IB Physics Topic 2.5: Current and circuits | Notes & Flashcards | Aimnova

IB Physics — Current and circuits

Topic 2.5 of IB Physics covers Current and circuits, which is part of Unit 2: The particulate nature of matter. Students explore key concepts including Current, charge and potential difference, Resistance, Ohm's law and resistivity, Series and parallel circuits, and more. A strong understanding of current and circuits is essential for IB Physics exams and builds the foundation for connected topics across the syllabus.

Key concepts in Current and circuits

Key Idea: This topic builds a complete picture of an electric circuit: charge flowing as a current, voltage as the energy each coulomb carries, resistance opposing the flow, and power as the energy delivered each second. It finishes with real cells, which lose a little voltage inside themselves. It is examined on both papers — quick Paper 1A multiple-choice (rank resistor combinations, read a resistance off an I–V graph, why terminal p.d. drops) and longer Paper 2 structured questions ('show that' a current, an equivalent resistance, a power ratio, or an internal resistance).

📋 Key formulas

Almost all of these are given in the data booklet (look for the booklet badge). The only one you must build yourself is the terminal-p.d. form, which drops straight out of ε = I(R + r).

I = \frac{\Delta q}{\Delta t}

Current = charge ÷ time. Current is the rate of flow of charge. Given in the data booklet.

$I$: electric current (A, amperes)
$\Delta q$: charge that flows past a point (C, coulombs)
$\Delta t$: time for that charge to flow (s)

V = \frac{W}{q}

Potential difference = energy ÷ charge — the energy given to each coulomb. Given. So 1 V = 1 J per coulomb.

$V$: potential difference / voltage (V, volts)
$W$: energy given to the charge (J, joules)
$q$: amount of charge moved (C, coulombs)

V = IR

Ohm's law (given as R = V ÷ I). Resistance is the voltage across a component divided by the current through it.

$V$: potential difference across the component (V)
$I$: current through the component (A)
$R$: resistance (Ω, ohms)

R = \frac{\rho L}{A}

Resistance of a wire (given as ρ = RA ÷ L). Longer wire → more resistance; thicker wire (bigger area A) → less.

$R$: resistance of the wire (Ω)
$\rho$: resistivity of the material (Ω m) — a property of the substance
$L$: length of the wire (m)
$A$: cross-sectional area of the wire (m²)

R_s = R_1 + R_2 + \ldots

Resistors in SERIES: the resistances simply add. Same current everywhere; the p.d.s add up.

$R_s$: total (equivalent) resistance of resistors in series (Ω)
$R_1, R_2, \ldots$: the individual resistances (Ω)

\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \ldots

Resistors in PARALLEL: the reciprocals add. Same voltage across each; the branch currents add. Remember to FLIP at the end.

$R_p$: total (equivalent) resistance of resistors in parallel (Ω)
$R_1, R_2, \ldots$: the individual resistances (Ω)

P = IV = I^{2}R = \frac{V^{2}}{R}

Electrical power — three equivalent given forms. Pick the one matching the quantities you already know.

$P$: electrical power — energy transferred per second (W, watts)
$I$: current through the component (A)
$V$: potential difference across the component (V)
$R$: resistance of the component (Ω)

E = Pt

Electrical energy = power × time. Use kW and hours to get kilowatt-hours (kWh) for an electricity bill.

$E$: electrical energy transferred (J; or kWh for bills)
$P$: power (W; or kW for bills)
$t$: time the component runs (s; or hours for bills)

\varepsilon = I(R + r)

A real cell drives current through BOTH the load R and its own internal resistance r. Given.

$\varepsilon$: emf of the cell (V) — the energy it gives each coulomb
$I$: current drawn from the cell (A)
$R$: external (load) resistance (Ω)
$r$: internal resistance of the cell (Ω)

V = \varepsilon - Ir

Terminal p.d. = emf minus the 'lost volts' Ir. Not given separately — multiply out ε = I(R + r) = IR + Ir.

