2.6 Current, potential difference and resistance
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AS content for Unit 2 (Waves and Electricity). Series and parallel circuits and I–V characteristics are on 2.7, resistivity and potential dividers on 2.8, e.m.f. on 2.9, and temperature and light effects on 2.10.
What this page covers
6 points
- Current as the rate of flow of charge; the coulomb and the ampere; counting electrons
- Potential difference as energy transferred per unit charge; the volt
- Resistance \(R = V/I\), the ohm, and Ohm's law as a special case
- Electrical power and energy: \(P = VI\), \(W = VIt\), \(P = I^2R\), \(P = V^2/R\), kilowatt-hours
- Drift velocity and the derivation of \(I = nqvA\)
- How the number density of charge carriers explains the resistivities of metals, semiconductors and insulators
In one minute
6 points
- Current is the rate of flow of charge, \(I = \Delta Q/\Delta t\), with 1 A = 1 C s⁻¹, and charge comes in multiples of \(e = 1.60 \times 10^{-19}\) C.
- Potential difference is the energy transferred per unit charge: \(V = W/Q\). 1 V = 1 J C⁻¹.
- Resistance is defined by \(R = V/I\). Ohm's law: for a metallic conductor at constant temperature, \(I \propto V\).
- Power \(P = VI = I^2R = V^2/R\); energy \(W = VIt\). 1 kWh = 3.6 × 10⁶ J.
- \(I = nqvA\): current depends on the number of carriers per m³, their charge, their drift velocity and the cross-sectional area.
- Metals have a huge \(n\) (low resistivity), semiconductors far fewer carriers, insulators almost none.
Contents
12 sections
- What is electric current?
- What is potential difference?
- What is resistance, and what does Ohm's law say?
- How do you calculate electrical power and energy?
- What is drift velocity, and where does I = nqvA come from?
- Why do metals, semiconductors and insulators have such different resistivities?
- Exam tips for 2.6
- Common misconceptions
- How this topic connects
- Key equations
- Check your understanding
- Specification checklist
What is electric current?
Every circuit question rests on three quantities: how much charge flows each second, how much energy each coulomb carries, and how hard it is to push the charge through. Current is the first.
This equation is in the formulae list supplied in the exam. For a steady current \(\Delta Q = I\Delta t\); for a changing current, the charge is the area under the current–time graph.
- Units: the ampere is an SI base unit. The coulomb is defined from it: 1 C = 1 A s, the charge passing a point when 1 A flows for 1 s.
- Charge carriers: free (delocalised) electrons in metals; positive and negative ions in electrolytes and ionised gases; electrons and "holes" in semiconductors.
- Charge is quantised: any charge is a whole number of electron charges, \(Q = Ne\), so the number of electrons passing is \(N = Q/e\) (learn this; it is not given).
- Charge is conserved: it is not used up in a component, so the current into a component equals the current out of it. Measure current with an ammeter in series; an ideal ammeter has zero resistance.
Conventional current is drawn from the positive terminal to the negative terminal round the external circuit. In a metal the electrons actually move the opposite way; the convention was fixed before the electron was discovered.
A phone charger delivers a steady current of 1.8 A for 45 minutes. \(e = 1.60 \times 10^{-19}\) C. Calculate (a) the charge that flows and (b) the number of electrons this represents.
Time in seconds: \(t = 45 \times 60 = 2700\) s.
(a) \(\Delta Q = I\Delta t = 1.8 \times 2700 =\) 4.9 × 10³ C (4860 C).
(b) \(N = Q/e = 4860 \div 1.60 \times 10^{-19} =\) 3.0 × 10²² electrons.
Leaving the time in minutes or hours. \(I = \Delta Q/\Delta t\) gives coulombs only if the current is in amperes and the time in seconds; \(1.8 \times 45 = 81\) is not a charge in any standard unit.
What is potential difference?
This equation is in the formulae list supplied in the exam. Across a component such as a resistor or lamp, energy is transferred from the charge carriers to the component, for example as thermal energy.
- The volt: 1 V = 1 J C⁻¹. A p.d. of 1 V means 1 J is transferred for every coulomb that passes.
