DP Physics HL B.5 Current and Circuits

DP Physics HL B.5 Current and Circuits - Full notes

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DP Physics HL | B.5 Current and Circuits

DP PHYSICS · THEME B: THE PARTICULATE NATURE OF MATTER

B.5

Current and Circuits

Full notes for HL · IB-style

Contents

1	Charge, current and potential difference
2	Resistance
3	Series and parallel circuits
4	Electrical power and energy
5	Cells and internal resistance
6	Potential dividers and sensors
7	Exam technique
8	Data-based questions and further practice
9	Common misconceptions and links
10	Exam-style question with markscheme
11	Check your understanding
12	Summary and topic map

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and do every Quick check before moving to the next section.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
define current, pd and emf and use I = nAvq
use R = V/I and interpret I–V characteristics
use resistivity R = ρL/A and describe its measurement
analyse series, parallel and combined circuits using Kirchhoff's laws
calculate electrical power and energy
use emf and internal resistance, including V–I graphs
analyse potential dividers with LDRs and thermistors

There is no additional HL content in this subtopic. SL and HL students study the same material.

Guiding questions

How do charges move through circuits and transfer energy? How do the properties of components determine current and pd in a circuit? How are sensors used in potential divider circuits?

How this topic is examined

B.5 appears in Papers 1 and 2: circuit calculations, I–V graphs, resistivity experiments, internal resistance graphs and potential dividers with sensors.

Draw circuit diagrams with standard IB symbols and label currents and pds.

Command term	What the examiner wants
State	a brief answer with no explanation
Outline	a brief account or summary
Describe	a detailed account without reasons
Explain	a detailed account including causes, reasons or mechanisms
Calculate	a numerical answer with working and correct units
Determine	obtain the only possible answer, often from a graph or data
Show that	obtain the given result, showing every step (give one more significant figure)
Estimate	an approximate value using reasonable assumptions
Sketch	a graph or diagram showing the general shape and key features
Deduce / Suggest	reach a conclusion from given information; propose a reasonable explanation

How to use these notes

Each section teaches one part of the syllabus in full. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section
Key point	the one idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
In the real world	where the physics shows up outside the classroom
Try this	something quick you can do yourself to test the idea
Worked example	full solutions with IB-style marks (M1 method, A1 answer, R1 reasoning)
Quick check	short questions to test yourself at the end of each section

1 Charge, current and potential difference

The basic quantities of electric circuits.

Think about it

When you switch on a light, the bulb glows instantly, yet electrons drift through the wire at less than a millimetre per second. How?

Electric circuits transfer energy from a source to components. This section defines the quantities used to describe that transfer: charge, current, potential difference and emf.

1.1 Electric charge

Charge is measured in coulombs (C). It comes in units of the elementary charge e = 1.60 × 10⁻¹⁹ C. In metals the charge carriers are free electrons; in electrolytes and gases, ions. Charge is conserved in all circuits.

WORKED EXAMPLE 1 Number of electrons (basic)

How many electrons make up a charge of 1.0 C?

Solution

N = 1.0/1.60 × 10⁻¹⁹M1

= 6.25 × 10¹⁸A1

1.2 Electric current

Electric current is the rate of flow of charge: I = Δq/Δt, measured in amperes (1 A = 1 C s⁻¹). Conventional current flows from positive to negative terminal; electrons flow the other way.

WORKED EXAMPLE 2 Charge (standard)

A current of 0.40 A flows for 5.0 minutes. Find the charge that passes.

Solution

q = 0.40 × 300M1

= 120 CA1

WORKED EXAMPLE 3 Electrons per second (standard)

Find the number of electrons passing a point per second in a 2.0 A current.

Solution

2.0/1.60 × 10⁻¹⁹M1

= 1.25 × 10¹⁹ s⁻¹A1

1.3 Charge from a current–time graph

The area under a current–time graph equals the charge that has flowed. This is useful when the current changes, for example as a capacitor or battery discharges.

Figure 1. A decaying current; the shaded area is the charge delivered.

WORKED EXAMPLE 4 Estimating charge (exam level)

Estimate the total charge delivered in the first 10 s.

Solution

Area ≈ 0.50 × 3 × (1 − e−10/3)M1

≈ 1.4 CA1

1.4 Drift velocity

Free electrons move randomly at high speeds but drift slowly along the wire when a pd is applied. The current is I = nAvq, where n is the number of charge carriers per unit volume, A the cross-sectional area, v the drift speed and q the charge on each carrier.