$V$: terminal p.d. — the voltage actually delivered to the circuit (V)
$\varepsilon$: emf of the cell (V)
$Ir$: 'lost volts' used up inside the cell (V)

⚖️ The five ideas side by side

🔀 Series vs parallel — the rules in full

After adding the reciprocals you have 1/Rₚ, not Rₚ. You must flip (take the reciprocal) for the final answer. Sanity check: a parallel combination is always smaller than the smallest resistor in it. Two equal resistors in parallel give exactly half of one.

✏️ Worked exam-style questions

IB-style question — resistance of a wire, then make it thicker

A copper wire has length 15 m and cross-sectional area 2.0 × 10⁻⁶ m². Copper has resistivity 1.7 × 10⁻⁸ Ω m. (a) Find the wire's resistance. (b) State the resistance of an identical wire whose cross-sectional area is doubled.

Solution:

(a) Start with the given resistivity equation:
$R = \frac{\rho L}{A}$
(a) Put in the numbers (ρ = 1.7 × 10⁻⁸, L = 15, A = 2.0 × 10⁻⁶):
$R = \frac{(1.7\times10^{-8})(15)}{2.0\times10^{-6}}$
(a) Work it out — keep the unit:
$R = 0.13\ \Omega$
(b) R is inversely proportional to area A, so doubling A halves R:
$R = \frac{0.13}{2} = 0.064\ \Omega$

Final answer:

(a) R ≈ 0.13 Ω. (b) R ≈ 0.064 Ω — a thicker wire (bigger area) carries current more easily, so it has less resistance.

IB-style question — current in a series-parallel network

A 12 V supply (negligible internal resistance) drives a 30 Ω resistor in series with two 60 Ω resistors that are in parallel with each other. (a) Find the total resistance. (b) Find the current drawn from the supply. (c) Find the current in each 60 Ω branch.

Solution:

(a) Combine the parallel pair first with the given rule:
$\frac{1}{R_p} = \frac{1}{60} + \frac{1}{60} = \frac{2}{60} \;\Rightarrow\; R_p = 30\ \Omega$
(a) Now add the 30 Ω in series with the given series rule:
$R_s = 30 + R_p = 30 + 30 = 60\ \Omega$
(b) Use the given Ohm's law, rearranged for current:
$I = \frac{V}{R_s} = \frac{12}{60} = 0.20\ \text{A}$
(c) The p.d. across the parallel pair is Vₚ = I × Rₚ = 0.20 × 30 = 6.0 V; each branch feels this same 6.0 V:
$I_{branch} = \frac{V_p}{60} = \frac{6.0}{60} = 0.10\ \text{A}$

Final answer:

(a) 60 Ω. (b) 0.20 A from the supply. (c) 0.10 A in each branch — the 0.20 A splits equally between the two equal branches.

IB-style question — power ratio of two heating wires

A heater is a wire of resistance R connected across the mains, dissipating 1200 W at 230 V. A second heater uses the same wire material and the same cross-sectional area, but with three times the length, connected across the same 230 V. Determine the power of the second heater.

Solution:

The voltage is the same for both, so use the given power form with V and R:
$P = \frac{V^{2}}{R}$
Same material and area, triple the length ⇒ triple the resistance (R ∝ L), so R₂ = 3R₁:
$\frac{P_2}{P_1} = \frac{V^{2}/R_2}{V^{2}/R_1} = \frac{R_1}{R_2} = \frac{R_1}{3R_1} = \frac{1}{3}$
Scale the first heater's power down by 3:
$P_2 = \frac{1200}{3} = 400\ \text{W}$

Final answer:

P₂ = 400 W. At the same voltage, power is inversely proportional to resistance — and a longer wire has more resistance, so it dissipates less power.

IB-style question — internal resistance of a cell

A cell of emf 1.5 V drives a current of 0.30 A through a 4.5 Ω resistor. (a) Find the internal resistance of the cell. (b) Find the terminal potential difference across the cell.