- P.d. is measured with a voltmeter in parallel with the component; an ideal voltmeter has infinite resistance, so it draws no current.
- For a single electron, the energy transferred is \(W = eV\), which gives the electronvolt used on 2.5. The energy given to each coulomb by a source is its e.m.f. (2.9).
The p.d. across the lamp of an LED torch is 3.6 V and the current in it is 0.25 A. The torch is switched on for 20 minutes. \(e = 1.60 \times 10^{-19}\) C. Calculate (a) the charge that passes through the lamp, (b) the energy transferred to the lamp and (c) the energy transferred by each electron.
(a) \(t = 20 \times 60 = 1200\) s, so \(\Delta Q = 0.25 \times 1200 =\) 300 C.
(b) \(W = VQ = 3.6 \times 300 =\) 1.1 × 10³ J (1080 J).
(c) The same equation for one electron's charge: \(W = 3.6 \times 1.60 \times 10^{-19} =\) 5.8 × 10⁻¹⁹ J (= 3.6 eV).
What is resistance, and what does Ohm's law say?
\(R = V/I\) is in the formulae list supplied in the exam. It defines resistance for any component at any instant: use the values of \(V\) and \(I\) at the same operating point. Resistance measures how much p.d. is needed per ampere: the larger \(R\), the smaller the current for a given p.d.
- Ohm's law is a special case: an ohmic conductor has a resistance that stays the same whatever the current, and its I–V graph is a straight line through the origin.
- \(R = V/I\) is not Ohm's law. It holds for every component; Ohm's law adds the claim that \(R\) is constant.
- The condition matters: if a wire heats up, its resistance rises and \(I\) is no longer proportional to \(V\). Filament lamps, thermistors and diodes are non-ohmic (2.7).
A student uses the circuit above to measure the current in a coil of constantan wire for several p.d.s. The coil is kept in a beaker of water at room temperature.
| V / V | 0.50 | 1.00 | 1.50 | 2.00 | 2.50 | 3.00 |
|---|---|---|---|---|---|---|
| I / A | 0.042 | 0.083 | 0.126 | 0.167 | 0.208 | 0.250 |
Show that the coil obeys Ohm's law and determine its resistance.
V/I for every row: 11.9, 12.0, 11.9, 12.0, 12.0, 12.0 Ω. All values agree to within 1%, so \(V/I\) is constant: \(I \propto V\) at constant temperature, and Ohm's law is obeyed.
Resistance: the mean of the ratios, or the inverse gradient of the line: R = 12 Ω (12.0 Ω).
The water bath keeps the temperature constant, and small currents also limit heating. Plotting the graph is the stronger test: a straight line through the origin shows proportionality, not just a straight line.
Using the gradient of a curved I–V graph as the resistance. The resistance at a point is \(V/I\) read from that point; the inverse gradient equals the resistance only for a straight line through the origin.
How do you calculate electrical power and energy?
\[P = VI \qquad W = VIt \qquad P = I^2R \qquad P = \frac{V^2}{R}\]Here \(P\) is power (W), \(W\) energy transferred (J) and \(t\) time (s); all four are in the formulae list supplied in the exam. They follow from the definitions: \(W = VQ\) and \(Q = It\), so \(W = VIt\) and \(P = W/t = VI\). Substituting \(V = IR\) gives \(I^2R\); substituting \(I = V/R\) gives \(V^2/R\).
- Use \(P = I^2R\) when you know the current (components in series share a current); use \(P = V^2/R\) when you know the p.d. (appliances in parallel on the same supply).
- All the quantities must refer to the same component: the p.d. across it, the current in it and its resistance.
- Kilowatt-hours: energy in kWh = power in kW × time in hours; 1 kWh = 3.6 × 10⁶ J (1000 W × 3600 s; learn it).
A fan heater is rated at 2.5 kW when connected to a 230 V supply. It runs for 3.5 hours. Calculate (a) the current, (b) the resistance of the heater when operating and (c) the energy transferred, in kWh and in J.
(a) \(I = P/V = 2500 \div 230 =\) 11 A (10.87 A).
(b) \(R = V^2/P = 230^2 \div 2500 =\) 21 Ω (21.2 Ω).