Figure 2. Electrons drift opposite to the conventional current.

WORKED EXAMPLE 5 Copper wire (exam level)

A copper wire of area 1.0 mm² carries 3.0 A. Copper has 8.5 × 10²⁸ free electrons per m³. Find the drift speed.

Solution

v = I/nAq = 3.0/(8.5 × 10²⁸ × 1.0 × 10⁻⁶ × 1.6 × 10⁻¹⁹)M1

= 2.2 × 10⁻⁴ m s⁻¹A1

Quick check

1. If the same current flows through a thinner section of wire, what happens to the drift speed?

Answers: 1. it increases (v ∝ 1/A)

1.5 Potential difference and emf

Potential difference V is the energy transferred per unit charge between two points: V = W/q (1 V = 1 J C⁻¹). The emf ε of a source is the energy transferred into electrical energy per unit charge passing through it. Emf describes energy given to charges; pd describes energy taken from them in components.

WORKED EXAMPLE 6 Energy transferred (standard)

A 9.0 V battery drives 0.50 A for 2.0 minutes. Find the energy transferred.

Solution

q = 60 C; W = 9.0 × 60M1

= 540 JA1

WORKED EXAMPLE 7 Electronvolt (standard)

Find the energy gained by an electron accelerated through 2.5 kV in J and eV.

Solution

W = qV = 1.6 × 10⁻¹⁹ × 2500 = 4.0 × 10⁻¹⁶ JA1

= 2.5 keVA1

1.6 Measuring current and pd

Ammeters are connected in series and should have negligible resistance; voltmeters are connected in parallel and should have very high (ideally infinite) resistance so they draw almost no current.

Figure 3. An ammeter in series and a voltmeter in parallel with a resistor.

Common mistake

Connecting a voltmeter in series. Its very high resistance would stop the current.

2 Resistance

How components oppose current.

2.1 Resistance and Ohm's law

Resistance R = V/I, measured in ohms (Ω). Ohm's law states that for a conductor at constant temperature, the current is proportional to the pd; such a conductor is ohmic and has constant resistance.

WORKED EXAMPLE 8 Resistance (basic)

A component has 6.0 V across it and 0.25 A through it. Find its resistance.

Solution

R = 6.0/0.25M1

= 24 ΩA1

2.2 I–V characteristics

The I–V graph of a component reveals how its resistance behaves. Resistance at any point is V/I (not the gradient).

•Ohmic resistor: straight line through the origin; R constant.
•Filament lamp: curve flattens; as current rises the filament heats up and its resistance increases.
•Diode: almost no current in reverse; conducts above about 0.6 V forward.
•NTC thermistor: resistance falls as temperature rises.

Figure 4. I–V characteristics of four components.

Common mistake

Using the gradient of an I–V graph to find resistance. R = V/I at that point, not 1/gradient (except for a straight line through the origin).

2.3 Investigating I–V characteristics

Connect the component in series with an ammeter, with a voltmeter across it. Supply the pd from a potential divider (potentiometer) so the pd can be varied smoothly from zero. Reverse the connections to obtain negative values. For a diode, include a protective series resistor.

Figure 5. Supplying a variable pd from 0 V using a potentiometer.

WORKED EXAMPLE 9 Diode protection (exam level)

A diode is to be tested with a 6.0 V supply. The maximum safe current is 20 mA and the diode drops 0.70 V when conducting. Find the minimum series resistance.

Solution

R = (6.0 − 0.70)/0.020M1

= 265 ΩA1

WORKED EXAMPLE 10 Controlling heating (standard)

Why should readings be taken quickly and the circuit switched off between readings when testing a resistor?

Solution

Current heats the resistorR1

Its resistance would change, so it would not stay at constant temperatureR1

2.4 Filament lamp in detail

As a lamp filament heats up, metal ions vibrate more, so electrons collide with them more often and resistance increases. The lamp's resistance when hot can be ten times its cold value.

Figure 6. I–V graph for a 12 V, 6 W lamp.

WORKED EXAMPLE 11 Resistance from the graph (exam level)

Use the graph to find the lamp's resistance at 2.0 V and at 12 V.

Solution

At 2.0 V: I = 0.17 A; R = 12 ΩA1

At 12 V: I = 0.50 A; R = 24 ΩA1

2.5 Resistivity

Resistance of a uniform wire depends on its material, length and area: R = ρL/A, where ρ is the resistivity (Ω m). Doubling the length doubles R; doubling the area halves R.