Solution:

(a) Start with the given emf equation:
$\varepsilon = I(R + r)$
(a) Substitute (ε = 1.5, I = 0.30, R = 4.5) and divide through by I:
$\frac{1.5}{0.30} = 4.5 + r \;\Rightarrow\; 5.0 = 4.5 + r$
(a) Rearrange for r — keep the unit:
$r = 5.0 - 4.5 = 0.50\ \Omega$
(b) Terminal p.d. = emf minus the lost volts:
$V = \varepsilon - Ir = 1.5 - (0.30)(0.50) = 1.35\ \text{V}$

Final answer:

(a) r = 0.50 Ω. (b) V = 1.35 V — equal to IR = 0.30 × 4.5 across the external resistor, a little less than the 1.5 V emf because 0.15 V is lost inside the cell.

🧠 Quick self-check

Tap each card to reveal the answer.

🎯 Exam tips

Exam Tips

Keep the two definitions straight: current = charge ÷ time (I = Δq/Δt, measured THROUGH, ammeter in series); voltage = energy ÷ charge (V = W/q, measured ACROSS, voltmeter in parallel).
Read resistance off an I–V graph as V ÷ I at a point. A straight line through the origin is ohmic (constant R); a curve (e.g. a filament lamp) is non-ohmic.
In R = ρL/A: longer ⇒ more resistance (R ∝ L), thicker ⇒ less (R ∝ 1/A). Resistivity ρ is a property of the material, not the shape.
Series adds resistances and shares the voltage at one current; parallel adds reciprocals and shares the current at one voltage. After 1/Rₚ, always FLIP — a parallel total is smaller than the smallest resistor.
Choose the power form by the quantities given: P = IV, P = I²R, or P = V²/R. For ratio questions, write the formula for each case and divide so the common factors cancel.
Electricity bills use kilowatt-hours: power in kW × time in HOURS = energy in kWh; cost = kWh × unit price. Convert minutes/seconds before using hours.
A real cell obeys ε = I(R + r). Don't forget r: the terminal p.d. is V = ε − Ir, always a little below the emf. Only when r ≈ 0 does V ≈ ε.

IB Physics — Current and circuits

Key concepts in Current and circuits

Key Idea: This topic builds a complete picture of an electric circuit: charge flowing as a current, voltage as the energy each coulomb carries, resistance opposing the flow, and power as the energy delivered each second. It finishes with real cells, which lose a little voltage inside themselves. It is examined on both papers — quick Paper 1A multiple-choice (rank resistor combinations, read a resistance off an I–V graph, why terminal p.d. drops) and longer Paper 2 structured questions ('show that' a current, an equivalent resistance, a power ratio, or an internal resistance).

📋 Key formulas

Almost all of these are given in the data booklet (look for the booklet badge). The only one you must build yourself is the terminal-p.d. form, which drops straight out of ε = I(R + r).

I = \frac{\Delta q}{\Delta t}

Current = charge ÷ time. Current is the rate of flow of charge. Given in the data booklet.

$I$: electric current (A, amperes)
$\Delta q$: charge that flows past a point (C, coulombs)
$\Delta t$: time for that charge to flow (s)

V = \frac{W}{q}

Potential difference = energy ÷ charge — the energy given to each coulomb. Given. So 1 V = 1 J per coulomb.

$V$: potential difference / voltage (V, volts)
$W$: energy given to the charge (J, joules)
$q$: amount of charge moved (C, coulombs)

V = IR

Ohm's law (given as R = V ÷ I). Resistance is the voltage across a component divided by the current through it.

$V$: potential difference across the component (V)
$I$: current through the component (A)
$R$: resistance (Ω, ohms)

R = \frac{\rho L}{A}

Resistance of a wire (given as ρ = RA ÷ L). Longer wire → more resistance; thicker wire (bigger area A) → less.

$R$: resistance of the wire (Ω)
$\rho$: resistivity of the material (Ω m) — a property of the substance
$L$: length of the wire (m)
$A$: cross-sectional area of the wire (m²)

R_s = R_1 + R_2 + \ldots

Resistors in SERIES: the resistances simply add. Same current everywhere; the p.d.s add up.