(c) \(E = 2.5\ \text{kW} \times 3.5\ \text{h} = 8.75\ \text{kWh}\); \(\times 3.6 \times 10^{6}\): 8.8 kWh = 3.2 × 10⁷ J.
Check: \(I^2R = 10.87^2 \times 21.16 = 2500\) W, the rated power.
A workshop takes 25 kW from a supply through a cable whose total resistance is 0.15 Ω. Compare the power wasted in the cable when the power is delivered at 240 V and when it is delivered at 2400 V.
At 240 V: \(I = P/V = 25\,000 \div 240 = 104.2\) A. The cable's current is known, so use \(I^2R\): \(P = 104.2^2 \times 0.15 = 1.6\) kW (1628 W, 6.5% of 25 kW).
At 2400 V: \(I = 10.42\) A; \(P = 10.42^2 \times 0.15 = 16\) W (0.065% of 25 kW).
Ten times the p.d. → one tenth of the current → one hundredth of the power wasted (1.6 kW → 16 W).
The p.d. across the cable at 240 V is only \(IR = 15.6\) V. Using \(240^2/0.15 = 384\) kW mixes the supply p.d. with the cable's resistance, an impossible answer.
Saying "a bigger resistance always gets hotter". With the same current (series), \(P = I^2R\), so yes. With the same p.d. (parallel), \(P = V^2/R\), so the smaller resistance transfers more power. Say which quantity is fixed.
What is drift velocity, and where does I = nqvA come from?
The free electrons in a metal move randomly at high speed, colliding with the vibrating ions of the lattice; with no p.d. there is no net movement. A p.d. creates an electric field in the wire. Superimposed on the random motion, the electrons gain a small average velocity along the wire: the drift velocity \(v\). The number density \(n\) is the number of charge carriers per unit volume (m⁻³).
Deriving I = nqvA
The volume of wire swept through the cross-section in \(\Delta t\) is \(A \times v\Delta t\). It contains \(nAv\Delta t\) carriers, so the charge passing is \(\Delta Q = nAv\Delta t \times q\). Dividing by \(\Delta t\):
\[I = \frac{\Delta Q}{\Delta t} = nqvA\]Here \(n\) is in m⁻³, \(q\) the charge on each carrier (C), \(v\) the drift velocity (m s⁻¹) and \(A\) the cross-sectional area (m²). The equation is in the formulae list supplied in the exam, but be ready to derive it as above.
- In a series circuit the current is the same everywhere, so where a wire is thinner the electrons drift faster: \(v \propto 1/A\).
- Drift velocities in metal wires are tiny, typically fractions of a millimetre per second, because \(n\) is enormous.
- A lamp still lights almost at once when you close a switch: the electric field is set up round the whole circuit almost instantly, so the free electrons everywhere start drifting together. No electron has to travel from the switch to the lamp.
A copper connecting lead has a diameter of 0.80 mm and carries a current of 2.5 A. Copper has \(8.5 \times 10^{28}\) free electrons per m³. \(e = 1.60 \times 10^{-19}\) C. Calculate (a) the drift velocity of the electrons and (b) the time for an electron to drift along the 1.5 m length of the lead.
Area: radius \(0.40\ \text{mm} = 0.40 \times 10^{-3}\) m, so \(A = \pi(0.40 \times 10^{-3})^2 = 5.03 \times 10^{-7}\ \text{m}^2\).
(a) \(v = I/nqA = 2.5 \div (8.5 \times 10^{28} \times 1.60 \times 10^{-19} \times 5.03 \times 10^{-7}) =\) 3.7 × 10⁻⁴ m s⁻¹.
(b) \(t = 1.5 \div 3.66 \times 10^{-4} =\) 4.1 × 10³ s (about 68 minutes).
Using the diameter as the radius, or converting mm² with \(10^{-3}\). Halve the diameter first, convert mm to m, then square; or remember 1 mm² = \(10^{-6}\) m².
Why do metals, semiconductors and insulators have such different resistivities?