Material	Resistivity / Ω m
silver	1.6 × 10⁻⁸
copper	1.7 × 10⁻⁸
aluminium	2.8 × 10⁻⁸
nichrome	1.1 × 10⁻⁶
graphite	about 10⁻⁵
silicon	about 10³
glass	10¹⁰–10¹⁴

WORKED EXAMPLE 12 Nichrome wire (standard)

Find the resistance of 1.5 m of nichrome wire of diameter 0.40 mm.

Solution

A = π(0.20 × 10⁻³)² = 1.26 × 10⁻⁷ m²M1

R = 1.1 × 10⁻⁶ × 1.5/1.26 × 10⁻⁷ = 13 ΩA1

WORKED EXAMPLE 13 Stretching a wire (exam level)

A wire is stretched to twice its length with constant volume. By what factor does its resistance change?

Solution

L doubles, A halvesM1

R × 4A1

2.6 Measuring resistivity

Measure the resistance of different lengths of a wire using an ammeter and voltmeter (or an ohmmeter), and the diameter with a micrometer at several points. A graph of R against L has gradient ρ/A.

Figure 7. Resistance against length for a nichrome wire of diameter 0.40 mm.

WORKED EXAMPLE 14 From the gradient (exam level)

Find the resistivity from the graph.

Solution

Gradient = 8.8 Ω m⁻¹; A = 1.26 × 10⁻⁷ m²M1

ρ = 8.8 × 1.26 × 10⁻⁷ = 1.1 × 10⁻⁶ Ω mA1

Try this

Use crocodile clips on a wire taped to a metre rule. Keep the current small and switch off between readings so the wire does not heat up.

2.7 Superconductors

Some materials lose all electrical resistance below a critical temperature. Superconductors carry large currents with no heating loss and are used in MRI scanner magnets, particle accelerators and some power cables; most must be cooled with liquid helium or nitrogen.

Figure 8. Resistance of mercury near its critical temperature.

Quick check

1. Why are superconductors not used for household wiring?

Answers: 1. they need very low temperatures, which is expensive

3 Series and parallel circuits

Combining components.

Think about it

Why are the lights in your house connected in parallel rather than in series?

3.1 Series circuits

In series, the same current flows through every component; the pds add up to the emf; total resistance RT = R₁ + R₂ + …

Figure 9. Two resistors in series with a 12 V supply.

WORKED EXAMPLE 15 Series (standard)

For the circuit shown (negligible internal resistance), find the current and the pd across each resistor.

Solution

RT = 6.0 Ω; I = 12/6.0 = 2.0 AM1 A1

V₂ = 4.0 V; V₄ = 8.0 VA1

3.2 Parallel circuits

In parallel, each branch has the same pd; the currents in the branches add up to the total current; 1/RT = 1/R₁ + 1/R₂ + … The total resistance is less than the smallest branch resistance.

Figure 10. Two resistors in parallel across a 6.0 V supply.

Key point

Two equal resistors in parallel have half the resistance of one.

WORKED EXAMPLE 16 Parallel (standard)

For the circuit shown, find the total resistance and the current from the supply.

Solution

1/R = 1/3 + 1/6 = 1/2: R = 2.0 ΩM1 A1

I = 6.0/2.0 = 3.0 A (2.0 A + 1.0 A)A1

3.3 Combined networks

Simplify step by step: combine parallel groups into a single equivalent resistor, then add series parts. Work back out to find individual currents and pds.

WORKED EXAMPLE 17 Network (exam level)

A 4.0 Ω resistor is in series with a parallel pair of 6.0 Ω and 12 Ω, across a 12 V supply. Find the total current and the current in the 12 Ω resistor.

Solution

Parallel: 4.0 Ω; total 8.0 Ω; I = 1.5 AM1 A1

pd across pair = 6.0 V; I₁₂ = 0.50 AA1

3.4 Meter readings in a network

To predict meter readings, find the total resistance and current, then work through the circuit using V = IR and Kirchhoff's laws. Ideal ammeters have zero resistance; ideal voltmeters draw no current.

Figure 11. A series–parallel network across a 12 V supply.

WORKED EXAMPLE 18 Readings (exam level)

For the network shown, find the current in each resistor and the pd across the 4.0 Ω resistor.

