Q: What does A-Level Physics: 16) Circuits Guide cover? A: From resistor networks to exponential RC transients, this post unpacks Topic 16 of the 2026 H2 Physics syllabus for IP students and parents.
TL;DR Circuits is not “just Ohm's Law” - it is the control panel behind practical Paper 4 and every data-logger question. This guide turns the SEAB bullet-points into classroom-tested check-lists, sensor hacks and WA timing tricks.
Concrete example: how to use this page
For a mixed circuit, redraw it in stages. Combine obvious series or parallel parts first, then apply Kirchhoff rules only where the circuit cannot be simplified cleanly.
Keep the full circuits + electromagnetism toolkit handy via the H2 Physics notes hub; it links this post with the preceding Currents/Electric Fields chapters and the upcoming electromagnetism topics.
Circuit-solving decision map
Use this map before substituting numbers. Most lost marks happen when students choose the right formula too early but apply it to the wrong part of the circuit.
What the diagram is asking
First move
Main rule
Misconception check
Total resistance or current
Collapse obvious series and parallel groups first.
Series resistances add; parallel branches add by reciprocals.
Do not add parallel resistances directly. The equivalent resistance must be smaller than the smallest branch.
Unknown branch current or p.d.
Mark junctions and closed loops before writing equations.
Current is conserved at a junction; p.d. changes around a closed loop sum to zero.
A branch with larger resistance does not automatically have larger current. Check whether branches share the same p.d.
Sensor output voltage
Identify which resistor is the output leg.
The output takes the same fraction of supply as that output resistance takes of total series resistance.
LDR and NTC behaviour only matters after you know whether the sensor is the top or bottom resistor.
Capacitor charging or discharging
Decide whether the value starts at zero or starts at its initial maximum.
Charging uses the approach-to-maximum curve; discharging uses the decay curve.
The time constant τ=RC is not the time to become fully charged. It marks about 63 percent charged or 37 percent remaining.
1 Circuit symbols & diagrams
Memorise SEAB's full symbol set - cell, switch, fixed/variable resistor, LDR, NTC thermistor, diode, capacitor, ammeter and voltmeter. Symbols must be drawn with a ruler in Paper 2 for the mark.
Parent tip: have your teen print the symbol sheet and stick it on the inside cover of their graph-book.
1.1 Mini-drill
Sketch a potential-divider with a fixed resistor and an LDR controlling Vout. Mark the sensing node. Time limit = 30 s.
2 Resistance, resistivity and internal resistance
2.1 Ohm's law refresher
The definition is R=IV(Ω). Use it only when the graph through the origin is linear.
2.2 Microscopic link
For a uniform wire R=ρAl,
where ρ is resistivity, l (m) is length and A (m2) is cross-sectional area. Double l → double R; halve A → double R.
2.3 Temperature stories
Metals (e.g. filament lamp): the number density of charge carriers is essentially fixed - nearly all conduction electrons are already free. As temperature rises, increased lattice vibrations scatter electrons more frequently, reducing the drift velocity for a given applied field. Lower drift velocity means lower current at the same voltage, so resistivity increases.
Semiconductors (e.g. NTC thermistor): thermal excitation promotes electrons across the band gap, sharply increasing the number density of charge carriers. This increase outweighs any reduction in drift velocity, so resistivity falls with rising temperature.
Exam cue: outcome (f) asks you to use the phrases "drift velocity" (metals) and "number density of charge carriers" (semiconductors) - include both explicitly for full marks.
2.4 Internal resistance (r)
Real cells obey V=ε−Ir,
so increasing load current drops terminal p.d.
Graph cue: gradient = −r, intercept = ε. Quote ε in V not eV.
2.5 Output power and internal resistance
The power delivered to the external load R is
Pout=I2R=(r+Rε)2R.
When R is very small, most of the e.m.f. drives current through r, so Pout→0. When R is very large, current becomes negligible, so Pout→0 again. Output power therefore peaks at an intermediate load value. The maximum-power-transfer condition is R=r: the load resistance equals the internal resistance.
Exam cue: questions on outcome (g) can ask you to state both effects - terminal p.d. falls as I increases, and output power varies with load, peaking when R=r. Quoting the formula above is sufficient; no calculus derivation is required.
Internal-resistance load-change checkpoint
When the external load R changes, track current, terminal p.d., and useful power separately. They do not all peak at the same load.
