B.5 Current and circuits – IB Physics exam questions with mark scheme

1Standard6 marks

A cell of emf 12 V and internal resistance $0.50\ \Omega$ is connected to the circuit shown. A $3.0\ \Omega$ resistor is in series with a parallel pair of $4.0\ \Omega$ and $6.0\ \Omega$ resistors.

(a) Show that the combined resistance of the parallel pair is $2.4\ \Omega$. [2]

(b) Calculate the current supplied by the cell. [2]

$1/R=1/4.0+1/6.0$ (= 0.4167) (allow product over sum $4.0\times6.0/(4.0+6.0)$)(1)
$R = 1/0.41\dot 6 = 24/10 = 2.4\ \Omega$ (result must follow from shown working; restating 2.4 Ω alone earns no credit)(1)
Total resistance $=0.50+3.0+2.4 = 5.9\ \Omega$ (internal resistance included)(1)
$I=\varepsilon/R_{tot} = 12/5.9 = 2.03$ A (2.0 A)(1)
p.d. across parallel pair $=IR = 2.03\times2.4=4.88$ V (4.9 V)(1)
$I=V/R = 4.88/6.0 = 0.81$ A (allow current-divider $2.03\times4.0/10.0$; allow 0.80 A if rounded 2.0 A and 4.8 V are used; allow ECF from (b))(1)
Answer: (a) 2.4 Ω (b) 2.0 A (c) 0.81 A
Examiner tip: Do not forget the cell's internal resistance when finding the total resistance.

2Core5 marks

A copper wire has length 2.5 m and diameter 0.40 mm. The resistivity of copper is $1.7\times10^{-8}\ \Omega\,\mathrm{m}$.

(a) Calculate the cross-sectional area of the wire. [1]

(b) Calculate the resistance of the wire. [2]

(d) A second wire has the same length and material but twice the diameter. State its resistance. [1]

$A=\pi d^2/4 = 1.26\times10^{-7}\ \mathrm{m^2}$ ($1.3\times10^{-7}$)(1)
$R=\rho L/A = 1.7\times10^{-8}\times2.5/1.26\times10^{-7}$(1)
$R=0.34\ \Omega$ (0.34)(1)
$P=I^2R = 3.0$ W (3.0 W; allow 3.1 W)(1)
Area is 4 times larger so $R = 0.34/4 = 0.085\ \Omega$ (0.085 Ω)(1)
Answer: (a) $1.3\times10^{-7}\ \mathrm{m^2}$ (b) 0.34 Ω (c) 3.0 W (d) 0.085 Ω
Examiner tip: Convert diameter in mm to metres and halve it to get the radius (or use $d^2/4$).

3Hard6 marks

A student measures the terminal potential difference $V$ of a cell for different currents $I$. The straight-line graph is shown. The line passes through $(0,\ 1.50\ \mathrm{V})$ and $(2.0\ \mathrm{A},\ 0.30\ \mathrm{V})$.

(a) State the emf of the cell. [1]

(b) Determine the internal resistance of the cell. [2]

(d) The cell is connected to a variable resistor. The power transferred to the resistor is a maximum when its resistance equals the internal resistance. Calculate this maximum power. [2]

emf = vertical intercept = 1.50 V(1)
$V=\varepsilon-Ir$ so $r=-$gradient (magnitude of gradient)(1)
$r=(1.50-0.30)/2.0 = 0.60\ \Omega$ (0.60 Ω)(1)
$I=\varepsilon/r = 1.50/0.60 = 2.5$ A (2.5 A)(1)
At maximum power, $R=r$ so $I=\varepsilon/2r = 1.25$ A; $P=I^2R$ (or $\varepsilon^2/4r$)(1)
$P=0.94$ W (0.94 W)(1)
Answer: (a) 1.50 V (b) 0.60 Ω (c) 2.5 A (d) 0.94 W
Examiner tip: The internal resistance is the magnitude of the gradient; give it without a negative sign.

Stuck on B.5 Current and circuits?

Get it sorted in a free 30-minute session

We work through your weakest questions on this topic on Zoom. I reply the same day.