Solution

Parallel 6 ‖ 12 = 4.0 Ω; total 8.0 Ω; I = 1.5 A through the 4.0 Ω (pd 6.0 V)M1 A1

pd across pair 6.0 V: I₆ = 1.0 A, I₁₂ = 0.50 AA1

Quick check

1. What happens to the total current if the 12 Ω resistor is removed?

Answers: 1. total R rises to 10 Ω; current falls to 1.2 A

3.5 Kirchhoff's laws

First law (charge conservation): the sum of currents into a junction equals the sum out. Second law (energy conservation): around any closed loop, the sum of emfs equals the sum of pds (Σε = ΣIR).

WORKED EXAMPLE 19 Junction (standard)

Currents of 2.0 A and 0.5 A enter a junction; one wire leaves carrying 1.8 A. Find the current in the other leaving wire.

Solution

2.5 = 1.8 + IM1

I = 0.7 AA1

WORKED EXAMPLE 20 Two cells (exam level)

Cells of emf 6.0 V and 2.0 V are connected in opposition in series with a 4.0 Ω resistor. Find the current.

Solution

Net emf = 4.0 VM1

I = 1.0 AA1

3.6 Comparing series and parallel

Feature	Series	Parallel
current	same through each component	splits between branches; branch currents add
potential difference	shared; pds add to the supply pd	same across each branch
total resistance	R₁ + R₂ + … (larger than any one)	1/R = 1/R₁ + 1/R₂ + … (smaller than any one)
if one lamp fails	all go out	others stay lit
adding a resistor	total current falls	total current rises

WORKED EXAMPLE 21 Adding branches (standard)

Explain why switching on more appliances in a house increases the current drawn from the mains.

Solution

Appliances are in parallel: each new branch lowers total resistanceR1

Same supply pd, so total current increasesR1

3.7 Why homes use parallel wiring

Each appliance in parallel receives the full mains voltage and can be switched independently; if one fails, the others keep working. The total current from the supply is the sum of the appliance currents, which is why too many appliances can overload a circuit.

WORKED EXAMPLE 22 Overload (standard)

A 230 V circuit has a 13 A fuse. Can it supply a 2.0 kW kettle and a 1.2 kW toaster at the same time?

Solution

I = 3200/230 = 13.9 AM1

Exceeds 13 A: the fuse would blowR1

4 Electrical power and energy

How fast circuits transfer energy.

4.1 Electrical power

Power is the rate of energy transfer: P = IV = I²R = V²/R. Energy transferred E = Pt = IVt.

WORKED EXAMPLE 23 Heater (standard)

A heater has resistance 24 Ω on a 230 V supply. Find its power and the current.

Solution

P = 230²/24 = 2.2 kWA1

I = 230/24 = 9.6 AA1

WORKED EXAMPLE 24 Cost (standard)

Find the cost of running a 2.2 kW heater for 3.0 hours at $0.15 per kWh.

Solution

E = 6.6 kWhM1

Cost = $0.99A1

4.2 Power in series and parallel

In series, the larger resistor dissipates more power (P = I²R, same I). In parallel, the smaller resistor dissipates more power (P = V²/R, same V).

Figure 12. Power dissipated against current: P = I²R is a parabola.

WORKED EXAMPLE 25 Two lamps (exam level)

Two lamps rated 60 W and 100 W for 230 V are connected in series to 230 V. Which is brighter?

Solution

R = V²/P: 60 W lamp has larger R (882 Ω vs 529 Ω)M1

Same current: the 60 W lamp dissipates more and is brighterR1

4.3 Fuses and circuit breakers

A fuse is a thin wire that melts if the current exceeds its rating, breaking the circuit and protecting the wiring from overheating. Circuit breakers do the same job using an electromagnet or bimetallic strip and can be reset. The heating effect P = I²R explains why thin, high-resistance fuse wire heats up first.

WORKED EXAMPLE 26 Fuse heating (exam level)

A fuse wire of resistance 0.020 Ω carries 13 A. Find the power dissipated in it and explain why it, rather than the cable, melts first.

Solution

P = 13² × 0.020 = 3.4 WA1

Thin wire with small mass and low melting point heats rapidlyR1

4.4 The heating effect in use

The heating effect of current is used in kettles, toasters, hair dryers and electric showers. Heating elements use high-resistivity alloys such as nichrome, which do not oxidise at high temperatures.

