Notes

A filament lamp is a common non-ohmic example — as current increases, the filament heats up, its resistance rises, and the $V$ – $I$ graph curves away from a straight line rather than staying straight.
A semiconductor diode is sharply non-ohmic in a different way: it conducts easily in forward bias but allows only a very small current in reverse bias, giving a strongly asymmetric $I$ – $V$ curve rather than a symmetric curved one.
On a curved graph, resistance at a chosen point is still $R = V / I$ using that point's own coordinates — a genuinely different quantity from the tangent gradient at that point, which instead gives the graph's local dynamic slope, not resistance.
Confusing 'gradient of the graph' with 'resistance' is a common and costly exam error on any non-ohmic $I$ – $V$ curve.

Resistivity 3.5.1.3

Definitions

Resistivity, $ρ$ : a material property (depending only on the material and, for most materials, temperature) linked to resistance by $R = \frac{ρ L}{A}$ .

Method

Required practical 5: measure the test wire's diameter with a micrometer at several points along its length and average, then calculate its cross-sectional area $A = \frac{π d ^{2}}{4}$ .
For each of several lengths $l$ between fixed contacts, use a small current (to limit self-heating) to measure $V$ and $I$ , and calculate $R = \frac{V}{I}$ .
Plot $R$ against $l$ : the gradient of the resulting straight line is $\frac{ρ}{A}$ , so $ρ = gradient \times A$ .

In practiceA wire of mean diameter 0.32 mm gives a gradient of 6.2 Ω m $^{- 1}$ for $R$ against $l$ . Find its resistivity.

$A = \frac{π ( 0.32 \times 1 0 ^{- 3} ) ^{2}}{4} = 8.04 \times 1 0^{- 8} m^{2}$
$ρ = gradient \times A = 6.2 \times 8.04 \times 1 0^{- 8} = 5.0 \times 1 0^{- 7} Ω m$ close to constantan

Notes

A longer wire has more resistance ( $R \propto L$ ); a thicker wire has less ( $R \propto 1/ A$ ) — both follow directly from $R = ρ L / A$ , and are worth stating as two separate, independent proportionalities rather than one combined rule.
Resistance depends on both the material and its geometry, while resistivity depends only on the material.
Plotting $R$ against $l$ rather than reading a single measurement averages out random error across several lengths, and any genuine nonzero intercept is worth investigating rather than dismissing — it usually signals contact or lead resistance the fixed-length setup did not eliminate, not a reason to force a best-fit line through the origin.

Temperature and resistance 3.5.1.3

Definitions

Thermistor: a resistor whose resistance changes sharply and predictably with temperature — an NTC (negative temperature coefficient) thermistor's resistance falls as temperature rises.
Superconductor: a material whose resistivity drops to exactly zero at and below a material-specific critical temperature, $T_{c}$ .

Notes

In a metal conductor, resistance rises with temperature: the ions vibrate more, scattering the charge-carrying electrons more often and more strongly.
In an NTC thermistor, resistance falls as temperature rises instead, because more charge carriers become available to carry current as temperature increases — an effect that outweighs the same increased-scattering mechanism seen in metals.
This opposite temperature dependence — resistance rising for a metal but falling for a thermistor — is exactly why thermistors, not plain metal resistors, are used as the sensing element in a temperature-sensitive potential divider.
Superconductivity is the extreme case: below $T_{c}$ , resistivity is not merely small but genuinely zero, so current can flow with no energy dissipated as heat at all — the practical motivation for using superconductors in strong-field electromagnets and for reducing resistive losses in power transmission.
Critical temperature is a property of the specific material, not a universal constant — different superconductors have widely different $T_{c}$ values.

Series and parallel circuits 3.5.1.4

Definitions

Kirchhoff's first law (conservation of charge): the total current entering a junction equals the total current leaving it.
Kirchhoff's second law (conservation of energy): the sum of EMFs around any closed loop equals the sum of potential differences around that same loop.

Key results

Series resistance: $R_{total} = R_{1} + R_{2} + \dots$ (same current through every component).
Parallel resistance: $\frac{1}{R _{total}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \dots$ (every branch shares the same potential difference).
Identical cells in series (aiding): combined emf $= n ε$ , combined internal resistance $= n r$ .
Identical cells in parallel: combined emf $= ε$ (unchanged), combined internal resistance $= \frac{r}{n}$ .

