B.5 Current and circuits · Educerie Physics

Educerie · IB Diploma · Physics

Theme B The particulate nature of matter · B.5 Current and circuits

Level: SL and HL. Nothing here is HL only, so every section is examinable for both.
Themes (key concepts): particles, energy and forces. A current is moving charged particles, resistance is those particles colliding with the lattice they move through, and every circuit calculation is the bookkeeping of energy handed to charge by a cell and given up in components.
The question this unit answers: how do charged particles flow through materials, how do we measure how easily they flow, and what does resistance cost us?
Where it is examined: Paper 1A multiple choice (series and parallel combinations, meter readings, power, typically several questions per paper); Paper 1B, where the emf and internal resistance experiment or a resistivity experiment is a common data set (gradient, intercept, uncertainty); Paper 2 structured questions on circuits, worth 4 to 8 marks across their parts.

What you must be able to do

You must be able to	Level	What it looks like in the exam
Use I = Δq/Δt, and explain conductors and insulators by the mobility of charge carriers	SL, HL	"Calculate the number of electrons passing a point each second" (2 marks)
Use V = W/q, and distinguish potential difference from emf	SL, HL	"Distinguish between emf and potential difference" (2 marks)
Outline chemical and solar cells as energy sources, with advantages and disadvantages	SL, HL	"Outline one advantage and one disadvantage of a solar cell" (2 marks)
Draw and read circuit diagrams using the data booklet symbols, with meters placed correctly	SL, HL	"Draw a circuit that could be used to measure…" (3 marks)
Explain the origin of resistance; use R = V/I and ρ = RA/L	SL, HL	Paper 2 resistivity calculation (2 or 3 marks)
State Ohm's law and distinguish ohmic from non-ohmic behaviour, including the heating effect	SL, HL	"Sketch the I–V characteristic of a filament lamp and explain its shape" (4 marks)
Use P = IV = I²R = V²/R	SL, HL	1 or 2 marks, often inside a longer question
Combine resistors in series and in parallel and solve networks	SL, HL	Paper 1A, and Paper 2 multi-step questions
Use ε = I(R + r) for a cell with internal resistance, and analyse V–I data	SL, HL	Paper 1B graph: ε from the intercept, r from the gradient
Describe thermistors, LDRs and potentiometers, and use them in circuits	SL, HL	"Explain how the output pd changes when the temperature rises" (3 marks)

Before you start

You need energy and work from A.3 (a joule is a newton metre, and power is energy per second) and the particle model of solids from B.1. Electric charge is measured in coulombs, C; the charge on an electron is −1.60 × 10⁻¹⁹ C, and the size of that charge is written e. You will meet electric fields properly in D.2; here it is enough that a potential difference across a wire pushes charge along it.

1The idea in one paragraph

An electric current is a flow of charge. In a metal wire the moving charges are free electrons, pushed along by the cell. The cell gives each coulomb of charge some energy, its emf; the charge gives that energy up as it passes through the components, and the energy given up per coulomb between two points is the potential difference between them. How much current flows for a given pd is set by the resistance, which comes from the electrons colliding with the vibrating ions of the metal and depends on the material (its resistivity), its length and its thickness. Every collision turns electrical energy into internal energy, which is the heating effect, and the rate of that transfer is the power. Resistors combine in series and in parallel by two simple rules, and even the cell itself has some resistance inside it, which is why its terminal pd falls when it delivers more current.

2Charge carriers and current

Current is the rate at which charge flows past a point:

I = Δq ÷ Δt

Its unit is the ampere, A, which is one coulomb per second. The charges that move are called charge carriers.

Whether a material conducts depends on whether it has charge carriers that are free to move, what the guide calls their mobility. Figure 1 shows the difference.

Figure 1 · Why a metal conducts and a plastic does not

In a conductor such as a metal, each atom gives up one or more outer electrons to a shared "sea" of free electrons. The positive ions stay fixed in a lattice. Apply a pd and the free electrons drift along the wire, giving a current. Metals have an enormous number of free electrons per cubic metre, so they conduct very well.
In an insulator such as plastic, glass or dry wood, every electron is bound to its own atom or molecule. There are almost no free charge carriers, so apply a pd and essentially no current flows.

Charge carriers need not be electrons. In a salt solution they are positive and negative ions moving in opposite directions; in a semiconductor there are fewer free carriers than in a metal, and their number can rise with temperature or light, which section 10 uses.