$R_s$: total (equivalent) resistance of resistors in series (Ω)
$R_1, R_2, \ldots$: the individual resistances (Ω)

\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \ldots

Resistors in PARALLEL: the reciprocals add. Same voltage across each; the branch currents add. Remember to FLIP at the end.

$R_p$: total (equivalent) resistance of resistors in parallel (Ω)
$R_1, R_2, \ldots$: the individual resistances (Ω)

P = IV = I^{2}R = \frac{V^{2}}{R}

Electrical power — three equivalent given forms. Pick the one matching the quantities you already know.

$P$: electrical power — energy transferred per second (W, watts)
$I$: current through the component (A)
$V$: potential difference across the component (V)
$R$: resistance of the component (Ω)

E = Pt

Electrical energy = power × time. Use kW and hours to get kilowatt-hours (kWh) for an electricity bill.

$E$: electrical energy transferred (J; or kWh for bills)
$P$: power (W; or kW for bills)
$t$: time the component runs (s; or hours for bills)

\varepsilon = I(R + r)

A real cell drives current through BOTH the load R and its own internal resistance r. Given.

$\varepsilon$: emf of the cell (V) — the energy it gives each coulomb
$I$: current drawn from the cell (A)
$R$: external (load) resistance (Ω)
$r$: internal resistance of the cell (Ω)

V = \varepsilon - Ir

Terminal p.d. = emf minus the 'lost volts' Ir. Not given separately — multiply out ε = I(R + r) = IR + Ir.

$V$: terminal p.d. — the voltage actually delivered to the circuit (V)
$\varepsilon$: emf of the cell (V)
$Ir$: 'lost volts' used up inside the cell (V)

⚖️ The five ideas side by side

🔀 Series vs parallel — the rules in full

After adding the reciprocals you have 1/Rₚ, not Rₚ. You must flip (take the reciprocal) for the final answer. Sanity check: a parallel combination is always smaller than the smallest resistor in it. Two equal resistors in parallel give exactly half of one.

✏️ Worked exam-style questions

IB-style question — resistance of a wire, then make it thicker

Solution:

(a) Start with the given resistivity equation:
$R = \frac{\rho L}{A}$
(a) Put in the numbers (ρ = 1.7 × 10⁻⁸, L = 15, A = 2.0 × 10⁻⁶):
$R = \frac{(1.7\times10^{-8})(15)}{2.0\times10^{-6}}$
(a) Work it out — keep the unit:
$R = 0.13\ \Omega$
(b) R is inversely proportional to area A, so doubling A halves R:
$R = \frac{0.13}{2} = 0.064\ \Omega$

Final answer:

(a) R ≈ 0.13 Ω. (b) R ≈ 0.064 Ω — a thicker wire (bigger area) carries current more easily, so it has less resistance.

IB-style question — current in a series-parallel network

Solution:

(a) Combine the parallel pair first with the given rule:
$\frac{1}{R_p} = \frac{1}{60} + \frac{1}{60} = \frac{2}{60} \;\Rightarrow\; R_p = 30\ \Omega$
(a) Now add the 30 Ω in series with the given series rule:
$R_s = 30 + R_p = 30 + 30 = 60\ \Omega$
(b) Use the given Ohm's law, rearranged for current:
$I = \frac{V}{R_s} = \frac{12}{60} = 0.20\ \text{A}$
(c) The p.d. across the parallel pair is Vₚ = I × Rₚ = 0.20 × 30 = 6.0 V; each branch feels this same 6.0 V:
$I_{branch} = \frac{V_p}{60} = \frac{6.0}{60} = 0.10\ \text{A}$

Final answer:

(a) 60 Ω. (b) 0.20 A from the supply. (c) 0.10 A in each branch — the 0.20 A splits equally between the two equal branches.

IB-style question — power ratio of two heating wires

Solution:

The voltage is the same for both, so use the given power form with V and R:
$P = \frac{V^{2}}{R}$
Same material and area, triple the length ⇒ triple the resistance (R ∝ L), so R₂ = 3R₁:
$\frac{P_2}{P_1} = \frac{V^{2}/R_2}{V^{2}/R_1} = \frac{R_1}{R_2} = \frac{R_1}{3R_1} = \frac{1}{3}$
Scale the first heater's power down by 3:
$P_2 = \frac{1200}{3} = 400\ \text{W}$

Final answer:

P₂ = 400 W. At the same voltage, power is inversely proportional to resistance — and a longer wire has more resistance, so it dissipates less power.

