By the end of this workbook you should be able to:
Explain that cells provide a source of emf, using chemical reactions or (for solar cells) light as the energy source.
Interpret circuit diagrams and describe direct current as a flow of charge carriers, I = ΔQ/Δt.
Define potential difference as the work done per unit charge, and apply Kirchhoff's two laws.
Explain the origin of electrical resistance in terms of charge-carrier mobility, and use R = V/I and resistivity ρ = RA/L.
Distinguish ohmic from non-ohmic behaviour, including the heating effect of resistors, and use P = IV.
Combine resistors in series and parallel, and use ε = I(R + r) for a cell with internal resistance.
Explain how a variable resistor can be used as a potentiometer to divide a potential difference.
1. Cells and Circuit Diagrams
Whenever you use an electronic device, something has to supply the energy that pushes charge around the circuit. That "something" is almost always an electrical cell — a single cell is often loosely called a battery, though strictly a battery is two or more cells joined together.
Key idea. A circuit transfers energy. That energy is provided by the cells as they transfer chemical energy (a battery) or solar panels transfer the energy from absorbed photons. If a power pack is used, electrical energy is provided to the circuit, but that energy originated in an energy transfer at a power station. The circuit takes in this energy from the cell and then transfers it out through the components (e.g. bulb, resistor, motor).
Other sources of electrical energy exist too — a generator converts kinetic energy (for example from a wind turbine or a bicycle dynamo) into a p.d. — but cells and batteries are by far the most common source built into a portable circuit. Choosing between them involves trade-offs: convenience, cost, whether the energy source is renewable, whether the supply is continuous (a solar cell only works in daylight), and — as you'll meet in section 9 — the cell's internal resistance.
Fig. 1.1 A simple series circuit: cell, switch, resistor and ammeter.
Circuit diagram: a diagram that represents the arrangement of components in a circuit using standardised symbols, rather than a realistic drawing of the physical components. Every physicist, anywhere in the world, draws a cell, a resistor or an ammeter the same way.
Check your understanding
1Give one advantage and one disadvantage of a solar cell compared with a chemical cell as the energy source for a circuit.
Advantage: a solar cell uses a renewable, freely available energy source and produces no pollution while running (chemical cells rely on materials that are mined/manufactured and eventually need disposal). Disadvantage: a solar cell only supplies a useful p.d. when illuminated, so it can't power a circuit continuously without extra storage (e.g. a rechargeable chemical cell) or backup, whereas a chemical cell works in the dark and can be made very compact and portable. (Any other reasonable, correctly justified pair of advantage/disadvantage is acceptable.)
2. Charge and Current
"The flow of electrons" is a common description of current, but it isn't quite precise enough to calculate with. In a metal, the charge carriers are free (delocalized) electrons, and current is defined as the rate at which charge passes a point in the circuit.
Key idea. Direct current (dc), I, is a flow of charge carriers, given by the rate of change of charge with time.
I = ΔQ / Δt
where I is current in amperes (A), ΔQ is the charge transferred in coulombs (C), and Δt is the time interval in seconds (s). Rearranged, this also gives ΔQ = IΔt.
Charge itself is quantised: every charge is a whole-number multiple of the elementary charge, e = 1.6 × 10−19 C. So the number of charge carriers passing a point is n = Q/e. A current of 1 A means 1 C of charge — about 6.25 × 1018 electrons — passes a point every second.
Charge carrier: a charged particle that is free to move (mobile). In a metal wire the charge carriers are delocalized electrons; in other conductors (solutions, ionised gases) they may be positive or negative ions.
Worked example 2.1
The current through an LED lamp is 50 mA. Calculate the charge that flows through the lamp in one minute.
Answer:
ΔQ = IΔt
ΔQ = (50 × 10−3) × 60
Check your understanding
2The current in a wire is 0.35 A. Calculate the charge that passes a point in the wire in 10 s.
ΔQ = IΔt = 0.35 × 10
3A charge of 210 C passes a point in a wire in 10 minutes. Calculate the average current.
Convert the time to seconds first: 10 minutes = 600 s. I = ΔQ/Δt = 210 / 600
4An electron beam carries a current of 1.2 mA. Calculate how many electrons pass along the beam each minute (e = 1.6 × 10−19 C).