WORKED EXAMPLE 27 Kettle element (standard)

A 2.4 kW kettle element runs on 230 V. Find its resistance and the length of nichrome wire (area 2.0 × 10⁻⁷ m²) needed.

Solution

R = V²/P = 22 ΩA1

L = RA/ρ = 22 × 2.0 × 10⁻⁷/1.1 × 10⁻⁶ = 4.0 mM1 A1

4.5 Power transmission losses

The power wasted in cables of resistance R is I²R. Transmitting at high voltage reduces the current for the same power, cutting losses dramatically.

WORKED EXAMPLE 28 Cable loss (exam level)

A 20 kW load is supplied through cables of total resistance 0.50 Ω. Compare losses at 230 V and at 11 kV.

Solution

230 V: I = 87 A; loss = 3.8 kWA1

11 kV: I = 1.8 A; loss = 1.7 WA1

5 Cells and internal resistance

Real sources lose some energy inside.

5.1 Internal resistance

Real cells have internal resistance r. Some of the emf is used to drive current through r, so the terminal pd is less than the emf: V = ε − Ir. The 'lost volts' Ir increase with current.

Figure 13. A cell of emf ε and internal resistance r driving a load R.

WORKED EXAMPLE 29 Terminal pd (standard)

A 12 V battery with r = 0.50 Ω drives a 5.5 Ω load. Find the current and terminal pd.

Solution

I = 12/6.0 = 2.0 AA1

V = 12 − 1.0 = 11 VA1

5.2 Measuring emf and internal resistance

Vary the load resistance and record terminal pd V and current I. A graph of V against I is a straight line: the y-intercept is ε and the gradient is −r.

Figure 14. Terminal pd against current for a cell.

WORKED EXAMPLE 30 From the graph (exam level)

Find the emf and internal resistance from the graph.

Solution

Intercept: ε = 1.50 VA1

Gradient = −0.80: r = 0.80 ΩA1

5.3 Maximum power transfer

The power delivered to the load is P = ε²R/(R + r)². It is greatest when R = r, but then only half the energy is delivered usefully (efficiency 50%).

Figure 15. Power delivered to a load by a 12 V source with r = 1.0 Ω.

WORKED EXAMPLE 31 Maximum power (exam level)

Find the maximum power the 12 V, 1.0 Ω source can deliver to a load.

Solution

R = r = 1.0 Ω; I = 6.0 AM1

P = 36 WA1

5.4 Cells in series and parallel

Identical cells in series add emfs and internal resistances. Identical cells in parallel keep the same emf but the internal resistance falls (they share the current), so they can deliver larger currents for longer.

WORKED EXAMPLE 32 Battery pack (exam level)

Four cells (each 1.5 V, 0.20 Ω) are connected (a) in series; (b) in parallel. Find the emf and internal resistance of each arrangement.

Solution

(a) 6.0 V, 0.80 ΩA1

(b) 1.5 V, 0.050 ΩA1

5.5 Comparing batteries

Different battery technologies trade off energy per kilogram, cost, rechargeability and lifetime.

Type	Cell emf / V	Rechargeable?	Typical use
alkaline	1.5	no	remote controls, toys
NiMH	1.2	yes	rechargeable AA cells
lead–acid	2.0	yes	car starter batteries
lithium-ion	3.7	yes	phones, laptops, electric cars

5.6 Battery capacity and discharge

Battery capacity is often given in ampere-hours (A h) or mA h: the charge it can deliver. Energy stored = capacity × emf. During discharge, the terminal pd of most cells stays nearly constant, then drops rapidly near the end.

Figure 16. Discharge curve of an AA cell at constant current.

WORKED EXAMPLE 33 Phone battery (standard)

A 3.7 V phone battery has a capacity of 4000 mA h. Find the energy stored.

Solution

Charge = 4.0 × 3600 = 14 400 CM1

E = 14 400 × 3.7 = 53 kJA1

6 Potential dividers and sensors

Controlling voltages.

6.1 The potential divider

Two resistors in series across a supply share the pd in proportion to their resistances: Vout = VinR₂/(R₁ + R₂). A variable resistor or potentiometer gives an adjustable output.

Figure 17. A potential divider; Vout is taken across R₂.

WORKED EXAMPLE 34 Divider (standard)

Find Vout for the circuit shown.

Solution

Vout = 9.0 × 1/3M1

= 3.0 VA1

6.2 Loading a potential divider

Connecting a load across R₂ puts it in parallel with R₂, lowering the effective resistance and hence Vout. A voltmeter of resistance comparable with R₂ gives a reading lower than expected.