Load situation
Current I
Terminal p.d. V=ε−Ir
Output power Pout=I2R
Common trap
R is very small
Large
Small, because most of the e.m.f. is lost across r.
Small, because the external load is tiny.
Thinking high current always means high useful power.
R=r
Moderate
ε/2
Maximum
Forgetting that half the e.m.f. is then lost inside the source.
R is very large
Small
Close to ε
Small, because current is tiny.
Thinking high terminal p.d. means high useful power.
A V-against-I graph is given
Read from the x-axis.
Use y-intercept =ε and gradient =−r.
Calculate from VI at the chosen point.
Quoting the intercept as output power.
Worked check: a cell has ε=6.0V and r=2.0Ω. With R=2.0Ω, I=6.0/(2.0+2.0)=1.5A, so V=IR=3.0V and Pout=I2R=4.5W. With R=8.0Ω, I=0.60A and V=4.8V, but Pout=2.9W. The higher terminal p.d. case does not give more useful power.
Misconception check: maximum terminal p.d. is not maximum output power. In an open-circuit limit, terminal p.d. is close to ε, but current is close to zero, so useful power is close to zero.
3 Resistors in series & parallel
Arrangement
Combined resistance
Series
Rtot=R1+R2+…
Parallel
Rtot1=R11+R21+…
Series-parallel reduction checkpoint
Before using a final current or p.d. formula, reduce only the parts that are genuinely in series or genuinely in parallel. Work from the simplest hidden group outward.
Circuit clue
What you may combine
Quick test
Common trap
Components share one unbranched current path
Add their resistances directly.
The same current must pass through every component in that chain.
Calling two components series just because they are drawn in a row before a junction.
Components connect across the same two nodes
Add reciprocals to find the equivalent resistance.
The same p.d. is across every branch.
Adding the branch resistances directly.
One branch is already a small series chain
Combine that branch first, then treat it as one parallel branch.
The branch equivalent can be compared with the other branches.
Mixing a series resistor into a parallel formula before simplifying the branch.
Final equivalent for a parallel group
It must be smaller than the smallest branch resistance.
If it is larger than a branch, recheck the reciprocal step.
Forgetting to invert after adding reciprocals.
Worked check: a 4.0Ω resistor is in series with a parallel pair of 6.0Ω and 3.0Ω. First reduce the parallel pair:
Rparallel1=6.01+3.01=2.01,
so Rparallel=2.0Ω. Then add the series resistor: Rtotal=4.0Ω+2.0Ω=6.0Ω.
Misconception check: the 4.0Ω resistor is not in parallel with the two-branch group. It sits before the split, so all current passes through it before the current divides.
Be ready to spot hidden series chains in messy WA diagrams.
Kirchhoff setup checkpoint
Use Kirchhoff's laws when the circuit cannot be reduced cleanly into one series-parallel equivalent. The first mark is usually for defining current directions and loop signs consistently.
Before writing equations
What to mark
Equation move
Common trap
Junction with several branches
Arrow each branch current.
Currents entering = currents leaving.
Assuming the largest resistor must have the smallest current before checking branch p.d.
Closed loop with a cell
Choose a clockwise or anticlockwise loop direction.
Crossing a cell from negative to positive terminal is a potential rise.
Changing loop direction halfway through the equation.
Resistor in the loop
Mark the current through that resistor.
Moving with the current gives a potential drop IR.
Giving every resistor the same current in a multi-branch circuit.
Shared resistor
Decide which loop currents pass through it.
Use the net current through the resistor before writing IR.
Counting the same current twice or with the wrong sign.
Worked check: for a single-loop circuit with e.m.f. ε, internal resistance r, and external resistor R, moving around the loop with the current gives
+ε−Ir−IR=0,
so ε=I(R+r) and the terminal p.d. across the external resistor is V=IR=ε−Ir.
Misconception check: Kirchhoff equations are bookkeeping, not a new formula set. If the chosen current direction turns out wrong, the solved current will be negative; the circuit is not invalid.
3.1 Timing hack
Write the total resistance before inserting numbers; algebra first reduces keypad slips.
4 Potential divider circuits
4.1 Core formula
Vout=VinR1+R2R2.
4.2 Potential-divider output checkpoint
Before deciding whether Vout rises or falls, mark the output leg in the diagram. The output voltage is the p.d. across the component connected between the output node and the reference line.
