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.
§1 — Current and Charge Topic B.5
Definition
- Direct current (dc): charge carriers flow in one direction only.
- Conventional current: defined as positive-charge flow, opposite in direction to electron drift.
§2 — EMF and Potential Difference Topic B.5
Both are measured in volts ($1\,\text{V} = 1\,\text{J C}^{-1}$).
§3 — Resistance, Resistivity & Ohm's Law Topic B.5
Definitions
§4 — I-V Characteristics Topic B.5
| Component | Behaviour |
|---|---|
| Ohmic resistor | 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 |
§5 — Electrical Power Topic B.5
$$\boxed{P = IV = I^2 R = \dfrac{V^2}{R}} \quad [\text{W}]$$
§6 — Series and Parallel Circuits Topic B.5
| Series | Parallel | |
|---|---|---|
| Current | $I = I_1 = I_2$ | $I = I_1 + I_2$ |
| Voltage | $V = V_1 + V_2$ | $V = V_1 = V_2$ |
| Resistance | $R_s = R_1 + R_2$ | $\dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2}$ |
§7 — EMF and Internal Resistance Topic B.5
$$\boxed{\varepsilon = I(R + r) = V + Ir}$$
Terminal p.d.: $V = \varepsilon - Ir$ ("lost volts" $= Ir$).
On a $V$ vs $I$ graph the $y$-intercept is $\varepsilon$ and the gradient is $-r$.
§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 $V_\text{supply}$ |
§9 — Exam Attack Plan All sections
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)
- Apply $\varepsilon = I(R + r)$ with $R = 4.0\,\Omega$: $1.5 = I(4.0 + 0.5) = 4.5 I$. (M1)
- Solve: $I = 1.5/4.5 \approx 0.333\,\text{A}$. (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.?
Why isn't a filament lamp ohmic?
How do I add resistors in parallel quickly?
Which power formula should I pick?
Why is a thermistor used as a temperature sensor?
Get the printable PDF version
Same cheatsheet, formatted for A4 print — keep it next to your study desk. Free for signed-in users.