Conventional current is defined as the direction positive charge would flow: out of the positive terminal of a cell, round the circuit, and into the negative terminal. In a metal the electrons actually move the other way. Both descriptions give the same answers; just use conventional current on circuit diagrams.

Worked example 1. A charge of 360 C passes through a heater in 2.0 minutes. Find the current and the number of electrons passing any point each second.

I = Δq ÷ Δt = 360 ÷ 120 = 3.0 Atime in seconds

electrons per second = 3.0 ÷ (1.60 × 10⁻¹⁹) = 1.9 × 10¹⁹

3Potential difference, emf and cells

The potential difference V between two points is the work done (energy transferred) per unit charge in moving a positive charge between them, along the path of the current:

V = W ÷ q

Its unit is the volt, V, which is one joule per coulomb. If 24 J of energy is transferred to a lamp as 2.0 C of charge passes through it, the pd across the lamp is 24 ÷ 2.0 = 12 V.

The emf, ε, of a source is the same kind of quantity seen from the other side: the energy the source transfers to each coulomb of charge, from some other form (chemical, light) into electrical energy. So pd is energy per coulomb given up by charge in a component; emf is energy per coulomb given to charge by a source. Both are measured in volts, and a good exam answer to "distinguish" says exactly that.

The guide names two sources. A chemical cell turns chemical energy into electrical energy through reactions at its electrodes. A solar cell (photovoltaic cell) turns light energy directly into electrical energy in a semiconductor. The guide requires the advantages and disadvantages of these sources.

Source	Advantages	Disadvantages
Primary chemical cell (single use)	portable; ready to use; stores energy for years; cheap	used once, then waste; some contain toxic metals; emf falls as it runs down
Rechargeable chemical cell	portable; reused hundreds of times, so less waste and lower running cost	more expensive; capacity falls with age; must be charged from another source
Solar cell	renewable; no fuel and no emissions while operating; little maintenance	no output at night and less in cloud; small emf per cell, so many are needed; large area; needs storage for when it is dark

The linking questions ask about the advantages of cells more widely: a cell is portable and delivers direct current with no moving parts, which is why phones, watches and spacecraft use them.

4Circuit diagrams, and where the meters go

A circuit diagram shows how components are connected, not where they sit on the bench. The guide says to use the symbols in the physics data booklet; Figure 2 shows the ones this subtopic needs. Wires are straight lines with right-angled corners, and a dot marks a junction where wires join.

Figure 2 · The circuit symbols this subtopic uses

Two meters, two rules. Figure 3 shows both.

Figure 3 · Ammeter in series, voltmeter in parallel

An ammeter measures current, so all the current must pass through it: it goes in series. An ideal ammeter has zero resistance, so it does not change the current it measures.
A voltmeter measures the pd between two points, so it connects across them, in parallel with the component. An ideal voltmeter has infinite resistance, so no current flows through it.

Unless a question says otherwise, meters are ideal. When they are not, the guide says their resistance is treated as a constant, and you include it like any other resistor (Q6 does this).

5Resistance, and where it comes from

The resistance R of a component is the ratio of the pd across it to the current through it:

R = V ÷ I

Its unit is the ohm, Ω, one volt per ampere.

Where resistance comes from. As the free electrons drift through a metal they keep colliding with the positive ions of the lattice. Each collision transfers some of the energy they gained from the pd to the ions, which vibrate more. So resistance is the obstruction the lattice presents to the drift of charge, and the energy lost in the collisions ends up as internal energy of the metal: the conductor warms. That is the heating effect of a current. It is also why the resistance of a metal rises with temperature: hotter ions vibrate more widely and are harder to get past.

Resistivity. Two wires of the same metal can have very different resistances. A longer wire has more lattice to cross, so R ∝ L. A thicker wire offers more paths side by side, so R ∝ 1/A, where A is the cross-sectional area. The material itself is described by its resistivity ρ:

ρ = RA ÷ L, so R = ρL ÷ A

Resistivity is a property of the material, not of the particular wire, and its unit is the ohm metre, Ω m. Copper is about 1.7 × 10⁻⁸ Ω m; typical insulators are more than 10¹⁰ Ω m.

Worked example 2: a copper wire. Find the resistance of 20 m of copper wire of diameter 0.50 mm.

