IB Physics HL Current and Circuits — Complete Cheatsheet
Every formula, definition, trick, and trap you need for IB Physics HL Topic B.5 — current, EMF, resistance, I-V characteristics, electrical power, series & parallel networks, internal resistance.
Current and Circuits closes out Theme B and is the foundation for everything in Theme D — Fields: induction, capacitance, electromagnetic motion, and magnetic forces all assume confident handling of $V$, $I$, and $R$. The topic also returns directly in Paper 3 practical-skills questions, where examiners love a potential-divider sensor or an internal-resistance experiment. Master B.5 now and the rest of the electricity block becomes routine algebra rather than a fresh worry.
This cheatsheet condenses every formula, definition, trick, and trap from Topic B.5 into one revisable page. Each section pairs the formula box with the trick that saves time and the trap that quietly costs marks. The worked example walks through an internal-resistance experiment in IB mark-scheme rhythm, and the FAQ block addresses the conceptual questions our small-group students bring up most often.
Direct current (dc): charge carriers flow in one direction only.
Conventional current: defined as positive-charge flow, opposite in direction to electron drift.
TrickTo find the number of electrons that flow: $n = \Delta q / e$, with $e = 1.60 \times 10^{-19}\,\text{C}$.
TrapCurrent is the same at every point in a series circuit — charge is not "used up" as it passes through a resistor. Energy is dissipated, not charge.
From the Photon question bank
What examiners actually test on this concept:
I = Δq/ΔtConventional vs electron flown = Δq/eCurrent same in series
Current is the same everywhere in a series loop — charge isn't "used up", energy is. frequent
Conventional current is positive-charge flow, opposite in direction to the electron drift.
Electrons that pass: $n = \Delta q/e = I\Delta t/e$, with $e = 1.60\times10^{-19}$ C.
§2 — EMF and Potential Difference Topic B.5
EMF:$\varepsilon = \dfrac{W}{q}$ — energy supplied per unit charge by the source
Potential diff:$V = \dfrac{W}{q}$ — energy transferred per unit charge between two points
Both are measured in volts ($1\,\text{V} = 1\,\text{J C}^{-1}$).
NoteEMF is the "total" voltage of the source; p.d. is what appears across the external circuit when current flows. The two only agree at $I = 0$ (open circuit), because no voltage is dropped across the internal resistance.
§3 — Resistance, Resistivity & Ohm's Law Topic B.5
Ohm's law:$V = IR$ when $R$ is constant ($V \propto I$ at constant temperature)
TrickFor a circular wire $A = \pi r^2 = \pi (d/2)^2$. Watch for diameter vs radius and convert mm → m before substituting into $R = \rho L / A$.
Trap$R = V/I$ is always valid (it is the definition). Ohm's law additionally requires $R$ to be constant. A filament lamp obeys $R = V/I$ at any instant, but it does not obey Ohm's law.
Resistivity of a wire. Resistance grows with length and shrinks with cross-sectional area: $R = \rho L/A$ — so a long, thin wire resists most.
From the Photon question bank
What examiners actually test on this concept:
R = V/IR = ρL/AOhm's law needs constant RDiameter vs radius
$R=V/I$ is always the definition; Ohm's law ADDS that $R$ is constant.very frequent A lamp obeys $R=V/I$ at any instant but not Ohm's law.
$R = \rho L/A$ with $A = \pi(d/2)^2$; convert mm → m before substituting — a classic slip.
Straight line through origin; $R$ constant (metal at constant temperature)
Filament lamp
Curve flattens at high $V$; $R$ rises with temperature
Diode
Conducts forward above ≈ 0.6 V; very high $R$ in reverse
NoteThe heating effect of current explains why a filament lamp is non-ohmic: increased temperature ⇒ more lattice vibrations ⇒ more electron–lattice collisions ⇒ higher $R$.
$I$–$V$ characteristics. An ohmic resistor is a straight line; a filament lamp curves as it heats ($R$ rises); a diode conducts only above ~0.6 V forward and blocks reverse current.
TrickFor two resistors in parallel: $R_p = \dfrac{R_1 R_2}{R_1 + R_2}$ (product over sum).