WORKED EXAMPLE 35 Loaded divider (exam level)

A 2.0 kΩ load is connected across R₂ in the divider above. Find the new Vout.

Solution

R₂ ‖ 2 kΩ = 0.67 kΩM1

Vout = 9.0 × 0.67/2.67 = 2.25 VA1

6.3 Sensors: LDRs and thermistors

The resistance of a light-dependent resistor (LDR) falls as light intensity increases. The resistance of an NTC thermistor falls as temperature increases. In a potential divider they produce an output pd that depends on light or temperature, used in automatic lights, thermostats and alarms.

Figure 18. A light sensor: LDR in a potential divider.

Figure 19. Resistance of an NTC thermistor against temperature.

WORKED EXAMPLE 36 Dusk switch (exam level)

In the LDR circuit, the LDR's resistance is 1.0 kΩ in daylight and 100 kΩ in darkness. Find Vout in each case and state when a lamp switched by Vout > 3 V comes on.

Solution

Day: 6.0 × 1/11 = 0.55 VA1

Dark: 6.0 × 100/110 = 5.5 V: the lamp comes on in darknessA1

6.4 A thermostat circuit

A thermistor in a potential divider can switch a heater: as the room warms, the thermistor's resistance falls, changing the output pd. A variable resistor sets the switching temperature.

Figure 20. A temperature-sensing potential divider.

WORKED EXAMPLE 37 Switching point (exam level)

In the circuit, the output is taken across the variable resistor R. The switch operates when Vout reaches 6.0 V. The thermistor has 2.2 kΩ at 20 °C. What value of R makes the switch operate at 20 °C?

Solution

Equal pd across each: R = thermistor resistanceM1

R = 2.2 kΩA1

6.5 Potentiometers

A potentiometer is a variable resistor with a sliding contact. Used as a potential divider, it gives a continuously variable output from 0 V to the supply pd, unlike a variable resistor in series, which cannot reduce the pd to zero.

Figure 21. A potentiometer used as a variable potential divider.

Quick check

1. Why is a potentiometer better than a series rheostat for obtaining I–V characteristics from 0 V?

Answers: 1. it can provide pds all the way down to zero

7 Exam technique

Solving circuit problems systematically.

7.1 Method checklist

•Redraw complicated circuits and label every current and pd.
•Combine parallel resistors first, then series ones.
•Use the same current for series components and the same pd for parallel branches.
•Include internal resistance: ε = I(R + r).
•Check that pds around each loop add to the emf and currents at each junction balance.

7.2 Multi-step problem: sensor circuit

WORKED EXAMPLE 38 Frost alarm (exam level)

A thermistor (resistance 1.5 kΩ at 10 °C, 6.0 kΩ at 0 °C) is in series with a 3.0 kΩ resistor across a 9.0 V supply. The output is taken across the thermistor. Find Vout at both temperatures and explain how the circuit could trigger a frost alarm.

Solution

10 °C: 9.0 × 1.5/4.5 = 3.0 VA1

0 °C: 9.0 × 6.0/9.0 = 6.0 VA1

Output rises as temperature falls; an alarm set to trigger above about 5 V sounds near freezingR1

7.3 Multi-step problem: lamps and battery

WORKED EXAMPLE 39 Torch (exam level)

A torch uses two 1.5 V cells (each r = 0.30 Ω) in series with a lamp of resistance 4.4 Ω. Find (a) the current; (b) the lamp's power; (c) the efficiency of energy transfer to the lamp.

Solution

(a) I = 3.0/(4.4 + 0.60) = 0.60 AA1

(b) P = 0.60² × 4.4 = 1.6 WA1

8 Data-based questions and further practice

Applying circuit ideas.

8.1 Experimental data: a filament lamp

A student measures I and V for a lamp.

V / V	0	1.0	2.0	4.0	6.0	8.0	12.0
I / A	0	0.11	0.17	0.26	0.33	0.39	0.50

WORKED EXAMPLE 40 Resistance trend (exam level)

Calculate R at 1.0 V and 12 V and explain the trend.

Solution

R = 9.1 Ω and 24 ΩA1

Filament temperature rises; ions vibrate more; more collisionsR1

8.2 Graded practice

WORKED EXAMPLE 41 Charge and energy (standard)

A 1.5 V cell delivers 0.20 A for 2.0 hours. Find the charge and energy delivered.