Resistivity is a property of the material (the resistance of a 1 m cube of it); its equation \(R = \rho l/A\) is on 2.8. Here the question is why it varies so much. Rearranging \(I = nqvA\): for a given current in a given sample, the fewer carriers there are, the faster each must drift, and a larger p.d. is needed to drive them. So a material with a small \(n\) has a large resistance and a large resistivity.
- Metals: roughly one or more free electrons per atom, \(n \sim 10^{28}\)–\(10^{29}\) m⁻³, so very low resistivity.
- Semiconductors (silicon, germanium): far fewer free carriers at room temperature, so a much higher resistivity. Their \(n\) rises steeply as temperature rises or light falls on them, the basis of thermistors and LDRs (2.10).
- Insulators: almost all electrons are bound to atoms; \(n\) is negligibly small, so the resistivity is enormous.
A rectangular semiconductor sample is 4.0 mm wide and 0.50 mm thick, and carries a current of 15 mA. The number density of charge carriers in it is \(2.0 \times 10^{22}\) m⁻³, each with charge \(1.60 \times 10^{-19}\) C. For copper, \(n = 8.5 \times 10^{28}\) m⁻³. (a) Calculate the drift velocity in the sample. (b) Calculate the drift velocity in a copper strip of the same cross-section carrying the same current. (c) Explain what the comparison shows.
Area: \(A = 4.0 \times 10^{-3} \times 0.50 \times 10^{-3} = 2.0 \times 10^{-6}\ \text{m}^2\).
(a) \(v = I/nqA = 0.015 \div (2.0 \times 10^{22} \times 1.60 \times 10^{-19} \times 2.0 \times 10^{-6}) =\) 2.3 m s⁻¹ (2.34 m s⁻¹).
(b) The same equation with copper's \(n\): 5.5 × 10⁻⁷ m s⁻¹.
(c) The ratio \(2.34 \div 5.51 \times 10^{-7} = 4.25 \times 10^{6}\) equals \(8.5 \times 10^{28} \div 2.0 \times 10^{22}\). With millions of times fewer carriers, each must drift millions of times faster; driving them needs a far larger p.d., which is why the semiconductor's resistivity is so much greater.
Exam tips for 2.6
mA → × 10⁻³, μA → × 10⁻⁶, kW → × 10³, minutes → × 60, mm² → × 10⁻⁶ m². A prefix slip loses credit just as surely as a physics error.
"Current is directly proportional to potential difference" needs the condition "for a metallic conductor at constant temperature" (constant physical conditions). Writing \(V = IR\) alone is the definition of resistance, not the law.
Read a pair of values off the curve and divide. Only for a straight line through the origin does the gradient give the resistance (as 1/gradient on an I–V graph).
Same current through several components: compare with \(I^2R\). Same p.d. across them: compare with \(V^2/R\). Always use the p.d. across, and the current in, the component you are asking about.
Start with "the current is the same at every point in series" or "the material, so \(n\), is the same", then deduce how \(v\) must change. Each link in the chain is a separate statement.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| Current is used up as it passes through a lamp. | Charge is conserved; the current into the lamp equals the current out. | Energy is transferred from the charge carriers to the lamp; the charge carries on. |
| Electrons travel round a circuit at nearly the speed of light. | Drift velocities in metal wires are typically below 1 mm s⁻¹. | The electric field is established almost instantly, so all the electrons start drifting at once. |
| \(R = V/I\) is Ohm's law. | It applies to every component, ohmic or not. | \(R = V/I\) defines resistance; Ohm's law says \(R\) is constant for a metal at constant temperature. |
| Resistance is the gradient (or inverse gradient) of an I–V graph at a point. | For a curved graph the tangent gives neither \(V\) nor \(I\) at that point. | Resistance at a point = \(V/I\) for that point. |
| Conventional current shows the direction in which electrons move. | Electrons are negative, so they drift from − to + round the external circuit. | Conventional current is the direction positive charge would flow: + to −. |
| A larger resistance always transfers more power. | On a fixed p.d., \(P = V^2/R\) falls as \(R\) rises. | Larger \(R\) → more power only when the current is fixed (\(P = I^2R\)). |
How this topic connects
Unit 1 power and efficiency
\(P = W/t\) and the kilowatt-hour are the same ideas as before; \(P = VI\) is the electrical input power to motors, heaters and lamps whose efficiency you calculate.
