IB Diploma Physics HL · Theme B.5 Current and circuits
Mini-Lesson
Electric Current & Circuits
This mini-lesson covers Theme B.5 — Current and circuits: current, potential difference and resistance, Ohm's law, electrical power, series and parallel networks, and emf with internal resistance.
Work through each screen, answer the questions as you go (some are reasoning, some are calculations) and collect ⭐ stars. Watch for the HL flag on higher-level extensions. Press Start when you're ready.
B.5 · basics
Current, pd and resistance
Current I is the rate of flow of charge; potential difference V is the energy transferred per coulomb; resistance R opposes the flow.
I = Δq ÷ Δt · V = IRcurrent (A) · charge (C) · pd (V) · resistance (Ω)
An ohmic conductor at constant temperature has V ∝ I (a straight line through the origin). A filament lamp is non-ohmic: it gets hotter and its resistance rises, so the graph curves.
Quick check
Quick check
?The current–voltage graph for a metal wire at constant temperature is a straight line through the origin. What does this show?
Calculate
Calculate
#A resistor of 48 Ω carries a current of 0.25 A. Find the potential difference across it.
V
Hint: V = IR = 0.25 × 48.
Calculate
Calculate
#A current of 2.0 A flows for 5.0 minutes. Find the charge that passes (in coulombs).
C
Hint: Q = It, with t = 5.0 × 60 = 300 s → 2.0 × 300.
B.5 · power
Electrical power
Electrical power is the rate of energy transfer. Three equivalent forms (combine with V = IR):
P = VI = I²R = V² ÷ Rpower in watts (W)
Worked example — a kettle
A kettle runs at 230 V drawing 3.0 A.
P = VI = 230 × 3.0 = 690 W
Calculate
Calculate
#A heater operates at 230 V and draws 3.0 A. Find its power.
W
Hint: P = VI = 230 × 3.0.
B.5 · networks
Series and parallel
Two arrangements, with opposite rules:
Series: same current everywhere; pds add; resistances add: R = R₁ + R₂ + …
Parallel: same pd across each branch; currents add at junctions; 1/R = 1/R₁ + 1/R₂ + …
Adding resistors in parallel gives a total resistance smaller than the smallest branch — there are more paths for charge to flow.
Calculate
Calculate
#Three resistors of 10 Ω, 20 Ω and 30 Ω are connected in series. Find the total resistance.
Ω
Hint: in series, R = R₁ + R₂ + R₃ = 10 + 20 + 30.
Calculate
Calculate
#Two 6.0 Ω resistors are connected in parallel. Find the total resistance.
Ω
Hint: 1/R = 1/6 + 1/6 = 2/6, so R = 3.0 Ω.
Sort it
Series, parallel, or always true?
Tap an item, then tap the group it belongs to.
🔗 True in series
🌿 True in parallel
✔️ Always true
B.5 · internal resistance
EMF and internal resistance
A real cell has an internal resistance r, so some energy is lost inside it. Its emf ε is the total energy per coulomb; the terminal pd is what the external circuit gets.
ε = I(R + r) = V + Iremf = terminal pd + "lost volts" across r
Worked example — a loaded cell
A cell of emf 12 V and internal resistance r = 1.0 Ω drives a 5.0 Ω resistor.
I = ε ÷ (R + r) = 12 ÷ 6.0 = 2.0 A
Terminal pd = IR = 2.0 × 5.0 = 10 V
Calculate
Calculate
#A cell of emf 12 V and internal resistance 1.0 Ω is connected to a 5.0 Ω resistor. Find the terminal pd across the resistor.
V
Hint: I = 12 ÷ (5.0+1.0) = 2.0 A; terminal pd = IR = 2.0 × 5.0.
Match it
Match the equation to its meaning
Tap a statement on the left, then its match on the right.
Statement
Answer
Recap
The big ideas to know
Basics: I = Δq/Δt; V = IR; ohmic ⇒ constant R
Power: P = VI = I²R = V²/R
Series: same current; resistances add
Parallel: same pd; 1/R = Σ1/Rᵢ; total < smallest
Real cells: ε = I(R + r); terminal pd = ε − Ir
That completes Electric Current & Circuits for IB Diploma Physics HL. Press Finish to see your score.
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