Notes

Adding a parallel branch always reduces the total resistance, because it opens an additional path for current.
Together, Kirchhoff's two laws are sufficient to solve any circuit, however complex — the first is just conservation of charge applied at a point, the second just conservation of energy applied around a loop, neither a new physical law beyond what was already introduced earlier in the course.
For a complex circuit with multiple loops and sources, a systematic approach is to assign a current direction to each branch, write one Kirchhoff's-first-law equation per junction and one Kirchhoff's-second-law equation per independent loop, then solve the resulting simultaneous equations.
Connecting identical cells in series multiplies the available emf but also the total internal resistance — useful for driving a higher-voltage load, but it does not by itself increase the maximum current the combination can supply into a very low-resistance load.
Connecting identical cells in parallel leaves the emf unchanged (every branch sits at the same potential) but reduces the combined internal resistance, letting the combination supply more current for the same terminal-voltage drop — the arrangement used when a circuit demands more current than one cell can comfortably provide.

Potential dividers 3.5.1.5

Key results

Potential divider: $V_{out} = V_{in} \times \frac{R _{2}}{R _{1} + R _{2}}$ (voltage across $R_{2}$ ).

Notes

A potential divider splits a supply voltage between two or more resistors in series in proportion to their resistance, since the same current flows through each.
Replacing one fixed resistor with a sensor (a thermistor or an LDR) turns this into a sensing circuit, where output voltage varies continuously with temperature or light level.
Which arm the sensor sits in sets the direction of the response: with the sensor as $R_{1}$ (the upper resistor), $V_{out}$ falls as the sensor's resistance rises (e.g. cooler, for a thermistor); moving the same sensor to the $R_{2}$ position exactly reverses that response.
Common error: forgetting that connecting a load (a voltmeter with finite resistance, or another circuit) across the output draws additional current and changes the ratio the divider was set up to produce — an 'ideal' voltmeter is assumed to have infinite resistance specifically to avoid this problem.
A potentiometer (a three-terminal variable resistor used as a divider) lets output voltage be swept continuously from 0 up to the full supply voltage by moving a slider, unlike a two-terminal fixed/sensor pair, which only reaches the ratio the two components' resistances allow.

EMF and internal resistance 3.5.1.6

Definitions

EMF (electromotive force): the total energy transferred to charge per unit charge by a source, including energy subsequently dissipated inside the source itself due to its internal resistance $r$ .

Key results

EMF equation: $ε = I (R + r) = I R + I r$ .
Terminal potential difference: $V = ε - I r$ .
Maximum power transfer to an external resistor occurs when $R = r$ .

Method

Required practical 6: connect a cell (of unknown emf and internal resistance) to a variable load resistor, with an ammeter in series and a voltmeter across the cell's terminals.
For each of several load-resistance settings, record the current $I$ and the terminal potential difference $V$ , disconnecting the circuit between readings to limit heating that would change $r$ .
Plot $V$ against $I$ : since $V = ε - I r$ , the graph is a straight line whose vertical intercept gives $ε$ and whose gradient gives $- r$ .

In practiceA $V$ – $I$ graph for a cell has intercept 1.52 V and gradient $- 0.80$ V A $^{- 1}$ . Find $ε$ , $r$ , and the terminal p.d. at 0.50 A.

$V = ε - I r$ a straight line in $V$ against $I$
$ε = 1.52 V, r = - gradient = 0.80 Ω$
$I = 0.50 A : V = 1.52 - 0.50 (0.80) = 1.12 V$ 0.40 V is lost inside the cell

Notes

Terminal potential difference — what you would actually measure across the terminals — is less than EMF whenever current flows, because some energy per unit charge is lost driving current through $r$ .
Terminal voltage drops as current drawn increases, which is why a battery's voltage sags under heavy load.
It only reads close to its full EMF when almost no current is drawn (for example, on an open circuit, or through a voltmeter of very high resistance).
A short circuit (external $R \approx 0$ ) draws the maximum possible current from a source, $I_{m a x} = ε / r$ — internal resistance is the only thing limiting current in this extreme case.

Worked examples

Worked example 3.5.1 · 4 marks

A phone charger delivers a constant current of $1.5 A$ to a battery for $2.0$ hours.

Calculate

(a) the charge transferred, and

(b) the number of electrons that flow, given the electron charge:

e = 1.60 \times 1 0^{- 19} C

Show worked solution

Converting time to seconds:

Δ t = 2.0 \times 3600 = 7200 s

Charge:

Δ Q = I Δ t = 1.5 \times 7200 = 1.08 \times 1 0^{4} C

Number of electrons:

n = \frac{Δ Q}{e} = \frac{1.08 \times 1 0 ^{4}}{1.60 \times 1 0 ^{- 19}} \approx 6.75 \times 1 0^{22}

Mark scheme · 4 marks

Converts the time to seconds, $Δ t = 7200 s$ 1 mark
Uses $Δ Q = I Δ t$ to find $Δ Q = 1.08 \times 1 0^{4} C$ 1 mark
Uses $n = Δ Q / e$ with the correct electron charge 1 mark
Calculates $n \approx 6.75 \times 1 0^{22}$ electrons 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.5.1 · 4 marks

A copper wire of cross-sectional area $2.0 \times 1 0^{- 6} m^{2}$ carries a current of $3.0 A$ .