Free 30-min session on this topic Send me 5 questions a day

4Core4 marks

A wireless earphone case is charged with a constant current of 0.25 A for 12 minutes. The potential difference across the battery in the case during charging is 3.7 V.

(a) Calculate the charge that flows into the battery. [2]

(b) Calculate the number of electrons that pass a point in the charging cable in this time. [1]

Converts time to seconds: $\Delta t = 12\times60 = 720$ s(1)
$\Delta q = I\Delta t = 0.25\times720 = 180$ C (award [2] for a bare correct answer; using 12 min to get 3.0 C scores [0])(1)
$N=\Delta q/e = 180/1.60\times10^{-19} = 1.1\times10^{21}$ (accept $1.13\times10^{21}$; allow ECF from (a))(1)
$W=qV = 180\times3.7 = 670$ J (accept 666 J; allow ECF from (a))(1)
Answer: (a) 180 C (b) $1.1\times10^{21}$ (c) 670 J (666 J)
Examiner tip: Convert minutes to seconds before using $I=\Delta q/\Delta t$; answers in coulomb-minutes score nothing.

5Core6 marks

The graph shows the current–voltage ($I$–$V$) characteristic of a filament lamp.

(a) State what is meant by an ohmic conductor. [1]

(b) Use the graph to calculate the resistance of the lamp at 2.0 V and at 6.0 V. [2]

(c) Explain, in terms of the charge carriers in the filament, why the resistance changes as the potential difference increases. [2]

(d) A student calculates the resistance at 6.0 V from the gradient of the tangent to the curve. Outline why this method is incorrect. [1]

(a) A conductor in which current is (directly) proportional to potential difference / resistance is constant, at constant temperature (accept: obeys Ohm's law with a statement of what Ohm's law is)(1)
(b) At 2.0 V: $I=0.20$ A so $R = 2.0/0.20 = 10\ \Omega$(1)
At 6.0 V: $I=0.30$ A so $R=6.0/0.30 = 20\ \Omega$ (allow readings of current ±0.005 A)(1)
(c) Larger current / greater power dissipated raises the temperature of the filament(1)
the (metal) lattice ions vibrate with greater amplitude, so electrons (charge carriers) collide with them more often, reducing the (drift) speed of the electrons for a given pd / so resistance increases (ignore references to 'more electrons')(1)
(d) Resistance is defined as $R=V/I$ (the ratio at that point), not $\Delta V/\Delta I$; the inverse gradient of the tangent gives a different (larger, about 56 Ω) value (accept: the graph is not a straight line through the origin, so the gradient does not equal $I/V$)(1)
Answer: (b) 10 Ω and 20 Ω; (c) temperature rises, more frequent electron–ion collisions; (d) $R=V/I$ not 1/gradient
Examiner tip: Resistance is always $V/I$ at the point; using the gradient of a curved $I$–$V$ graph is a classic lost mark.

6Standard6 marks

A temperature sensor uses a negative temperature coefficient (NTC) thermistor T in series with a fixed resistor of $2.2\ \mathrm{k\Omega}$. The supply has emf 9.0 V and negligible internal resistance. The output potential difference $V_{\text{out}}$ is taken across the fixed resistor.

(a) At 10 °C the resistance of the thermistor is $4.7\ \mathrm{k\Omega}$. Calculate $V_{\text{out}}$. [2]

(b) Explain what happens to $V_{\text{out}}$ as the temperature of the thermistor increases. [2]

(c) $V_{\text{out}}$ is used to switch on a cooling fan when it reaches 4.5 V. Determine the resistance of the thermistor at which the fan switches on. [2]

(a) Current $= 9.0/(4.7+2.2)\times10^3 = 1.30$ mA or ratio $V_{\text{out}} = 9.0\times\dfrac{2.2}{2.2+4.7}$(1)
$V_{\text{out}} = 2.9$ V (accept 2.87 V)(1)
(b) Resistance of the (NTC) thermistor decreases as temperature increases (so total resistance falls / current increases)(1)
so the thermistor takes a smaller share of the 9.0 V / pd across the fixed resistor $(=IR)$ increases, so $V_{\text{out}}$ increases (no credit for 'increases' with no reasoning)(1)
(c) pd across thermistor $=9.0-4.5 = 4.5$ V or current $=4.5/2200 = 2.05\times10^{-3}$ A(1)
$R_T = 2.2\ \mathrm{k\Omega}$ (equal pd, so equal resistance) (accept 2200 Ω)(1)
Answer: (a) 2.9 V (b) increases (c) 2.2 kΩ
Examiner tip: Say which resistor gets the bigger share of the pd; 'the voltage increases' without naming which one earns nothing.