A = π r² = π × (0.25 × 10⁻³)² = 1.96 × 10⁻⁷ m²radius, in metres

R = ρL ÷ A = 1.7 × 10⁻⁸ × 20 ÷ (1.96 × 10⁻⁷)

R = 1.7 Ω

Worked example 3: finding ρ from a measurement. A wire 0.80 m long with a diameter of 0.30 mm has a resistance of 5.4 Ω. Find the resistivity of its material.

A = π × (0.15 × 10⁻³)² = 7.07 × 10⁻⁸ m²

ρ = RA ÷ L = 5.4 × 7.07 × 10⁻⁸ ÷ 0.80

ρ = 4.8 × 10⁻⁷ Ω m

Two traps live in these examples. Use the radius, not the diameter, in πr². And convert millimetres to metres before squaring: 0.25 mm is 0.25 × 10⁻³ m, and its square is 6.25 × 10⁻⁸ m², not 6.25 × 10⁻⁵.

6Ohm's law, and ohmic and non-ohmic behaviour

Ohm's law states that the current through a conductor is directly proportional to the potential difference across it, provided its temperature (and other physical conditions) stay constant. A component that obeys it is ohmic; its resistance is constant. The guide says that a metal conductor at constant temperature is to be treated as ohmic.

Note what the law is not. R = V/I is a definition that holds for every component at every instant. Ohm's law is the extra claim that R stays the same as V changes. A filament lamp has a resistance at every point, but it is not ohmic.

The way to see the difference is an I–V characteristic: a graph of current against pd. Figure 4 compares the two cases the guide cares about.

Figure 4 · Ohmic and non-ohmic: I–V characteristics

Ohmic resistor at constant temperature: a straight line through the origin. Reverse the pd and the current reverses with the same size, so the line continues into the third quadrant.
Filament lamp: the current rises less and less steeply as the pd increases. More current means more collisions per second, so more heating; the filament's temperature rises, its ions vibrate more, and its resistance goes up. The same happens in reverse, so the curve has the same shape in the third quadrant.

Read resistance from V ÷ I at a point, never from the gradient of the tangent. A lamp rated "6.0 V, 3.0 W" has R = V²/P = 36 ÷ 3.0 = 12 Ω when lit; its cold resistance is much lower, which is why lamps usually fail at the moment they are switched on.

7Electrical power and the heating effect

Energy transferred per second in a component is its power. Since V is energy per coulomb and I is coulombs per second, their product is energy per second:

P = IV = I²R = V²/R

The second and third forms come from substituting V = IR. Choose the one that uses the quantities you know. I²R is useful for a series circuit, where I is the same everywhere; V²/R suits parallel branches, which share the same V. The energy transferred in time t is Pt.

Worked example 4: a kettle. A kettle is rated 2.3 kW at 230 V. Find the current, its resistance, and the energy it transfers in 3.0 minutes.

I = P ÷ V = 2300 ÷ 230 = 10 A

R = V² ÷ P = 230² ÷ 2300 = 23 Ω

E = Pt = 2300 × 180 = 4.1 × 10⁵ J

In a kettle the heating effect is the point. In power cables it is waste: for a given power, a higher transmission voltage means a smaller current, and the I²R loss falls with the square of the current. One of the guide's linking questions asks how the heating of a resistor can be explained from other areas of physics: it is the random kinetic energy of the lattice ions rising, which is internal energy from B.1.

8Series and parallel combinations

Two rules underlie every circuit: charge is conserved (whatever flows into a junction flows out) and energy is conserved (round any loop, the energy given to each coulomb equals the energy it gives up). Figure 5 turns them into the formulas in the data booklet.

Figure 5 · Series and parallel: what is shared and what is the same

In series there is one path. The same current flows through every component, and the pd of the source is shared between them:

I = I₁ = I₂ · V = V₁ + V₂ · Rₛ = R₁ + R₂ + …

In parallel each branch has the same pd across it, and the current divides between the branches:

I = I₁ + I₂ · V = V₁ = V₂ · 1/Rₚ = 1/R₁ + 1/R₂ + …

Two checks save marks. Adding a resistor in series always increases the total resistance. Adding a resistor in parallel always decreases it, because it opens another path: the combined resistance is less than the smallest single branch. For two equal resistors R in parallel the result is R/2.

Worked example 5: a network. Figure 6 shows a 9.0 V cell of negligible internal resistance connected to a 6.0 Ω resistor in series with a 12 Ω and a 4.0 Ω resistor in parallel. Find the current from the cell and the current in each branch.