Solution

q = 0.20 × 7200 = 1440 CA1

E = 1.5 × 1440 = 2.2 kJA1

WORKED EXAMPLE 42 Fuse rating (standard)

Choose a 3 A, 5 A or 13 A fuse for a 1.1 kW appliance on 230 V.

Solution

I = 1100/230 = 4.8 AM1

Use the 5 A fuseA1

WORKED EXAMPLE 43 Bridge network (exam level)

Two potential dividers share a 12 V supply: 4 kΩ over 8 kΩ, and 6 kΩ over 6 kΩ. Find the pd between their midpoints.

Solution

Midpoints at 8.0 V and 6.0 V (above the negative rail)M1

pd = 2.0 VA1

8.3 Graded practice: internal resistance

WORKED EXAMPLE 44 Car battery (exam level)

A 12 V car battery with r = 0.020 Ω supplies 150 A to a starter motor. Find the terminal pd and the power wasted inside the battery.

Solution

V = 12 − 150 × 0.020 = 9.0 VA1

P = I²r = 450 WA1

WORKED EXAMPLE 45 Two loads (exam level)

A cell gives 1.40 V across a 6.0 Ω load and 1.20 V across a 2.0 Ω load. Find its emf and internal resistance.

Solution

I₁ = 0.233 A, I₂ = 0.60 A; ε = 1.40 + 0.233r = 1.20 + 0.60rM1

r = 0.55 Ω; ε = 1.53 VA1

8.4 Graded practice: resistivity and power

WORKED EXAMPLE 46 Graphite pencil line (exam level)

A pencil line 0.10 m long, 2.0 mm wide and 1.0 × 10⁻⁵ m thick has resistance 400 Ω. Find the resistivity of the graphite.

Solution

A = 2.0 × 10⁻³ × 1.0 × 10⁻⁵ = 2.0 × 10⁻⁸ m²M1

ρ = RA/L = 400 × 2.0 × 10⁻⁸/0.10 = 8.0 × 10⁻⁵ Ω mA1

WORKED EXAMPLE 47 Extension lead (exam level)

An extension lead has two 10 m copper cores of area 1.0 mm² and carries 10 A. Find the power wasted in the lead.

Solution

R = 1.7 × 10⁻⁸ × 20/1.0 × 10⁻⁶ = 0.34 ΩM1

P = 10² × 0.34 = 34 WA1

WORKED EXAMPLE 48 Parallel heaters (standard)

Two 1.0 kW, 230 V heaters are connected in parallel to 230 V. Find the total current.

Solution

Each draws 4.35 AM1

Total = 8.7 AA1

8.5 Graded practice: circuits

WORKED EXAMPLE 49 Lamps in parallel (standard)

Three identical lamps in parallel draw 1.2 A from a 6.0 V supply. Find the resistance of each lamp.

Solution

Each draws 0.40 AM1

R = 6.0/0.40 = 15 ΩA1

WORKED EXAMPLE 50 Adding a resistor (exam level)

A 10 Ω resistor is connected across a 12 V battery with r = 2.0 Ω. A second 10 Ω resistor is then connected in parallel. Find the terminal pd before and after.

Solution

Before: I = 1.0 A; V = 10 VA1

After: R = 5.0 Ω; I = 1.71 A; V = 8.6 VM1 A1

8.6 Key terms

Term	Meaning
current	rate of flow of charge (A)
potential difference	energy transferred per unit charge in a component (V)
emf	energy given per unit charge by a source (V)
resistance	V/I (Ω)
resistivity	ρ = RA/L (Ω m)
internal resistance	resistance inside a source causing lost volts
ohmic conductor	I ∝ V at constant temperature
potential divider	series resistors that split a pd

8.7 Self-test

Try these without notes, then check the answers.

Quick check

1. Find the current when 30 C passes in 1 minute.

2. Three 6 Ω resistors are in parallel. Total resistance?

3. A 2 kW heater runs on 230 V. Find the current.

4. A cell of emf 1.5 V has terminal pd 1.2 V at 0.6 A. Find r.

5. Find Vout for 10 V across 3 kΩ (top) and 2 kΩ (bottom).

6. Which has higher resistance: a 60 W or a 100 W lamp (same supply)?

Answers: 1. 0.50 A 2. 2 Ω 3. 8.7 A 4. 0.5 Ω 5. 4.0 V 6. 60 W

8.8 Common exam errors

•Using 1/gradient of an I–V curve as resistance.
•Forgetting to convert mm² to m² (1 mm² = 10⁻⁶ m²) or using diameter instead of radius.
•Adding parallel resistances directly.
•Ignoring internal resistance when it is given.
•Saying current is 'used up' in a component.