2.7 Series and parallel circuits; 2.8 Resistivity and potential dividers
Conservation of charge (currents at a junction) and of energy (p.d.s round a loop) build directly on the definitions of current and p.d. here, and I–V graphs for lamps, diodes and thermistors extend the resistance section. \(R = \rho l/A\) puts numbers on the material property explained by \(n\).
2.10 Resistance, temperature and light
Thermistors and LDRs are explained by changes in \(n\) and in lattice vibrations, using \(I = nqvA\).
Key equations
| Equation | Use | In the formulae list? |
|---|---|---|
| \(I = \Delta Q/\Delta t\) | current as rate of flow of charge; I in A, Q in C, t in s | Yes |
| \(V = W/Q\) | p.d. as energy per unit charge; V in V, W in J | Yes |
| \(R = V/I\) | definition of resistance; R in Ω | Yes |
| \(P = VI\) and \(W = VIt\) | electrical power and energy; P in W | Yes |
| \(P = I^2R\) and \(P = V^2/R\) | power when the current or the p.d. is known | Yes |
| \(I = nqvA\) | current in terms of drift velocity; n in m⁻³, v in m s⁻¹, A in m² | Yes |
| \(N = Q/e\) | number of electrons in a charge | No — learn it (charge is quantised) |
| \(I \propto V\) at constant temperature | Ohm's law | No — learn the statement |
| 1 kWh \(= 3.6 \times 10^{6}\) J | energy unit conversion | No — learn it (1000 W × 3600 s) |
Check your understanding
1. Show that 1 V = 1 J C⁻¹ is the same as 1 W A⁻¹.
1 W A⁻¹ = 1 J s⁻¹ ÷ 1 C s⁻¹ = 1 J C⁻¹. (This is \(V = P/I\) from \(P = VI\).)
2. Use the current–time graph in the first section to find the mean current over the 8.0 s.
Charge = area = 2.4 C; mean current = \(2.4 \div 8.0 =\) 0.30 A.
3. A 60 Ω resistor carries 0.20 A. Calculate the p.d. across it and the power it dissipates, two ways.
\(V = IR =\) 12 V; \(P = VI = 12 \times 0.20 =\) 2.4 W, and \(I^2R = 0.20^2 \times 60 = 2.4\) W.
4. Explain why a lamp lights almost immediately when a switch is closed, even though the drift velocity is less than 1 mm s⁻¹.
Closing the switch sets up an electric field throughout the circuit almost instantly. Free electrons are already present everywhere in the wires and the lamp filament, so they all start drifting at once, including those already in the filament.
5. Two wires of the same metal are joined in series. Wire B has half the diameter of wire A. Compare the currents and the drift velocities.
The currents are equal (series). B's area is one quarter of A's, and \(n\) and \(q\) are the same, so the drift velocity in B is 4 times that in A.
6. A 9.0 V battery drives 0.15 A through a lamp for 2.0 minutes. Calculate the energy transferred and the number of electrons that pass.
\(W = VIt = 9.0 \times 0.15 \times 120 =\) 162 J. \(Q = It = 18\) C, so \(N = 18 \div 1.60 \times 10^{-19} =\) 1.1 × 10²⁰ electrons.
Specification checklist
Numbers refer to the specification points.
- (64) I can describe current as the rate of flow of charged particles and use \(I = \Delta Q/\Delta t\), including finding charge from the area under a current–time graph.
- (64) I can convert a charge into a number of electrons using \(e\).
- (65) I can define potential difference as energy transferred per unit charge and use \(V = W/Q\).
- (66) I can use \(R = V/I\) and find a resistance from data or a graph.
- (66) I can state Ohm's law with its condition and explain why it is a special case of \(R = V/I\).
- (69) I can derive and use \(P = VI\), \(W = VIt\), \(P = I^2R\) and \(P = V^2/R\), and work with energies in kWh.
- (73) I can derive and use \(I = nqvA\) to find drift velocities and compare them in different conductors.
- (73) I can explain the large differences in resistivity between metals, semiconductors and insulators in terms of the number density of charge carriers.