TrapDon't forget to take the reciprocal after summing $1/R$ values. Also: $R_p$ is always smaller than the smallest individual resistor — use this as a sanity check.
Series vs parallel. Series components share one current and add their resistances; parallel branches share one voltage and combine as $1/R_p = \sum 1/R$.
From the Photon question bank
What examiners actually test on this concept:
Series adds RParallel 1/RProduct over sumSanity checks
Series: same current, voltages add, $R_s=\sum R$. Parallel: same voltage, currents add, $1/R_p=\sum 1/R$.very frequent
Two in parallel: $R_p = R_1R_2/(R_1+R_2)$, and $R_p$ is always smaller than the smallest resistor — a quick check.
Remember to take the reciprocal after summing the $1/R$ values.
On a $V$ vs $I$ graph the $y$-intercept is $\varepsilon$ and the gradient is $-r$.
TrapAt $I = 0$ (open circuit), $V = \varepsilon$. As external resistance $R \to 0$ (short circuit), $I_\text{max} = \varepsilon / r$. Confusing these two limits is a common Paper 1 trap.
EMF and internal resistance. The cell drives current through $R+r$, so the terminal p.d. is $V = \varepsilon - Ir$; a $V$–$I$ graph has intercept $\varepsilon$ and gradient $-r$.
From the Photon question bank
What examiners actually test on this concept:
ε = I(R + r)V = ε − IrV–I graph intercept/slopeOpen vs short circuit
Terminal p.d. $V = \varepsilon - Ir$ (lost volts $= Ir$); on a $V$–$I$ graph the intercept is $\varepsilon$ and the gradient is $-r$. very frequent
Open circuit ($I=0$): $V=\varepsilon$. Short circuit ($R\to0$): $I_{\max}=\varepsilon/r$. Don't confuse the two limits.
EMF and terminal p.d. agree only when no current flows.
§8 — Variable Resistors Topic B.5
Component
Behaviour
Thermistor (NTC)
$R$ decreases as temperature increases
LDR
$R$ decreases as light intensity increases
Potentiometer
Sliding contact varies effective resistance; output voltage is adjustable from 0 to $V_\text{supply}$
NoteAmmeters are treated as ideal (zero resistance) and voltmeters as ideal (infinite resistance) unless the question states otherwise. Non-ideal meters have constant resistance — model them as a small series resistor (ammeter) or a large parallel resistor (voltmeter).
Potential divider. The supply voltage splits across a fixed resistor and a sensor (thermistor or LDR); the output taps off one of them and changes as the sensor's resistance changes.
From the Photon question bank
What examiners actually test on this concept:
NTC thermistorLDRPotential dividerIdeal meters
NTC thermistor: $R$ falls as $T$ rises; LDR: $R$ falls as light rises — both drive sensor potential dividers. frequent
Potential-divider output $V_{\text{out}} = V_s\,R_2/(R_1+R_2)$; swap in the sensor so the output tracks temperature or light.
When you see this in the question — reach for that:
Question trigger
Reach for
"Charge that flows in time $t$"
$\Delta q = I \Delta t$ (then $n = \Delta q/e$ for electrons)
"Resistance of a wire / cable"
$R = \rho L / A$, $A = \pi(d/2)^2$ for a round wire
I–V graph for a filament lamp
Non-ohmic; $R$ rises with $T$ — quote heating effect
"Power dissipated in a resistor"
Pick the simplest of $P = IV$, $I^2 R$, $V^2/R$
Series circuit calculation
$R_s = \sum R_i$; same $I$ throughout; sum of $V$s
Parallel circuit calculation
$1/R_p = \sum 1/R_i$; same $V$ across each branch
"EMF / lost volts / internal resistance"
$\varepsilon = V + Ir$; gradient of $V$–$I$ graph $= -r$
"Reading on the voltmeter / ammeter"
Identify whether it reads terminal $V$ or $\varepsilon$; ideal meter assumptions
Sensor circuit (LDR / thermistor)
Potential divider with one fixed resistor; output $\propto R_\text{sensor}/R_\text{total}$
"Why is the lamp brighter when…"
Trace through Ohm's law and $P = V^2/R$; check what stays constant
Worked Example — IB-Style Internal Resistance
Question (HL Paper 2 style — 7 marks)
A student connects a battery of EMF $\varepsilon$ and internal resistance $r$ across a variable resistor $R$. The student plots terminal p.d. $V$ against current $I$ and obtains a straight line with $y$-intercept $1.5\,\text{V}$ and gradient $-0.5\,\Omega$. (a) State $\varepsilon$ and $r$. (b) Find the current when $R = 4.0\,\Omega$ is connected. (c) Find the power dissipated in $R$ at this current.