Figure 6 · A network to solve: 6.0 Ω in series with 12 Ω and 4.0 Ω in parallel

1/R_p = 1/12 + 1/4.0 = 1/12 + 3/12 = 4/12 → R_p = 3.0 Ω

R_total = 6.0 + 3.0 = 9.0 Ω

I = V ÷ R = 9.0 ÷ 9.0 = 1.0 Acurrent from the cell

pd across 6.0 Ω = 1.0 × 6.0 = 6.0 V

pd across the parallel pair = 9.0 − 6.0 = 3.0 V

I₁₂ = 3.0 ÷ 12 = 0.25 A, I₄ = 3.0 ÷ 4.0 = 0.75 A

Check: 0.25 + 0.75 = 1.0 A, which is the current that went in. Always finish a network with a check like this. Notice that the smaller resistor takes the larger share of the current.

9Emf and internal resistance

A real cell has some resistance inside it, its internal resistance r, from the materials the charge must pass through within the cell. Energy is transferred to internal energy there as well, so less reaches the external circuit. Figure 7(a) draws a real cell as an ideal emf ε in series with r.

Figure 7 · Emf and internal resistance: the circuit and the graph

Applying conservation of energy round the loop, the emf is shared between the external resistance R and the internal resistance r:

ε = I(R + r)

The pd measured across the cell's terminals, the terminal pd, is what reaches the external circuit: V = IR = ε − Ir. The term Ir is sometimes called the "lost volts". When no current flows, V = ε; the more current the cell delivers, the lower its terminal pd.

Worked example 6. A cell of emf 1.5 V and internal resistance 0.50 Ω is connected to a 2.5 Ω resistor. Find the current, the terminal pd and the power delivered to the resistor.

I = ε ÷ (R + r) = 1.5 ÷ (2.5 + 0.50) = 0.50 A

V = IR = 0.50 × 2.5 = 1.25 Vor ε − Ir = 1.5 − 0.25

P = I² R = 0.50² × 2.5 = 0.63 W

Measuring ε and r. This is a standard experiment and a favourite Paper 1B data set. Vary R, and record the terminal pd V and the current I. Rearranged, the equation is a straight line:

V = −r I + εcompare y = mx + c

Plot V against I, as in Figure 7(b). The vertical intercept is ε and the gradient is −r. For the invented data plotted there, the intercept is 1.50 V and the gradient is −0.72 V A⁻¹, so r = 0.72 Ω. Take readings quickly and open the switch between them: a cell delivering a large current warms up and its internal resistance changes.

10Resistors that change: thermistors, LDRs and potentiometers

The guide limits variable resistors to three kinds. Figure 8 shows how the first two respond to their surroundings.

Figure 8 · Two resistors that respond to their surroundings

A thermistor (the negative temperature coefficient type) is made of a semiconductor. As its temperature rises, more charge carriers are released, so its resistance falls, steeply. It is used to sense temperature.
A light-dependent resistor (LDR) is also a semiconductor. Light releases charge carriers, so its resistance falls as the light intensity increases: high in the dark, low in bright light.
A potentiometer is a resistor with a sliding contact. Used with all three terminals, it gives an adjustable fraction of the pd across it; used with two, it is simply a variable resistor.

These components earn their place in a potential divider: two resistors in series across a supply, with the output taken across one of them. Because the current is the same through both, the supply pd is shared in the ratio of the resistances. Figure 9 uses a thermistor to make the output depend on temperature.

Figure 9 · A potential divider that senses temperature

Worked example 7. In Figure 9, the supply is 6.0 V and the fixed resistor is 2.0 kΩ. Find the output pd across the thermistor when its resistance is 4.0 kΩ (cold) and 1.0 kΩ (warm).

V_out = V × R_therm ÷ (R_fixed + R_therm)

cold: V_out = 6.0 × 4.0 ÷ (2.0 + 4.0) = 4.0 V

warm: V_out = 6.0 × 1.0 ÷ (2.0 + 1.0) = 2.0 V

As the thermistor warms, its resistance falls, it takes a smaller share of the 6.0 V, and the output falls. Swap the thermistor and fixed resistor and the output rises with temperature instead; replace the thermistor with an LDR and the circuit senses light. A potentiometer across a supply does the same job by hand, which is how a volume control works.

