Chapter revision
Revision summary
Key ideas, equations and common mistakes. Open any topic below for the full explanation.
Identify the nodes, the component and the stated conditions. Then choose a relationship that uses the correct current, voltage and source model.
Connections and measurements
- Series components share current along an unbranched path; parallel components share both endpoints and therefore p.d.
- An ammeter measures current in its own path. A voltmeter measures p.d. between its own terminals. Pair component V with component I before using R = V/I.
- Check range, resolution, zero, polarity, loading and thermal conditions. Repeated readings assess scatter but do not remove a consistent wrong-endpoint or zero error.
Resistance and material behaviour
A = πd2/4
Rseries = R1 + R2 + …
1/Rparallel = 1/R1 + 1/R2 + …
Convert diameter before squaring it. At fixed material, temperature and length, resistance is proportional to 1/A; doubling diameter quarters resistance. On an I-against-V plot, only a straight line through the origin has gradient 1/R at all points.
- Ohmic resistor: proportional I and V under fixed physical conditions.
- Filament lamp: settled heating raises resistance, so the I-V curve becomes less steep at greater voltage magnitude.
- Semiconductor diode: forward conduction and small reverse current before breakdown; no universal forward threshold.
- NTC thermistor: warming lowers resistance. A self-heated voltage sweep differs from a fixed-temperature, low-field measurement.
For a typical metal at the same electric field, greater scattering lowers mean drift speed while carrier density changes little. For the stated NTC regime, increased carrier number density is the main explanation for falling resistivity. Random thermal motion is not the directed drift.
Sources, networks and outputs
I = Es/(Rexternal + r)
Pload = IVterminal
Pinternal = I2r
EsI = Pload + Pinternal
These source signs describe current delivered to the load. A terminal-V-against-I line has intercept Es and gradient -r. Maximum load power occurs at Rexternal = r for this fixed source model, where load and internal powers are equal; this is not maximum efficiency.
Reduce a network to find total current, then work back to its branch voltages and currents. After a connection changes, recalculate the complete network: another branch's voltage may change even if its resistance does not.
Vout = VsRlower/(Rupper + Rlower)
Use the actual supply-terminal p.d. The selected output determines the numerator. Work through sensor resistance, total current and output rather than memorising a direction. An output load is parallel with its selected arm and changes the effective divider.
Capacitors and switching
Cparallel = C1 + C2 + …
1/Cseries = 1/C1 + 1/C2 + …
Parallel capacitors share voltage. In the stated initially uncharged series arrangement, an isolated neutral middle node enforces equal charge magnitudes; its potential need not be zero. Keep any supplied initial node charge in a more general arrangement.
For the simple RC path, τ = RC. Take Q on the initially positive plate, V = Q/C and positive current towards that plate:
Q = CEs(1 - e-t/τ)
V = Es(1 - e-t/τ)
I = (Es/R)e-t/τ
Discharging from Q0 = CV0:
Q = Q0e-t/τ
V = V0e-t/τ
I = -(V0/R)e-t/τ
At one time constant, a rising quantity reaches 63.2% of its final value and a decaying quantity retains 36.8% of its initial value. The half-value time is τ ln 2, while 90% charging takes -τ ln(0.10).
Read discharge data
τ = -1/(log-plot gradient)
C = τ/R ΔQ = signed current-time area
Log ratios and exponential arguments are dimensionless. Fit ln(V/V0) against numeric time; the gradient has unit s-1. A coarse trapezium sum approximates a curved area, and rounded model values can move a fitted result slightly. Neither is evidence of measured scatter. Meter loading and the actual switched resistor path matter when interpreting a real record.
Quantity and unit reference
| Quantity | Symbol | SI unit |
|---|---|---|
| Current | I | A |
| Potential difference | V | V = J/C |
| Source e.m.f. | Es, E | V |
| Resistance; internal resistance | R; r | Ω = V/A |
| Resistivity | ρ | Ω m |
| Length; diameter | l; d | m |
| Cross-sectional area | A | m2 |
| Power | P | W = J/s |
| Charge | Q | C = A s |
| Time | t | s |
| Capacitance | C | F = C/V |
| Time constant | τ | s |
Read symbols in context: A as area differs from A as the ampere unit, and C as charge's unit differs from a node labelled C or the capacitance symbol C. Useful prefixes are micro (µ, 10-6), milli (m, 10-3), kilo (k, 103) and mega (M, 106). Squared lengths and areas require squared conversion factors.
For a calculation, draw or trace the nodes, choose reference directions and endpoints, state the component conditions, solve with units, then check current and energy accounts. For a measurement, retain the actual readings and instrument information as well as the inferred result.
Practise exam questions
Write your answer first, then tick the mark points it makes.
Exam-style question
Define the resistivity of a material. [2]
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Exam-style question
A wire of length 2.0 m has a diameter of 0.50 mm. The resistivity of its material is 1.1 x 10-6 Ω m. Calculate its resistance. [3]
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Exam-style question
Explain why the terminal potential difference of a cell is less than its e.m.f. when it supplies a current. [2]
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Exam-style question
A student varies the current from a cell and plots terminal p.d. V against current I. The line is straight, meets the V-axis at 1.50 V and passes through (0.50 A, 1.10 V). Use the graph to find the e.m.f. and the internal resistance of the cell. [3]
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Exam-style question
Data-based question: A defibrillator capacitor [20]
A defibrillator stops a dangerous heart rhythm by passing a short, large current through the patient's chest. A 32 µF capacitor is charged to 5.00 kV and then discharged through two pads on the chest. The chest acts as a resistor R. The capacitor's plates are separated by a 0.10 mm layer of insulating material, which breaks down if the electric field in it exceeds 2.0 x 108 V m-1.
| t / ms | 0 | 2.0 | 4.0 | 6.0 | 8.0 |
|---|---|---|---|---|---|
| V / kV | 5.00 | 3.03 | 1.84 | 1.12 | 0.68 |
- (a) Define capacitance. [1]
- (b) Calculate the energy stored in the capacitor at 5.00 kV. [2]
- (c) Use Table 1 to show that the discharge is exponential. [2]
- (d) Determine the time constant of the discharge, and hence the resistance of the chest. [3]
- (e) Calculate the initial current and the initial power delivered to the chest. [2]
- (f) Calculate the energy delivered to the chest in the first 4.0 ms. [2]
- (g) Calculate the electric field strength in the insulating layer at 5.00 kV, and comment on whether it is safe. [2]
- (h) A technician says that doubling the charging voltage would double the energy delivered. Explain whether this is correct. [2]
- (i) A guideline says at least 300 J must reach the chest within 5.0 ms. Evaluate whether this defibrillator meets the guideline for this patient and for a patient whose chest resistance is twice as large. [3]
- (j) State one assumption you made in (d) or (i). [1]
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Quick recall
R1 joins A to B, while R2 and R3 each join B to C. A source has external terminals A/C and a private internal node D before its internal resistance. Which redraw and terminal-voltage measurement preserve the intended circuit?
Review a topic
- Read connections and place meters
- Resistance, material and geometry
- I-V curves and temperature
- Sources with internal resistance
- Reduce a network and work back
- Potential dividers and sensors
- Capacitor combinations
- RC charging and discharging
- Infer a time constant from data