Solution
From $V = \varepsilon - Ir$: $y$-intercept = $\varepsilon$ and gradient = $-r$. Read off $\varepsilon = 1.5\,\text{V}$, $r = 0.5\,\Omega$. (R1)(A1)
Terminal p.d. across $R$: $V_R = IR = 0.333 \times 4.0 \approx 1.33\,\text{V}$. (M1)
Power dissipated in $R$: $P = V_R I = 1.33 \times 0.333 \approx 0.444\,\text{W}$ (or use $P = I^2 R = (0.333)^2 \times 4.0 \approx 0.444\,\text{W}$). (M1)(A1)
Examiner's note: Students often quote $P = \varepsilon I = 1.5 \times 0.333 \approx 0.5\,\text{W}$ — but that is the total power supplied by the battery, including the heat dissipated in the internal resistance ($I^2 r \approx 0.056\,\text{W}$). The question asks specifically for power in $R$, so use $V_R I$ or $I^2 R$. Always check which resistance the question is about.
Common Student Questions
What is the difference between EMF and terminal p.d.?
EMF ($\varepsilon$) is the energy supplied per unit charge by the source — it is the open-circuit voltage of the cell. Terminal potential difference $V$ is what appears across the external circuit when current is flowing: $V = \varepsilon - Ir$, where $r$ is the internal resistance and $Ir$ is the "lost volts" inside the cell. At $I = 0$ (open circuit) $V = \varepsilon$; at short circuit $V = 0$ and $I_\text{max} = \varepsilon/r$.
Why isn't a filament lamp ohmic?
A filament lamp's resistance increases with temperature. As current flows, the filament heats up, lattice vibrations increase, and electrons collide more often — raising $R$. So $V/I$ is not constant: the I–V graph is a curve that flattens at high $V$. $R = V/I$ is still a valid definition, but Ohm's law (which requires $R$ constant) is not satisfied. Be careful to distinguish "definition of $R$" from "Ohm's law" in your answers.
How do I add resistors in parallel quickly?
For two resistors use the product-over-sum shortcut: $R_p = R_1 R_2 / (R_1 + R_2)$. For more than two, take reciprocals of every $R$, sum them, then take the reciprocal of the result — never forget the final reciprocal step. The parallel combination is always smaller than the smallest individual resistor; use this as a quick sanity check.
Which power formula should I pick?
$P = IV$ is the master formula. Use $P = I^2 R$ when you know the current through the component and its resistance (the heating-effect form). Use $P = V^2/R$ when you know the voltage across the component and its resistance. Pick the formula that needs the two quantities you already have — converting through Ohm's law just propagates rounding errors.
Why is a thermistor used as a temperature sensor?
An NTC thermistor's resistance falls sharply as temperature rises. Wire it as one arm of a potential divider with a fixed resistor across a constant supply, and the output voltage shifts dramatically with temperature. The same trick is used with an LDR (resistance falls as light intensity rises) for light sensors. You should be able to draw and explain a potential-divider sensor circuit in any IB practical-skills question.
What's NOT in this cheatsheet
This page gives you the formulas and the traps. The full Photon Academy Current and Circuits library (only available to enrolled students or via the resource library subscription) adds:
Notes PDF — every concept worked through with full circuit diagrams and worked intuition.
Tutorial booklet — IB-style questions sequenced from foundation to full HL difficulty.
Tutorial Solutions — full mark-scheme-style solutions with R1/M1/A1 annotations.
Practice Solutions — extra past-paper-style problems with detailed walk-throughs.
Cheatsheet PDF — print-ready, brand-formatted, the same one our students take into mock exams.