11Where marks are lost

Putting the voltmeter in series or the ammeter in parallel. Ammeter in series, through it passes the whole current; voltmeter in parallel, across the component.
Confusing emf and pd. Emf is energy per coulomb given to charge by a source; pd is energy per coulomb transferred from charge to a component. "Emf is the voltage of the battery" scores nothing.
Using the diameter in A = πr², or squaring before converting mm to m. Halve first, convert to metres, then square.
Saying a filament lamp obeys Ohm's law "because R = V/I". R = V/I defines resistance for any component; Ohm's law requires R to be constant, and for a lamp it is not.
Reading resistance from the gradient of an I–V graph. Resistance at a point is V ÷ I at that point. The gradient of a curved I–V graph is not 1/R.
Adding parallel resistances like series ones. 1/Rₚ = 1/R₁ + 1/R₂; and remember to invert at the end. A parallel combination is always smaller than its smallest branch.
Forgetting internal resistance, or putting r in parallel. r is in series with the emf, so the current through it is the full circuit current.
"Electrons are used up in the lamp." Charge is conserved; the current is the same on both sides of a series component. It is energy that is transferred, not charge.

12Draw it right

Use the data booklet symbols exactly (Figure 2). A resistor is a rectangle, not a zigzag; the long plate of a cell is the positive terminal.
Draw wires straight, with right-angled corners, and put a dot at every junction.
Ammeters in series, voltmeters in parallel across the component they measure.
On an I–V graph, label the axes I / A and V / V. An ohmic resistor is a straight line through the origin; a filament lamp is a curve whose gradient decreases, with rotational symmetry through the origin.
For the internal resistance experiment, show the voltmeter across the cell's terminals, the ammeter in series and a variable resistor to change the current. On the V–I graph, mark the intercept as ε and the gradient as −r.
Thermistor and LDR graphs: resistance on the vertical axis, falling curves that never reach zero.

13Try it

Marks in brackets. Take e = 1.60 × 10⁻¹⁹ C. Meters are ideal unless stated. Answers and marker's notes are at the end.

Q1. (Paper 1A style) Three identical resistors, each of resistance R, are connected so that two of them are in parallel and that pair is in series with the third. What is the total resistance? 1 mark

A. R/3    B. 2R/3    C. 3R/2    D. 3R

Q2. A heating element is made from 2.0 m of wire with resistivity 1.1 × 10⁻⁶ Ω m. Its resistance is 14 Ω.

(a) Calculate the diameter of the wire. 3 marks

(b) The element is connected to a 230 V supply. Calculate the power it dissipates, assuming its resistance stays the same. 1 mark

Q3. (a) Sketch the I–V characteristic of a filament lamp, for positive and negative pd. 2 marks

(b) Explain, in terms of the particles in the filament, the shape of your graph. 3 marks

Q4. A cell of emf 6.0 V and internal resistance 1.0 Ω is connected to a 5.0 Ω resistor.

(a) Calculate the current and the terminal pd. 2 marks

(b) Calculate the power dissipated in the 5.0 Ω resistor and in the cell. 2 marks

(c) A second 5.0 Ω resistor is connected in parallel with the first. Determine the new terminal pd, and explain why it has changed. 3 marks

Q5. (Paper 1B style) A student measures the terminal pd V of a battery for several values of current I. The data are invented for this question.

I / A	0.10	0.20	0.30	0.40	0.50
V / V	2.83	2.63	2.47	2.27	2.11

(a) Explain why a graph of V against I is expected to be a straight line. 1 mark

(b) Using the first and last readings, determine the internal resistance and the emf of the battery. 3 marks

Q6. (a) Two 20 kΩ resistors are connected in series across a 12 V supply of negligible internal resistance. A voltmeter of resistance 20 kΩ is connected across one of them. Determine the reading on the voltmeter, and state what an ideal voltmeter would read. 3 marks

(b) Outline one advantage and one disadvantage of using solar cells rather than chemical cells to power a garden light. 2 marks

14In one breath

Current is the rate of flow of charge, I = Δq/Δt; metals conduct because they have free electrons, insulators do not because every electron is bound. Potential difference is energy transferred per coulomb, V = W/q; emf is the energy a source gives each coulomb, from chemical energy in a chemical cell or light in a solar cell. Ammeters go in series and voltmeters in parallel, and ideal meters do not disturb the circuit. Resistance, R = V/I, comes from free electrons colliding with lattice ions, which also heats the conductor; for a wire R = ρL/A, with ρ the resistivity of the material. Ohm's law says I ∝ V at constant temperature; a metal resistor obeys it, a filament lamp does not, because it heats and its resistance rises. Power is P = IV = I²R = V²/R. In series the current is the same everywhere and the pd is divided, Rₛ = R₁ + R₂; in parallel the pd is common and the current divides, 1/Rₚ = 1/R₁ + 1/R₂. A real cell has internal resistance: ε = I(R + r), so the terminal pd V = ε − Ir falls as current rises, and a V–I graph gives ε as the intercept and −r as the gradient. Thermistors and LDRs lose resistance as temperature or light rises, and in a potential divider they turn that change into a change of output pd.