First find the charge: ΔQ = IΔt = (1.2 × 10−3) × 60. Then the number of electrons, n = ΔQ / e, using e = 1.6 × 10−19 C.
3. Potential Difference and Kirchhoff's Laws
Current tells you how much charge is flowing; potential difference tells you how much energy each coulomb of that charge is transferring. It's the electrical equivalent of gravitational potential difference: a mass falls because of a difference in gravitational potential, transferring gravitational PE to kinetic energy. A charge carrier moves (or transfers energy) because of a difference in electric potential.
Key idea. The electric potential difference, V, is the work done per unit charge on moving a positive charge between two points along the path of the current.
V = W / q
where W is the work done (energy transferred, in joules) and q is the charge moved (in coulombs). One volt means one joule of energy is transferred by each coulomb of charge moving between two points: 1 V = 1 J C−1.
Fig. 3.1 A single series loop, the total change in potential around the closed loop must be zero.
Kirchhoff's voltage law (potential difference / loop rule). Energy is conserved, so the sum of the potential differences around any closed loop is zero (take pd added by a source as positive, and pd used up by components as negative).
Kirchhoff's current law (current / junction rule). This follows from conservation of charge: charged particles cannot be created or destroyed, so the current entering any junction must equal the current leaving it.
In a parallel circuit, every branch between the same two points sees the same pd — this is a direct consequence of the voltage law, since a loop through the supply and just one branch must still sum to zero.
Worked example 3.1
A power pack has a p.d. of 10 V. Calculate the energy transferred to one electron as it moves through this p.d. (e = 1.6 × 10−19 C).
Answer:
Energy transferred = charge × p.d.
W = qV = (1.6 × 10−19) × 10
Check your understanding
5A 12 V battery in a simple series loop has only one bulb in the circuit. Using Kirchhoff's voltage law, state the p.d. across the bulb, showing your reasoning.
Around the loop: +12 − Vbulb = 0, so Vbulb = 12 V — the bulb takes the full battery p.d, since it's the only component in the loop.
6Two identical bulbs are connected in series across a 12 V battery. Use symmetry and Kirchhoff's voltage law to find the p.d. across each bulb.
The two p.d.s must add to 12 V: 12 = V + V. Since the bulbs are identical, by symmetry they share this equally, so rearrange 2V = 12 to find V.
7At a junction, a current of 5 A flows in. One outgoing branch carries 3 A. Use Kirchhoff's current law to find the current in the other outgoing branch.
Current in = current out: 5 = 3 + I3, so rearrange to find I3.
4. Series and Parallel Circuits
Kirchhoff's two laws are all you need to analyse any circuit in this course — you don't need to memorise separate "rules" for series and parallel. But it's worth stating the patterns they produce explicitly, since you'll use them constantly:
Current in series is the same at every point (no junctions exist for charge to split at).
Current in parallel splits across each branch, in inverse proportion to resistance.
Potential difference in series splits across components, in proportion to resistance.
Potential difference in parallel is the same across every branch.
Fig. 4.1 Series circuit: ammeters A₁, A₂ and A₃ all read the same current — there are no junctions for charge to split at.Fig. 4.2 Parallel circuit: A₁ (main line) = A₂ + A₃ (branch currents), by Kirchhoff's current law.
Check your understanding
8Two identical bulbs are in series in a single loop, with an ammeter reading 0.7 A immediately after the battery. State the readings on ammeters placed (a) between the two bulbs and (b) after the second bulb, explaining your reasoning.
It's a single series loop, so every ammeter reads the same current — there are no junctions. Both ammeters read 0.7 A.
9Two identical bulbs are connected in parallel across a battery. The ammeter in one branch reads 0.2 A. Use Kirchhoff's current law to find the current in the main line supplying both branches.
Identical bulbs in parallel share the current equally, so the other branch also carries 0.2 A. Main-line current = sum of branch currents = 0.2 + 0.2.
10A circuit has a battery and ammeter A₁ in series with two parallel branches: branch 1 is a single bulb with ammeter A₂, branch 2 is two identical bulbs in series with ammeter A₃. If A₂ reads 0.3 A, and branch 2's total resistance is double branch 1's, find A₁ and A₃.