7Standard6 marks

A student is given a reel of uniform nichrome wire, a metre rule, a micrometer screw gauge, a cell, an ammeter, a voltmeter and crocodile clips.

Describe how the student could use this apparatus to determine the resistivity of nichrome. Your answer should include the measurements taken, how a graph is used to obtain the result, and steps taken to improve the accuracy and reliability of the result. [6]

Measure the diameter $d$ of the wire with the micrometer at several points / in different orientations along the wire and take a mean; check for zero error; $A=\pi d^2/4$(1)
Connect a length of wire between crocodile clips in series with the ammeter and the cell; voltmeter in parallel with the length of wire between the clips(1)
Measure the length $L$ between the clips with the metre rule; record $V$ and $I$ and calculate $R=V/I$(1)
Repeat for at least five (a range of) different lengths, e.g. 0.20 m to 1.00 m, repeating readings and averaging(1)
Plot $R$ against $L$: a straight line through the origin with gradient $=\rho/A$(1)
$\rho = \text{gradient}\times A$ (a systematic contact resistance shows up as a non-zero intercept and does not affect the gradient)(1)
Keep the current small / switch off between readings so the temperature of the wire (and hence its resistance) does not rise(1)
Keep the wire straight and taut (no kinks) when measuring length; read the metre rule at eye level to avoid parallax(1)

Level	Marks	What it looks like
Level 3	5–6	Clear, logically ordered method: correct circuit, measurement of diameter and length with stated instruments, range of lengths with $R=V/I$, graph of $R$ against $L$ with gradient correctly related to $\rho$ and $A$, and at least two valid accuracy/reliability steps (e.g. low current, mean diameter, repeats).
Level 2	3–4	A workable method with the correct circuit and measurements of $V$, $I$, $L$ and $d$; a graph is mentioned but the link between gradient and resistivity may be incomplete, or only one valid precaution is given.
Level 1	1–2	Some relevant statements, e.g. use of $R=V/I$ or $R=\rho L/A$, or measuring length or diameter, but no coherent method.
Level 0	0	No relevant content.

Answer: Measure $d$ (mean of several micrometer readings), find $R=V/I$ for several lengths $L$, plot $R$ against $L$, $\rho=$ gradient $\times\,\pi d^2/4$; keep current low to avoid heating.
Examiner tip: Students often forget that the gradient is $\rho/A$ and quote the gradient as the resistivity; always multiply by the cross-sectional area.

8Hard7 marks

(a) Outline the energy transfers that take place in (i) a chemical cell and (ii) a solar cell when each is used as a source of emf. [2]

Four identical chemical cells A, B, C and D, each of emf 1.5 V and internal resistance $0.40\ \Omega$, are connected as shown to a lamp of resistance $5.6\ \Omega$.

(b) Determine the emf and the internal resistance of the combination of cells. [2]

(d) Calculate the power dissipated in the internal resistance of the cells. [1]

(a)(i) Chemical (potential) energy is transferred to electrical energy (of the charge carriers)(1)
(ii) Energy of light / electromagnetic radiation (photons) is transferred to electrical energy(1)
(b) emf $=1.5+1.5 = 3.0$ V (parallel branches do not add emf)(1)
internal resistance: each branch $0.80\ \Omega$; two branches in parallel give $0.40\ \Omega$(1)
(c) $I = \varepsilon/(R+r) = 3.0/(5.6+0.40) = 0.50$ A(1)
$V = IR = 0.50\times5.6 = 2.8$ V (accept $V=\varepsilon - Ir = 3.0-0.20$; allow ECF from (b))(1)
(d) $P = I^2r = 0.50^2\times0.40 = 0.10$ W (accept $0.25\ \mathrm{A}$ per branch: $4\times0.25^2\times0.40$; allow ECF)(1)
Answer: (b) 3.0 V, 0.40 Ω (c) 2.8 V (d) 0.10 W
Examiner tip: Cells in parallel branches share the current but do not add their emfs; only series cells add.