The number density of free electrons in copper is $8.5 \times 1 0^{28} m^{- 3}$ and the electron charge is $1.60 \times 1 0^{- 19} C$ .

Calculate the mean drift velocity of the electrons.

Show worked solution

Rearranging $I = n A v q$ gives:

v = \frac{I}{n A q}

The denominator:

n A q = 8.5 \times 1 0^{28} \times 2.0 \times 1 0^{- 6} \times 1.60 \times 1 0^{- 19} \approx 2.72 \times 1 0^{4}

So:

v = \frac{3.0}{2.72 \times 1 0 ^{4}} \approx 1.1 \times 1 0^{- 4} m s^{- 1}

— only a fraction of a millimetre per second, as expected for drift velocity.

Mark scheme · 4 marks

Rearranges $I = n A v q$ to make $v$ the subject 1 mark
Substitutes the correct values for $n$ , $A$ and $q$ 1 mark
Calculates $n A q \approx 2.72 \times 1 0^{4}$ 1 mark
Calculates $v \approx 1.1 \times 1 0^{- 4} m s^{- 1}$ 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.5.1 · 5 marks

Three components — an ohmic resistor, a filament lamp and a semiconductor diode — are each tested on the same $I$ – $V$ apparatus.

Component A gives a straight line through the origin.

Component B gives a curve whose gradient decreases as $∣ V ∣$ increases, symmetric for positive and negative $V$ .

Component C conducts strongly only for positive $V$ , with a very small current for negative $V$ .

(a) Identify A, B and C.

(b) Explain, in terms of resistance, why component B's graph curves in this way as $V$ increases.

Show worked solution

A is the ohmic resistor (constant gradient, $I \propto V$ through the origin).

B is the filament lamp (a symmetric curve bending away from the current axis, since heating depends on $I^{2}$ regardless of current direction).

C is the semiconductor diode (strongly asymmetric, conducting mainly in forward bias).

For (b): as the current through the filament increases, the filament's temperature rises, so the ions of the lattice vibrate with greater amplitude; this increases the frequency of collisions between the vibrating ions and the charge-carrying electrons, so the filament's resistance $R = V / I$ increases.

A larger resistance means a larger increase in $V$ is needed to produce the same increase in $I$ , so the gradient of the $I$ – $V$ graph falls as $∣ V ∣$ increases — the curve bends away from the current axis, towards the voltage axis.

Mark scheme · 5 marks

Identifies A as the ohmic resistor 1 mark
Identifies B as the filament lamp 1 mark
Identifies C as the semiconductor diode 1 mark
States that increasing current heats the filament, increasing lattice ion vibration and the frequency of electron–ion collisions 1 mark
States that the resulting rise in resistance reduces the graph's gradient, so the curve bends away from the current axis 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.5.1 · 6 marks

l / m	R / Ω
0.200	1.01
0.400	1.97
0.600	2.93
0.800	3.89
1.000	4.85

A student carries out Required Practical 5 to find the resistivity of a metal wire of diameter $0.36 mm$ (measured with a micrometer).

They measure the resistance $R$ between fixed contacts for several lengths $l$ of wire, giving the results in the table.

(a) Explain why the student should take readings at several lengths and plot a graph, rather than calculate resistivity from a single length.

(b) Calculate the gradient of the best-fit line.

(d) The best-fit line does not pass through the origin.

Suggest a reason for this.

Show worked solution

(a) Plotting several lengths and drawing a line of best fit averages out random error across all the readings, giving a more reliable gradient than a single-length calculation, which would carry the full random error of just one measurement.

(b) Using the two end points: gradient:

= \frac{4.85 - 1.01}{1.000 - 0.200} = \frac{3.84}{0.800} = 4.80 Ω m^{- 1}

A = \frac{π d ^{2}}{4} = \frac{π ( 0.36 \times 1 0 ^{- 3} ) ^{2}}{4} \approx 1.02 \times 1 0^{- 7} m^{2}

Resistivity:

ρ = gradient \times A = 4.80 \times 1.02 \times 1 0^{- 7} \approx 4.90 \times 1 0^{- 7} Ω m

(d) A nonzero intercept suggests a fixed resistance not accounted for by the wire's length, such as the resistance of the connecting leads and contacts.