Both branches are directly across the battery, so they see the same p.d. Since branch 2 has double the resistance of branch 1, by V = IR it carries half the current: A₃ = A₂ / 2 = 0.3 / 2. Then A₁ = A₂ + A₃ (Kirchhoff's current law).
5. Conductors, Insulators and the Origin of Resistance
When the same p.d. is connected across different components, the currents produced vary hugely. This is because materials differ in how many mobile charge carriers they contain, and in how easily those carriers can move.
Key idea. A good electrical conductor has a large number of mobile charge carriers (usually free electrons) per unit volume, giving it low resistance. An insulator has very few mobile charge carriers, so a current through it is negligible — it has high resistance. A semiconductor (e.g. silicon) sits between the two, with a resistivity that changes a great deal with temperature and light.
Fig. 5.1 In a metal, positive ions are fixed in the lattice; free electrons move randomly at high speed but gain a small net drift in one direction from the electric field — this drift is the current.
Extension — the Drude model Paul Drude (1900) treated the free electrons in a metal like a gas: moving randomly at around 100,000 m s−1, far faster than their drift velocity due to the field (typically only around 0.0001 m s−1). Because the ion spacing limits how far an electron travels before it collides with an ion, electrons can't accelerate continuously — they repeatedly scatter, with an average time between collisions called the relaxation time. As temperature rises, the lattice ions vibrate more, scattering the drifting electrons more often. This reduces the relaxation time, which increases resistivity — the microscopic origin of a metal's resistance increasing with temperature. Resistance is, macroscopically, the result of charge carriers transferring energy to a vibrating lattice as they collide with it.
An ammeter (connected in series) needs almost zero resistance, or it would change the very current it's trying to measure. A voltmeter (connected in parallel) needs a very high resistance, so that almost no current takes the "shortcut" through the meter instead of through the component being measured.
Check your understanding
11State the ideal resistance of (a) an ammeter and (b) a voltmeter, explaining why each value is needed.
(a) An ideal ammeter has zero resistance — since it sits in series with the current it's measuring, any resistance of its own would reduce that current and give a false low reading. (b) An ideal voltmeter has infinite resistance — since it sits in parallel with the component, a finite resistance would let some current bypass the component through the meter, and would also alter the p.d. it's supposed to be measuring.
12Explain, using the Drude model, why a metal's resistance increases as its temperature rises.
As temperature rises, the positive ions in the lattice vibrate more vigorously. This means the drifting free electrons collide with the ions more often (the relaxation time between collisions falls), which resists their motion more strongly — increasing resistivity and therefore resistance.
6. Resistance, Ohm's Law and I–V Characteristics
Resistance measures how difficult it is to make a current pass through a component — it resists the flow of charge and so reduces the current for a given p.d.
Key idea. Electrical resistance, R, is defined as the ratio of the potential difference across a conductor to the current flowing through it.
R = V / I
The SI unit is the ohm, Ω (1 Ω = 1 V A−1). Georg Ohm investigated this relationship experimentally and found that, for a conductor at constant temperature, current is proportional to p.d.: V ∝ I, or V = IR. This is Ohm's law.
Ohmic behaviour: a component is described as ohmic if its current is proportional to the p.d. across it (at constant temperature) — its I–V graph is a straight line through the origin, and its resistance stays constant. Metal wires at constant temperature are ohmic. Non-ohmic behaviour means the resistance changes as the current changes — the I–V graph is curved.
1. Resistor (ohmic): I ∝ V, a straight line through the origin.2. Filament lamp (non-ohmic): as V rises the filament heats up, so R increases and the curve bends over.3. Diode (non-ohmic): almost zero current in reverse, then a sharp rise once forward-biased above a threshold.
The heating effect Whenever a current passes through a resistor, some electrical energy is transferred to internal (thermal) energy — this is why a filament lamp gets hot enough to glow, and why the lamp above is non-ohmic: as it heats up, its resistance changes. Resistors used deliberately for heating (e.g. a kettle element) are designed to make the most of this effect.
Worked example 6.1
The current through an electrical component is 0.78 A when a p.d. of 4.4 V is applied across it. Calculate its resistance.