Answers

Q1. C. The parallel pair has resistance R/2; adding the third in series gives R/2 + R = 3R/2. C only. A is all three in parallel, D all three in series, B two in series with one in parallel.

Q2. (a) A = ρL ÷ R = 1.1 × 10⁻⁶ × 2.0 ÷ 14 = 1.57 × 10⁻⁷ m². r = √(A/π) = 2.24 × 10⁻⁴ m, so the diameter is 4.5 × 10⁻⁴ m (0.45 mm). (b) P = V²/R = 230² ÷ 14 = 3.8 × 10³ W. (a) M1 for rearranging ρ = RA/L for A, M1 for r = √(A/π), A1 for 0.45 mm (giving the radius as the diameter loses the A1). (b) A1 for 3.8 kW.

Q3. (a) A curve through the origin whose gradient decreases as the size of the pd increases, the same shape in the third quadrant (rotational symmetry about the origin). (b) As the pd increases the current increases, so the free electrons collide more often with the lattice ions and transfer more energy to them each second. The filament's temperature rises, the ions vibrate with greater amplitude, and the electrons collide with them more often, so the resistance increases. The current therefore rises less than in proportion to the pd, and the graph curves towards the V axis. (a) 1 for the correct curvature, 1 for both quadrants with symmetry. (b) 1 for more current producing heating / higher temperature, 1 for greater lattice ion vibration causing more frequent collisions, 1 for linking the increased resistance to the decreasing gradient. "The lamp gets hot so the resistance goes up" scores 1.

Q4. (a) I = ε ÷ (R + r) = 6.0 ÷ 6.0 = 1.0 A; terminal pd = IR = 1.0 × 5.0 = 5.0 V. (b) In the resistor: I²R = 1.0² × 5.0 = 5.0 W. In the cell: I²r = 1.0² × 1.0 = 1.0 W. (c) External resistance = 2.5 Ω. I = 6.0 ÷ (2.5 + 1.0) = 1.71 A. Terminal pd = 1.71 × 2.5 = 4.3 V. The external resistance is lower, so the cell delivers more current; more pd is lost across the internal resistance (Ir = 1.7 V instead of 1.0 V), so less is left across the terminals. (a) 1 each. (b) 1 each. (c) M1 for 2.5 Ω and the new current, A1 for 4.3 V, R1 for the explanation in terms of a larger current and more lost volts.

Q5. (a) V = ε − Ir, which has the form y = c + mx with ε and r constant, so V against I is a straight line. (b) Gradient = (2.11 − 2.83) ÷ (0.50 − 0.10) = −1.8 V A⁻¹, so r = 1.8 Ω. Then ε = V + Ir = 2.83 + 0.10 × 1.8 = 3.0 V. (c) To stop the battery warming up while delivering current, which would change its internal resistance (and to stop it running down). (a) 1 for V = ε − Ir compared with a straight-line equation. (b) M1 for the gradient, A1 for r = 1.8 Ω (with the sign interpreted), A1 for ε = 3.0 V. (c) 1 for heating / changing r, or running down the battery.

Q6. (a) The voltmeter (20 kΩ) in parallel with one resistor (20 kΩ) gives 10 kΩ. The supply pd divides between 20 kΩ and 10 kΩ, so the reading is 12 × 10 ÷ 30 = 4.0 V. An ideal voltmeter, drawing no current, would read 6.0 V. (b) Advantage: any one of renewable, no running cost for fuel or replacement cells, no waste cells to dispose of. Disadvantage: any one of no output at night or less in cloud, so the light needs a rechargeable cell to store energy; higher initial cost; low emf per cell. (a) M1 for the 10 kΩ parallel combination, A1 for 4.0 V, A1 for 6.0 V. (b) 1 for a valid advantage, 1 for a valid disadvantage; each must compare with chemical cells, not just describe solar cells.

Educerie · written from the published IB Diploma Programme Physics guide, first assessment 2025, section B.5 Current and circuits. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.

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