9 Common misconceptions and links

Students often think…	Actually…
current is used up in components	current is the same everywhere in a series loop
emf and terminal pd are the same	terminal pd = emf − Ir
resistance is the gradient of an I–V graph	R = V/I at the point
electrons travel fast through wires	drift velocity is very small
parallel resistances add	reciprocals add

How B.5 connects to the rest of the course

Topic	Connection
D.2 Electric and magnetic fields	potential and charge
D.4 Induction (HL)	generators and transmission
A.3 Work, energy and power	power and efficiency
B.1 Thermal energy transfers	heating effect of current

10 Exam-style question with markscheme

Try this under exam conditions (about 20 minutes) before reading the markscheme.

A cell of emf 6.0 V and internal resistance 0.50 Ω is connected to a 4.0 Ω resistor in series with a parallel combination of two 6.0 Ω resistors.

(a) Calculate the total external resistance. [2]

(b) Calculate the current from the cell. [2]

(d) Calculate the power dissipated in the 4.0 Ω resistor. [2]

(e) One 6.0 Ω resistor is removed. State and explain the effect on the terminal pd. [3]

Markscheme

Part	Answer	Marks
(a)	parallel pair = 3.0 Ω	M1
	7.0 Ω	A1
(b)	I = 6.0/(7.0 + 0.50)	M1
	0.80 A	A1
(c)	V = 6.0 − 0.40 = 5.6 V	A1
(d)	P = I²R = 0.80² × 4.0	M1
	2.6 W	A1
(e)	external resistance increases to 10 Ω	R1
	current decreases	R1
	lost volts decrease, so terminal pd increases	R1

Examiner's comments. In (e), many said the terminal pd decreases; reasoning via lost volts (Ir) gives the correct answer.

11 Check your understanding

Section A: multiple choice

1. The unit of resistivity is

A. Ω m B. Ω m⁻¹ C. Ω D. Ω m²

2. In a parallel circuit, components have the same

A. current B. resistance C. potential difference D. power

3. Two 4 Ω resistors in parallel have total resistance

A. 8 Ω B. 2 Ω C. 4 Ω D. 0.5 Ω

4. The resistance of a filament lamp

A. decreases with current B. is constant C. is zero D. increases with current

5. Terminal pd equals emf when

A. current is maximum B. no current flows C. r is large D. R = r

6. An LDR's resistance decreases when

A. light intensity increases B. temperature falls C. it is dark D. current is zero

7. Power dissipated by a resistor carrying current I is

A. IR² B. I/R C. R/I D. I²R

8. Kirchhoff's first law is a consequence of conservation of

A. energy B. momentum C. charge D. mass

Section B: structured

9. Calculate the resistance of 2.0 m of copper wire (ρ = 1.7 × 10⁻⁸ Ω m) of diameter 0.50 mm. [3]

10. Describe how to determine the emf and internal resistance of a cell experimentally. [4]

11. A thermistor and a 5.0 kΩ resistor form a potential divider across 9.0 V. The output across the resistor is 6.0 V. Find the thermistor's resistance. [2]

12. Explain why the resistance of a filament lamp increases with current. [3]

Answers

1. A 2. C 3. B 4. D 5. B 6. A 7. D 8. C

9. A = 1.96 × 10⁻⁷ m²; R = 0.17 Ω.

10. Vary load; measure V and I; plot V against I; intercept = ε; gradient = −r.

11. pd across thermistor 3.0 V: R = 2.5 kΩ.

12. Temperature rises; ions vibrate more; more frequent collisions with electrons.

12 Summary and topic map

Idea	Remember
Current	I = Δq/Δt = nAvq
pd and emf	V = W/q
Resistance	R = V/I; R = ρL/A
Circuits	series R add; parallel 1/R add
Power	P = IV = I²R = V²/R
Internal resistance	ε = I(R + r)
Dividers	Vout = VinR₂/(R₁ + R₂)

Before the exam, make sure you…

•convert mm to m and use radius for area
•use R = V/I, not the gradient
•include internal resistance
•explain sensors using changing resistance

Current and Circuits on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 22. Topic map for Current and Circuits.

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