IB-style question — internal resistance of a cell

A cell of emf 1.5 V drives a current of 0.30 A through a 4.5 Ω resistor. (a) Find the internal resistance of the cell. (b) Find the terminal potential difference across the cell.

Solution:

(a) Start with the given emf equation:
$\varepsilon = I(R + r)$
(a) Substitute (ε = 1.5, I = 0.30, R = 4.5) and divide through by I:
$\frac{1.5}{0.30} = 4.5 + r \;\Rightarrow\; 5.0 = 4.5 + r$
(a) Rearrange for r — keep the unit:
$r = 5.0 - 4.5 = 0.50\ \Omega$
(b) Terminal p.d. = emf minus the lost volts:
$V = \varepsilon - Ir = 1.5 - (0.30)(0.50) = 1.35\ \text{V}$

Final answer:

(a) r = 0.50 Ω. (b) V = 1.35 V — equal to IR = 0.30 × 4.5 across the external resistor, a little less than the 1.5 V emf because 0.15 V is lost inside the cell.

🧠 Quick self-check

Tap each card to reveal the answer.

🎯 Exam tips

Exam Tips

Keep the two definitions straight: current = charge ÷ time (I = Δq/Δt, measured THROUGH, ammeter in series); voltage = energy ÷ charge (V = W/q, measured ACROSS, voltmeter in parallel).
Read resistance off an I–V graph as V ÷ I at a point. A straight line through the origin is ohmic (constant R); a curve (e.g. a filament lamp) is non-ohmic.
In R = ρL/A: longer ⇒ more resistance (R ∝ L), thicker ⇒ less (R ∝ 1/A). Resistivity ρ is a property of the material, not the shape.
Series adds resistances and shares the voltage at one current; parallel adds reciprocals and shares the current at one voltage. After 1/Rₚ, always FLIP — a parallel total is smaller than the smallest resistor.
Choose the power form by the quantities given: P = IV, P = I²R, or P = V²/R. For ratio questions, write the formula for each case and divide so the common factors cancel.
Electricity bills use kilowatt-hours: power in kW × time in HOURS = energy in kWh; cost = kWh × unit price. Convert minutes/seconds before using hours.
A real cell obeys ε = I(R + r). Don't forget r: the terminal p.d. is V = ε − Ir, always a little below the emf. Only when r ≈ 0 does V ≈ ε.

Key concepts in Current and circuits

📋 Key formulas

⚖️ The five ideas side by side

🔀 Series vs parallel — the rules in full

✏️ Worked exam-style questions

IB-style question — resistance of a wire, then make it thicker

IB-style question — current in a series-parallel network

IB-style question — power ratio of two heating wires

IB-style question — internal resistance of a cell

🧠 Quick self-check

🎯 Exam tips

Exam Tips

What you'll learn in Topic 2.5

Study resources — 2.5 Current and circuits

Current, charge and potential difference

Resistance, Ohm's law and resistivity

Series and parallel circuits

Electrical power and energy

EMF and internal resistance

Ready to study Current and circuits?

Ready to practice?

Key concepts in Current and circuits

📋 Key formulas

⚖️ The five ideas side by side

🔀 Series vs parallel — the rules in full

✏️ Worked exam-style questions

IB-style question — resistance of a wire, then make it thicker

IB-style question — current in a series-parallel network

IB-style question — power ratio of two heating wires

IB-style question — internal resistance of a cell

🧠 Quick self-check

🎯 Exam tips

Exam Tips

What you'll learn in Topic 2.5

Study resources — 2.5 Current and circuits

Current, charge and potential difference

Resistance, Ohm's law and resistivity

Series and parallel circuits

Electrical power and energy

EMF and internal resistance

Ready to study Current and circuits?

Ready to practice?