Mark scheme · 6 marks

Explains that repeating over several lengths and using a best-fit line averages out random error 1 mark
Calculates the gradient from two widely-spaced points, $\approx 4.80 Ω m^{- 1}$ 1 mark
Converts diameter to area, $A \approx 1.02 \times 1 0^{- 7} m^{2}$ 1 mark
Calculates $ρ = gradient \times A$ 1 mark
States $ρ \approx 4.90 \times 1 0^{- 7} Ω m$ with correct unit 1 mark
Suggests the nonzero intercept is due to resistance of the leads/contacts, not the wire itself 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.5.1 · 4 marks

An NTC thermistor and a length of copper wire are both heated from $20\,^\circ\text{C}$ to $60\,^\circ\text{C}$ .

State how the resistance of each changes, and explain the microscopic reason why their behaviour differs.

Show worked solution

The copper wire's resistance increases with temperature, while the NTC thermistor's resistance decreases with temperature.

In copper, heating increases the amplitude of vibration of the lattice ions, so charge-carrying electrons are scattered more often; this reduces their mean drift velocity for a given current, while the number density of charge carriers $n$ stays essentially constant, so overall resistance rises.

In the thermistor's semiconducting material, heating instead releases significantly more charge carriers into the material, increasing $n$ ; this increase in $n$ outweighs the same increased-scattering effect present in the metal, so overall resistance falls.

Mark scheme · 4 marks

States the copper wire's resistance increases with temperature 1 mark
States the thermistor's resistance decreases with temperature 1 mark
Explains that in copper, increased lattice ion vibration increases electron scattering while $n$ stays constant, raising resistance 1 mark
Explains that in the thermistor, heating increases the number density of charge carriers $n$ , and this outweighs the increased scattering, lowering resistance 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.5.1 · 5 marks

A $12 V$ battery of negligible internal resistance is connected to a $5.0 Ω$ resistor in series with a parallel combination of a $6.0 Ω$ resistor and a $3.0 Ω$ resistor.

Calculate

(a) the combined resistance of the parallel section,

(b) the total circuit resistance,

(d) the current through the $6.0 Ω$ resistor.

Show worked solution

(a):

\frac{1}{R _{p}} = \frac{1}{6.0} + \frac{1}{3.0} = \frac{1}{6.0} + \frac{2}{6.0} = \frac{3}{6.0}

so $R_{p} = 2.0 Ω$ . (b):

R_{total} = 5.0 + 2.0 = 7.0 Ω

(c):

I = \frac{V}{R _{total}} = \frac{12}{7.0} \approx 1.7 A

(d) The p.d. across the parallel section is:

V_{p} = I R_{p} = 1.7 \times 2.0 \approx 3.4 V

so the current through the $6.0 Ω$ resistor is:

I_{6} = \frac{V _{p}}{6.0} \approx 0.57 A

Mark scheme · 5 marks

Calculates the parallel combination $R_{p} = 2.0 Ω$ 1 mark
Calculates the total resistance $R_{total} = 7.0 Ω$ 1 mark
Calculates the total current $I \approx 1.7 A$ 1 mark
Calculates the p.d. across the parallel section $\approx 3.4 V$ 1 mark
Calculates the current through the $6.0 Ω$ resistor $\approx 0.57 A$ 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.5.1 · 3 marks

A potential divider consists of a $4.0 k Ω$ resistor $R_{1}$ and a $12 k Ω$ resistor $R_{2}$ in series across a $9.0 V$ supply, with $V_{out}$ taken across $R_{2}$ .

Calculate $V_{out}$ .

Show worked solution

V_{out} = V_{in} \times \frac{R _{2}}{R _{1} + R _{2}} = 9.0 \times \frac{12}{4.0 + 12} = 9.0 \times \frac{12}{16} = 6.75 V

Mark scheme · 3 marks

Uses the correct potential-divider formula with $R_{2}$ over $R_{1} + R_{2}$ 1 mark
Substitutes the given values correctly 1 mark
Calculates $V_{out} = 6.75 V$ 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.5.1 · 5 marks

In a light-sensing circuit, an LDR is connected as $R_{1}$ in series with a fixed $8.0 k Ω$ resistor $R_{2}$ , across a $6.0 V$ supply, with $V_{out}$ taken across $R_{2}$ .

In bright light the LDR's resistance is $2.0 k Ω$ ; in darkness it rises to $200 k Ω$ .

(a) Calculate $V_{out}$ in bright light.

(b) Calculate $V_{out}$ in darkness.