Answer:
R = V / I
R = 4.4 / 0.78
Check your understanding
13A 12 V p.d. is applied across a resistor and a current of 2.0 A flows. Calculate the resistance.
R = V / I = 12 / 2.0
14A student plots an I–V graph for a component and gets a curve, not a straight line. State whether the component is ohmic, and explain how you can tell.
The component is non-ohmic. A straight line through the origin on an I–V graph means constant resistance (I ∝ V), which is the definition of ohmic behaviour. A curve means the gradient (and so the ratio V/I) changes as the current changes — the resistance is not constant.
15A filament lamp has current 0.20 A when the p.d. across it is 1.0 V, and current 0.50 A when the p.d. is 4.0 V. Calculate its resistance at each p.d., and use your answers to explain why the lamp is non-ohmic.
R = V/I at each point: R₁ = 1.0 / 0.20 and R₂ = 4.0 / 0.50. Once evaluated, these two resistances are different — the resistance has increased at the higher p.d. (because the filament is hotter), which is exactly what "non-ohmic" means.
7. Resistivity
The resistance of a wire depends not just on the material it's made from, but on its shape. A long, thin wire has more resistance than a short, thick one of the same material — so it doesn't make sense to talk about "the resistance of copper" without specifying dimensions.
Fig. 7.1 A uniform wire of length L and cross-sectional area A.
For a given material at constant temperature, resistance is directly proportional to length (longer wires have more resistance) and inversely proportional to cross-sectional area (thicker wires have less resistance). Combining these, resistance is proportional to L/A:
R ∝ L / A
Key idea. R is proportional to L/A, and the constant of proportionality is different for every material — this constant is called the material's resistivity, ρ. Numerically, it's the resistance of a 1 m length of the material with a 1 m² cross-section.
R = ρL / A
The SI unit of resistivity is the ohm metre (Ω m) — note this is not "ohms per metre". Good conductors (copper, silver) have very low resistivities (∼10−8 Ω m); good insulators (glass, PTFE) have extremely high ones (>1012 Ω m). Rearranged, ρ = RA / L — often the more useful form when finding a material's resistivity from measured values of R, A and L.
Worked example 7.1
Determine the resistance of a nichrome wire (resistivity 1.1 × 10−6 Ω m) at 20 °C, if it has a length of 1.96 m and a radius of 0.21 mm.
Answer:
R = ρL / A, where A = πr² = π × (0.21 × 10−3)²
R = [1.1 × 10−6 × π × (0.21 × 10−3)²] / 1.96
Check your understanding
16A wire of uniform cross-sectional area 0.5 mm² has a length of 10 m and a resistance of 3 Ω. Calculate the resistivity of the material.
Convert the area to m² first: A = 0.5 mm² = 0.5 × 10−6 m². ρ = RA / L = (3 × 0.5 × 10−6) / 10
17What length of alloy wire (resistivity 5.0 × 10−7 Ω m, diameter 0.50 mm) is required to make a 6.0 Ω resistor?
Radius r = 0.25 mm = 2.5 × 10−4 m, so A = πr² = π × (2.5 × 10−4)². Rearranging R = ρL/A gives L = RA / ρ = (6.0 × A) / (5.0 × 10−7) — substitute your value of A.
18A copper track on a circuit board (resistivity of copper 1.69 × 10−8 Ω m) is 35 µm thick and 1.6 mm wide. Calculate the resistance of 1 cm of this track.
A = thickness × width = (3.5 × 10−5) × (1.6 × 10−3) m². R = ρL/A = [1.69 × 10−8 × 0.01] / A, using your value of A and L = 1 cm = 0.01 m.
8. Combining Resistors
Real circuits usually contain several resistors. Imagine two rope bridges crossing a gorge one after another: everyone must cross both, so adding a second bridge in the only available path makes the journey harder overall — this is the series case. Now imagine a second bridge built alongside the first: there are two available routes, so more people can cross per minute overall — this is the parallel case. Adding a resistor in series always increases total resistance; adding one in parallel always decreases it.
Fig. 8.1 Resistors in series: the same current I flows through each in turn.
Rseries = R₁ + R₂ + …
Fig. 8.2 Resistors in parallel: all three share the same p.d., and the total current splits between them.