Sensor position
What happens to sensor resistance
What happens to Vout
Reason
LDR as the lower output resistor
Brighter light decreases LDR resistance.
Vout decreases.
The lower resistor takes a smaller fraction of the supply.
LDR as the upper resistor
Brighter light decreases LDR resistance.
Vout increases.
The fixed lower resistor takes a larger fraction of the supply.
NTC as the lower output resistor
Higher temperature decreases NTC resistance.
Vout decreases.
The lower resistor takes a smaller fraction of the supply.
NTC as the upper resistor
Higher temperature decreases NTC resistance.
Vout increases.
The fixed lower resistor takes a larger fraction of the supply.
Misconception check: do not memorise "LDR in bright light means higher output voltage". The answer flips when the sensor moves from the lower leg to the upper leg.
4.3 Sensor combos
Brightness probe: replace R2 with LDR, so Vout increases in the dark.
Fire alarm: replace R1 with NTC, so Vout increases when hot.
Include a buffer op-amp in higher-ability tutorials to prevent loading.
5 I-V characteristics
Component
Key graph feature
Exam explanation
Ohmic resistor
Straight line through origin
Constant R
Filament lamp
Curve flattens as I↑
T↑⇒R↑ in tungsten
Diode
Conducts after ≈0.6V
Forward bias overcomes p−n barrier
NTC thermistor
Steep at low V
T↑⇒R↓ due to more carriers
Plot with current on the y-axis - SEAB marks for axis labels.
I-V graph resistance checkpoint
Before reading resistance from an I-V graph, check whether the component is ohmic and which axis is plotted vertically. Resistance is V/I, so the graph shape decides whether one constant gradient is enough.
Graph clue
What to calculate
Answer move
Common trap
Straight line through the origin with I on the y-axis
Constant resistance
Use R=V/I, or take the reciprocal of the I-against-V gradient.
Calling the gradient itself resistance when current is on the y-axis.
Straight line through the origin with V on the y-axis
Constant resistance
The V-against-I gradient is R.
Forgetting to check which axis is vertical.
Curved filament-lamp graph
Changing resistance
Use R=V/I at the chosen point for resistance at that operating point.
Treating the whole curve as one fixed resistance.
Curved graph asking for small-change resistance
Local resistance
Use the tangent gradient on a V-against-I graph, or the reciprocal tangent gradient on an I-against-V graph.
Using a chord from the origin when the question asks for gradient at a point.
Worked check: if an ohmic resistor's I-against-V graph has gradient 0.25A⋅V−1, then R=1/0.25=4.0Ω. If the axes were reversed, the same numerical gradient would instead mean 0.25Ω.
Misconception check: "gradient" is not automatically resistance. It is resistance only when voltage is on the vertical axis and current is on the horizontal axis.
6 Capacitors in series & parallel
Arrangement
Combined capacitance
Series
Ctot1=C11+C21+…
Parallel
Ctot=C1+C2+…
Mnemonic: series for resistors adds R; series for capacitors adds 1/C.
Capacitor combination checkpoint
After finding the equivalent capacitance, decide what stays the same in the actual circuit. This prevents mixing the resistor rules with the capacitor rules.
Arrangement
Quantity that is the same
Quantity that splits
Reasoning move
Common trap
Capacitors in series
Charge on each capacitor
P.d. across each capacitor
Use Q=C1V1=C2V2, so the smaller capacitance has the larger p.d.
Giving every capacitor the supply voltage.
Capacitors in parallel
P.d. across each capacitor
Charge stored on each capacitor
Use Q1=C1V and Q2=C2V
Worked check: two capacitors, 2.0uF and 6.0uF, are connected in series to a 12V supply. Their equivalent capacitance is (1/2.0+1/6.0)−1=1.5uF, so the series charge is Q=CtotV=18uC. The p.d. values are V1=Q/C1=9.0V and V2=Q/C2=3.0V, adding back to 12V.
Misconception check: in series, capacitors do not share the supply voltage equally unless their capacitances are equal. They share the same charge; the voltage divides in inverse proportion to capacitance.
7 RC circuits with d.c. source
7.1 Time constant
τ=RC.
At t=τ a discharging capacitor's Q,V or I falls to e1≈37% of its initial value.
7.2 Exponential laws
Charging: Q=Q0[1−e−t/τ],V=V0[1−e−t/τ].