9Standard6 marks

The heating element of an electric kettle is a nichrome wire of diameter 0.50 mm. The kettle is rated at 2.0 kW when connected to a 230 V supply. The resistivity of nichrome at its working temperature is $1.1\times10^{-6}\ \Omega\,\mathrm{m}$.

(a) Show that the resistance of the element at its working temperature is about $26\ \Omega$. [1]

(b) Calculate the length of wire needed for the element. [3]

(c) A manufacturer connects two identical elements in parallel across the same 230 V supply. Determine the total power output, and explain your answer. [2]

(a) $R = V^2/P = 230^2/2000 = 26.45\ \Omega$ (must see substitution and an answer to at least 3 s.f., or a valid reverse calculation)(1)
(b) $A = \pi(0.25\times10^{-3})^2 = 1.96\times10^{-7}\ \mathrm{m^2}$ (radius, not diameter, used)(1)
$L = RA/\rho = 26.45\times1.96\times10^{-7}/1.1\times10^{-6}$(1)
$L = 4.7$ m (accept 4.6–4.8 m, e.g. using 26 Ω gives 4.6 m)(1)
(c) Each element still has the full 230 V across it (combined resistance halves to 13 Ω)(1)
so power $=V^2/R$ doubles to 4.0 kW (accept $2\times2.0$ kW; no mark for 4.0 kW without reasoning)(1)
Answer: (a) 26.45 Ω (b) 4.7 m (c) 4.0 kW
Examiner tip: Area uses the radius: halve the diameter and convert mm to m before squaring.

10Stretch7 marks

Two $100\ \mathrm{k\Omega}$ resistors X and Y are connected in series to a supply of emf 12 V and negligible internal resistance. A voltmeter of resistance $200\ \mathrm{k\Omega}$ is connected across Y, as shown.

(a) State the potential difference across Y before the voltmeter is connected. [1]

(b) Calculate the reading on the voltmeter. [3]

(c) The voltmeter is replaced by one of resistance $R_V$. Determine the minimum value of $R_V$ for the reading to be within 1.0% of the value in (a). [3]

(a) 6.0 V(1)
(b) Combined resistance of Y and voltmeter: $\left(\dfrac{1}{100}+\dfrac{1}{200}\right)^{-1} = 66.7\ \mathrm{k\Omega}$(1)
Reading $= 12\times\dfrac{66.7}{100+66.7}$ (or $I = 12/166.7\ \mathrm{k\Omega} = 72\ \mathrm{\mu A}$, then $V=I\times66.7\ \mathrm{k\Omega}$)(1)
$= 4.8$ V(1)
(c) Minimum reading $= 0.99\times6.0 = 5.94$ V, so pd across X $=6.06$ V and parallel resistance $R_p = 100\times5.94/6.06 = 98.0\ \mathrm{k\Omega}$(1)
$\dfrac{1}{R_V} = \dfrac{1}{98.0}-\dfrac{1}{100}$(1)
$R_V = 4.95\times10^{3}\ \mathrm{k\Omega} \approx 5.0\ \mathrm{M\Omega}$ (accept 4.9–5.0 MΩ; premature rounding of $R_p$ to 98 kΩ gives 4.9 MΩ, accept)(1)
Answer: (a) 6.0 V (b) 4.8 V (c) $4.95\ \mathrm{M\Omega}\approx5\ \mathrm{M\Omega}$
Examiner tip: A real voltmeter in parallel lowers the resistance of that section; keep 3+ significant figures in (c) because the answer is very sensitive to rounding.

Stuck on this topic?

Book a free 30-minute session on WhatsApp and we'll find exactly which step is costing you the marks. Or build a timed paper on it.

Free 30-minute session