1 / Rparallel = 1/R₁ + 1/R₂ + …
Rparallel is always smaller than the smallest individual resistance in the combination — every extra branch gives charge another route through.
Worked example 8.1
A 5000 Ω resistor and an 8000 Ω resistor are connected in series. Calculate their combined resistance.
Answer:
R = R₁ + R₂
R = 5000 + 8000
Check your understanding
19Two resistors of 6 Ω and 34 Ω are connected in series. Calculate the total resistance.
RT = R₁ + R₂ = 6 + 34
20Two resistors of 8 Ω and 6 Ω are connected in parallel. Calculate the total resistance.
1/RT = 1/8 + 1/6. Combine the fractions over a common denominator, then invert to find RT — remember your final answer must be smaller than 6 Ω.
21A 40 Ω resistor is in series with two 50 Ω resistors that are connected in parallel with each other. Calculate the total resistance of the network.
First combine the parallel pair: 1/Rp = 1/50 + 1/50, giving Rp = 25 Ω. Then add the series resistor: RT = 40 + Rp.
9. Emf and Internal Resistance
A real cell isn't a perfect energy source — it's made from materials and chemicals that themselves have some resistance, called the cell's internal resistance, r. Some of the energy the cell puts in gets taken straight back out again heating the cell itself, before it ever reaches the rest of the circuit.
Fig. 9.1 A cell of emf ε and internal resistance r (dashed box) connected to an external resistor R.
Key idea. Cells provide a source of emf, ε (the total energy transferred, per coulomb, that passes through the source). Electric cells are characterized by their emf and their internal resistance, r.
ε = I(R + r)
The same current I flows through R and r, so ε = IR + Ir. The term IR is the terminal p.d. — what's actually available to the rest of the circuit; Ir is the "lost volts" dissipated inside the cell itself. When the switch is open (I = 0), an ideal voltmeter across the cell reads the full emf, since no p.d. is lost internally.
Worked example 9.1
A battery with an emf of 1.5 V and internal resistance 0.82 Ω is connected in a circuit with a 5.6 Ω fixed resistor. Calculate the current in the circuit.
Answer:
I = ε / (R + r)
I = 1.5 / (5.6 + 0.82)
Check your understanding
22A cell has emf 12 V and internal resistance 1 Ω, connected to a 23 Ω resistor. Calculate (a) the current, and (b) the terminal p.d. across the 23 Ω resistor.
(a) I = ε/(R + r) = 12/(23 + 1). (b) V = IR, using the current from (a) and R = 23 Ω.
23A battery has emf 12.0 V and internal resistance 1.5 Ω. Calculate the terminal p.d. when it is supplying a current of 3.0 A.
V = ε − Ir = 12.0 − (3.0 × 1.5)
24A high-resistance voltmeter reads 12.5 V across a battery's terminals with no current flowing. When connected to a lamp, a current of 2.5 A flows and the voltmeter reading falls to 11.8 V. Calculate the internal resistance of the battery.
With no current, the voltmeter reads the emf: ε = 12.5 V. "Lost volts" Vr = ε − Vterminal = 12.5 − 11.8. Then r = Vr / I, using I = 2.5 A.
10. Electrical Power
If a current of 3 A flows through a resistor with a p.d. of 6 V across it, then 3 C of charge passes every second, and each coulomb transfers 6 J — so energy is being transferred at 3 × 6 = 18 joules every second (18 W). More generally, the rate of energy transfer (power) dissipated by a resistor is:
Key idea. Electrical power, P, dissipated by a resistor.
P = IV
Since V = IR, this can also be written two other useful ways:
P = I²R = V² / R
To find the total energy transferred over a time t, use energy = power × time, so electrical energy = VIt. Whenever current passes through resistance, some of that energy becomes internal (thermal) energy — the heating effect discussed in section 6, used deliberately in kettles, heaters and irons, and unavoidably in every resistor that carries a current.
Worked example 10.1
An electric iron is labelled 230 V, 1100 W. Calculate the resistance of its heating coil.
Answer:
P = V² / R
1100 = 230² / R
Check your understanding
25A 12 V p.d. is applied across a 240 Ω resistor. Calculate (a) the current and (b) the power dissipated.