Discharging: Q=Q0e−t/τ,V=V0e−t/τ,I=I0e−t/τ.
Plot lnV vs t to obtain a straight line of gradient −1/RC - a favourite Paper 4 practical.
RC linear-plot checkpoint
Use a logarithmic plot only after deciding which exponential form you have. For a discharging capacitor, V=V0e−t/RC, so
lnV=lnV0−RCt.
Graph plotted
Straight-line gradient
What it gives
Common trap
lnV against t for discharge
−RC1
RC=−gradient1
Forgetting the gradient is negative.
ln(V0−V) against t for charging
−RC1
RC=−gradient1
Worked check: if a discharge graph of lnV against t has gradient −0.50s−1, then
RC=−−0.501=2.0s.
Common trap: the y-intercept gives lnV0, not V0. Undo the logarithm before quoting the initial voltage.
7.3 Variation with time - key checkpoints
The table below summarises how charge Q, voltage V, and current I behave at notable time points. All values are fractions of the initial (discharging) or maximum (charging) quantity.
Time
Charging Q(t)/Qmax
Discharging Q(t)/Q0
t=0
0
1
t=τ
1−1/e≈0.63
1/e≈0.37
t=2τ
≈0.86
≈0.14
t=3τ
≈0.95
≈0.05
t→∞
1
0
The same fractions apply to V throughout. For current: during discharging, I follows the same decaying exponential shape as Q and V. During charging, I does the opposite - it starts at its maximum value I0=ε/R and decays to zero as the capacitor approaches full charge. In other words, the charging current curve has the same shape as a discharging Q curve, but flipped in the context of the circuit.
Revision cue: "one time constant = 63% charged (or 37% remaining)" is the most tested number - learn it as a reflex.
8 Three WA timing rules (Circuits edition)
Label units first for every numerical answer - avoids unit-free slips.
Sketch a quick circuit even if not asked; you see hidden series legs faster.
Log-log check: if your answer for R or C is < 0.1 or > 10 Ω/F, re-read prefixes.
Need structured practice on Circuits? Our H2 Physics tuition programme covers this topic with weekly problem sets and Paper 4 practical drills.
Comprehensive revision pack
9478 Section V, Topic 16 Syllabus outcomes
Candidates should be able to:
(a) recall and use appropriate circuit symbols.
(b) draw and interpret circuit diagrams containing sources, switches, resistors (fixed and variable), ammeters, voltmeters, lamps, thermistors, light-dependent resistors, diodes, capacitors and any other type of component referred to in the syllabus.
(c) define the resistance of a circuit component as the ratio of the potential difference across the component to the current in it, and solve problems using the equation V=IR.
(d) recall and solve problems using the equation relating resistance to resistivity, length and cross-sectional area, R=Aρl.
(e) sketch and interpret the I-V characteristics of various electrical components in a d.c. circuit, such as an ohmic resistor, a semiconductor diode, a filament lamp and a negative temperature coefficient (NTC) thermistor.
(f) explain the temperature dependence of the resistivity of typical metals (e.g. in a filament lamp) and semiconductors (e.g. in an NTC thermistor) in terms of the drift velocity and number density of charge carriers respectively.
(g) show an understanding of the effects of the internal resistance of a source of e.m.f. on the terminal potential difference and output power.
(h) solve problems using the formula for the combined resistance of two or more resistors in series.
(i) solve problems using the formula for the combined resistance of two or more resistors in parallel.
(j) solve problems involving series and parallel arrangements of resistors for one source of e.m.f., including potential divider circuits which may involve NTC thermistors and light-dependent resistors.
(k) solve problems using the formulae for the combined capacitance of two or more capacitors in series and in parallel.
(l) describe and represent the variation with time, of quantities like current, charge and potential difference, for a capacitor that is charging or discharging through a resistor, using equations of the form x=x0e−t/τ or x=x0[1−e−t/τ]
Concept map (in words)
Start with circuit symbols to communicate clearly. Use Ohm's law and resistivity for basic components. Combine resistors/capacitors systematically. Potential dividers convert sensor resistance to voltage signals. RC circuits introduce exponential behaviour governed by time constant RC.