Explain, without further calculation, how the LDR and fixed resistor should be rearranged to achieve this.

Show worked solution

(a):

V_{out} = 6.0 \times \frac{8.0}{2.0 + 8.0} = 6.0 \times 0.80 = 4.8 V

(b):

V_{out} = 6.0 \times \frac{8.0}{200 + 8.0} = 6.0 \times \frac{8.0}{208} \approx 0.23 V

To make $V_{out}$ rise as light falls, the LDR should instead be placed in the $R_{2}$ position (the one $V_{out}$ is taken across): in darkness the LDR's resistance is then large compared with the fixed resistor, so it takes a larger share of $V_{in}$ , and $V_{out}$ rises as light level falls.

Mark scheme · 5 marks

Calculates $V_{out} \approx 4.8 V$ in bright light 1 mark
Calculates $V_{out} \approx 0.23 V$ in darkness 1 mark
States that in the given arrangement $V_{out}$ falls as light falls, not rises 1 mark
States the LDR must instead be placed in the $R_{2}$ position 1 mark
Explains that with the LDR as $R_{2}$ , its larger resistance in darkness takes a larger share of $V_{in}$ , so $V_{out}$ rises as light falls 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.5.1 · 6 marks

I / A	V / V
0.20	1.40
0.40	1.30
0.60	1.20
0.80	1.10
1.00	1.00

A student carries out Required Practical 6 to find the EMF and internal resistance of a cell, using a variable resistor, an ammeter and a voltmeter across the cell's terminals.

The results are shown in the table.

(a) Explain why the circuit should only be connected briefly for each reading.

(b) Calculate the gradient of the $V$ – $I$ graph.

(d) Calculate the current that would flow if the cell's terminals were short-circuited.

Show worked solution

(a) Keeping the circuit connected only briefly for each reading limits self-heating of the cell and its internal components, which would otherwise change the internal resistance $r$ during the measurement.

(b) Using the two end points: gradient:

= \frac{1.00 - 1.40}{1.00 - 0.20} = \frac{- 0.40}{0.80} = - 0.50 V A^{- 1}

(c) Since $V = ε - I r$ , the $y$ -intercept gives $ε \approx 1.50 V$ and the gradient gives $- r$ , so $r \approx 0.50 Ω$ .

(d) On short circuit, $R \approx 0$ , so:

I_{m a x} = \frac{ε}{r} = \frac{1.50}{0.50} = 3.0 A

Mark scheme · 6 marks

Explains that brief connection limits self-heating that would change $r$ during the measurement 1 mark
Calculates the gradient from two data points, $\approx - 0.50 V A^{- 1}$ 1 mark
States EMF is given by the $y$ -intercept, $ε \approx 1.50 V$ 1 mark
States internal resistance is given by the magnitude of the gradient, $r \approx 0.50 Ω$ 1 mark
Uses $I_{m a x} = ε / r$ for the short-circuit condition 1 mark
Calculates $I_{m a x} = 3.0 A$ 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.

Worked example 3.5.1 · 5 marks

A cell has EMF $3.0 V$ and internal resistance $2.0 Ω$ .

(a) State the value of external resistance $R$ at which the power delivered to $R$ is a maximum, and calculate this maximum power.

(b) Explain why making $R$ much larger or much smaller than this value both deliver less power to $R$ .

Show worked solution

(a) Maximum power transfer occurs when $R = r = 2.0 Ω$ .

Then:

I = \frac{ε}{R + r} = \frac{3.0}{4.0} = 0.75 A

so:

P = I^{2} R = 0.7 5^{2} \times 2.0 \approx 1.1 W

(b) If $R$ is much smaller than $r$ , the current is close to its maximum, but $R$ itself is small, so $P = I^{2} R$ is small because most of the potential difference is dropped across $r$ rather than $R$ .

If $R$ is much larger than $r$ , the current $I = ε / (R + r)$ falls substantially as $R$ increases, and this fall in $I^{2}$ outweighs the increase in $R$ , so $P = I^{2} R$ again falls.

The maximum occurs at $R = r$ , where these two competing effects balance.

Mark scheme · 5 marks

States that maximum power transfer occurs when $R = r = 2.0 Ω$ 1 mark
Calculates the current at $R = r$ , $I = 0.75 A$ 1 mark
Calculates the maximum power $P \approx 1.1 W$ 1 mark
Explains that for small $R$ , power is limited because most p.d. is dropped across $r$ rather than $R$ 1 mark
Explains that for large $R$ , the fall in current outweighs the rise in $R$ , so power again falls 1 mark

Do not count matching words alone — ask whether your answer actually makes the same claim.