(a) I = V/R = 12/240. (b) P = IV, using your value of I and V = 12 V (or directly, P = V²/R = 12²/240).
26A 2.00 kW household water heater has a resistance of 24.3 Ω. Calculate the current that flows through it, and state the mains voltage it must be designed for.
Use P = I²R, rearranged: I² = P/R = 2000/24.3, then take the square root to find I. Once you have I, find the voltage from V = IR (or V = P/I).
11. Variable Resistors and Potentiometers
Not every resistor has a fixed value. A variable resistor (sometimes called a rheostat when used to control current) lets you change the resistance in a circuit continuously, usually via a sliding or rotating contact.
Potentiometer: a three-terminal variable resistor connected so that its sliding contact taps off a fraction of the full supply p.d. Moving the slider from one end to the other varies the output p.d. continuously from 0 V up to the full supply p.d. — this is the best way to vary the p.d. across a component under investigation.
Live simulation: potentiometer as a potential divider
Slide the wiper
Vin = 12.0 V → Vout = 6.0 V
Fig. 11.1 A potentiometer supplied with Vin = 12 V (unloaded). Vout is measured between the wiper (a floating lead that moves with the slider) and the 0 V end of the track (a fixed lead), and varies from 0 V to the full 12 V as the wiper slides.
Worked example 11.1
A rheostat can vary from 0 to 48 Ω and is in series with a fixed 24 Ω component, across a constant 12 V supply of negligible internal resistance. Calculate the current when the rheostat is set to 24 Ω.
Answer:
I = V / Rtotal
I = 12 / (24 + 24)
Check your understanding
27Using the same rheostat and component as Worked example 11.1 (12 V supply, 24 Ω fixed component, rheostat 0–48 Ω), calculate the p.d. across the fixed component when the rheostat is set to 0 Ω, and again when it is set to 48 Ω.
At 0 Ω: I = 12/(24+0) = 0.50 A, so Vcomponent = IR = 0.50 × 24. At 48 Ω: I = 12/(24+48), so Vcomponent = I × 24 using this new current.
28A potentiometer of total resistance 48 Ω is connected across a 12 V supply and used, unloaded, as a potential divider. The output is taken from a point one-quarter of the way up the track from the bottom (0 V) end. Calculate Vout.
With no load, the p.d. divides in direct proportion to the resistance of each section of track. One-quarter of the way up means one-quarter of the total p.d.: Vout = (1/4) × 12.
Glossary
Cell
A device that transfers energy from another source (chemical reaction, or light for a solar/photovoltaic cell) to charge carriers passing through it.
Emf, ε
The total energy transferred, per unit charge, by a source of electrical energy — the theoretical maximum p.d. it can supply.
Charge carrier
A charged particle free to move; in a metal, delocalized (free) electrons.
Direct current (dc)
A flow of electric charge that is always in the same direction, I = ΔQ/Δt.
Potential difference, V
The work done per unit charge on moving a positive charge between two points along the path of the current, V = W/q.
Kirchhoff's voltage law
The sum of the potential differences around any closed loop is zero.
Kirchhoff's current law
The current entering any junction equals the current leaving it.
Conductor
A material with many mobile charge carriers, and therefore low resistance.
Insulator
A material with very few mobile charge carriers, and therefore very high resistance.
Resistance, R
The ratio of p.d. across a conductor to the current through it, R = V/I. Unit: ohm (Ω).
Ohm's law
For a conductor at constant temperature, current is proportional to the p.d. across it.
Ohmic / non-ohmic
Ohmic components have constant resistance (I ∝ V); non-ohmic components' resistance changes with current, e.g. a filament lamp as it heats up.
Resistivity, ρ
A material property equal to RA/L; the resistance of a 1 m length with 1 m² cross-sectional area. Unit: ohm metre (Ω m).
Internal resistance, r
The resistance of the materials inside a cell or battery itself.
Terminal p.d.
The p.d. actually available to the external circuit, equal to the emf minus the "lost volts" due to internal resistance.
Electrical power, P
The rate of energy transfer in a circuit, P = IV = I²R = V²/R.
Rheostat
A variable resistor used (with two terminals in use) to control current.
Potentiometer
A three-terminal variable resistor used with a sliding contact to produce a varying output p.d.