Key relations
Quantity / concept
Expression / highlight
Ohm's law
V=IR
Resistivity relation
R=ρAl
Series resistors
Rtot=∑Ri
Parallel resistors
Rtot1=∑Ri1
Potential divider
Vout=VinR1+R2R2
Series capacitors
Ctot1=∑Ci1
Parallel capacitors
Ctot=∑Ci
RC charging equation
V(t)=V0(1−e−t/(RC))
RC discharging equation
V(t)=V0e−t/RC,;I(t)=I0e−t/RC
Derivations & reasoning to master
Potential divider: derive ratio using loop current, or use voltage drop proportionality.
RC exponential forms - use, don't derive: RC charging and discharging are assessable under 9478 Topic 16 outcome (l). Candidates are expected to use the given forms x=x0e−t/τ and x=x0[1−e−t/τ] with τ=RC; the underlying derivation from dtdQ=RV−Q/C is not required. Skip the derivation unless you want extra calculus practice.
Sensor behaviour: analyse how LDR/NTC resistance changes Vout and relate to control systems.
ln-linearisation: rearrange V=V0e−t/RC to lnV=lnV0−t/RC
Worked example 1 - potential divider sensor
Design a circuit that outputs 3.0V when an LDR (resistance 12kOhm in dark, 2.0kOhm in bright light) is exposed to daylight using a 9.0V supply. Determine the fixed resistor value and predict the output in darkness.
Method: solve Vout=VinRfixed+RLDRRLDR for the bright condition; check the dark condition to ensure the alarm threshold.
A 47kOhm resistor and 100uF capacitor form a delay circuit. (a) Find the time constant. (b) How long until the capacitor voltage reaches 90 percent of the supply? (c) If used with 5V logic, what is the voltage at 3τ?
Solution: The time constant is 4.7s(τ=RC).
V=V0(1−e−t/(RC))
Use this relation to solve for t and the specific voltages.
For 90% charging: 0.90=1−e−t/τ⇒t=τln10=2.303τ≈10.8s.
At 3τ: V=V0(1−e−3)=0.950V0≈4.75V for V0=5V.
Worked example 3 - combined capacitance
Three capacitors: C1=6.0uF, C2=3.0uF, and C3=4.0uF. C1 and C2 are connected in parallel; this parallel combination is then connected in series with C3. Find the total capacitance.
Step 1 - parallel combination of C1 and C2:
CP=C1+C2=6.0+3.0=9.0uF.
Step 2 - CP in series with C3:
Ctot1=CP1+C31=9.01+4.01=364.0+9.0=3613.
Ctot=1336≈2.8uF.
Note: the series step always gives a result smaller than either branch - 2.8uF<4.0uF. Use this as a quick sanity check.
Practical & data tasks
Build potential divider with light sensor; logVout under different lux levels and fit calibration curve.
Record capacitor discharge using Logger Pro; plot lnV vs t to extract −1/RC.
Investigate loading effect by attaching low-resistance voltmeter to a divider; observe output change.
Common misconceptions & exam traps
Forgetting to convert mm2 to m2 when using resistivity equation.
Mixing up series/parallel rules for capacitors vs resistors.
Ignoring meter resistance when measuring delicate dividers (loading).
Failing to state exponential behaviour explicitly in written explanations.
Quick self-check quiz
In a potential divider, what happens to Vout if R2 decreases? - _It decreases (assuming R1 is fixed).
How do you halve the time constant without changing capacitance? - Halve the resistance (since τ=RC).
What is the gradient of lnV vs t for a discharging capacitor? - −RC1.
Are thermistors ohmic? - No; resistance changes with temperature causing non-linear I−V behaviour.
Suggest one way to buffer a sensor output against loading. - Use an op-amp voltage follower.
Revision workflow
Redraw standard sensor-circuit templates and annotate expected Vout behaviour.
Practise using the given exponential equations and ln-linearisation (lnV vs t, gradient −1/RC) for RC data. Deriving the exponentials from dQ/dt is enrichment, not assessed.
Run a mock Paper 4 analysis on sample RC data to stay familiar with gradient extraction.
Practice Quiz
Test yourself on the key concepts from this guide.
Parents: book a 60-min Circuit Masterclass four weeks before WA 2 - most careless marks hide in potential-divider algebra.
Students: screenshot the RC graphs above and recreate them without notes tomorrow.
Last updated 20 Oct 2026. Check SEAB's syllabus list for school candidates for your examination year: 2026 / 2027.
, so the larger capacitance stores more charge.
Giving every capacitor the same charge.
Plotting lnV for a charging curve and expecting a straight line.
