D.C. circuits: components, networks and transients
Key idea: Read topology before calculating, distinguish component and source behaviour, and use shared-current, shared-p.d., shared-charge and exponential models only in the arrangements where they apply.
Before you start: Current Electricity objective chainElectric Fields objective chain
By the end, you can
- Use standard circuit symbols and interpret electrical topology with correct meter placement.
- Apply resistance and resistivity while explaining component I–V and temperature behaviour microscopically.
- Analyse terminal p.d. and power when a source has internal resistance.
- Solve resistor networks and sensor potential dividers.
- Combine capacitors using shared-charge and shared-potential-difference reasoning.
- Represent capacitor charging and discharging with the correct exponential and time constant.
Starting-point self-check
1. Check your starting point
Attempt all six groups without notes and mark the first diagram, material, source, network, capacitance or transient decision you cannot justify. Use the recorded topic diagnostic above when you want scoring and a personalised repair plan.
Resistance, resistivity, I–V characteristics and temperature 16(c)–(f)
Question 1
A 1.5 m metal wire of area 0.20 mm² carries 0.40 A at 2.4 V. Find R and ρ, then state how heating changes its I–V response.
Check the model response
R = V/I = 6.0 Ω. A = 2.0 × 10⁻⁷ m², so ρ = RA/l = 8.0 × 10⁻⁷ Ω m. Heating increases lattice vibration, reduces drift velocity for a given field and raises resistance, so current grows less than proportionally with p.d.
repair
2. Repair the common breaks
Use only the correction matching an error, then retry the corresponding diagnostic.
Resistance, resistivity, I–V characteristics and temperature 16(c)–(f)
Check this idea
Misconception: Resistance is always the gradient of an I–V graph.
Repair: Resistance is V/I at an operating point. Graph-gradient interpretation depends on which quantity is plotted vertically and on linearity.
Check this idea
Misconception: Metals and NTC thermistors change resistance for the same microscopic reason.
Repair: Metal heating reduces drift velocity through greater scattering; NTC heating greatly increases carrier number density.
worked example
3. Follow six worked models
Follow how each solution fixes nodes, axes, source boundary, network reduction, shared capacitor quantity or initial condition before calculating.
Resistance, resistivity, I–V characteristics and temperature 16(c)–(f)
Model 1
Compare the I–V shapes of an ohmic resistor, filament lamp, semiconductor diode and NTC thermistor.
Check the model response
At constant temperature an ohmic resistor gives a straight line through the origin. A lamp's gradient decreases in an I-against-V graph as heating raises metal resistivity. A diode has negligible reverse current and a steep forward rise after its turn-on region. An NTC thermistor heats as current rises; greater carrier number density lowers resistance, making the curve progressively steeper.
guided practice
4. Guided practice
Use each hint only to select the correct topology, material model, combination rule or exponential.
Resistance, resistivity, I–V characteristics and temperature 16(c)–(f)
Question 1
A wire's length doubles and diameter halves at constant ρ. State the resistance factor.
Hint: Apply the diameter change to area before using R = ρl/A.
Check the model response
Area is proportional to diameter squared, so it becomes one quarter. R = ρl/A therefore changes by 2/(1/4) = 8.
independent practice
5. Independent practice
Solve without repair notes and state graph axes, ideal-meter assumptions, source model and RC initial/final conditions.
Resistance, resistivity, I–V characteristics and temperature 16(c)–(f)
Question 1
Explain temperature effects in a metal and NTC semiconductor using the specific microscopic quantities required by the syllabus.
Check the model response
In a metal, higher temperature increases lattice scattering, so carrier drift velocity for a given field falls and resistivity rises. In an NTC semiconductor, heating releases many more charge carriers, increasing number density enough to lower resistivity despite increased scattering.
Practice exit check
6. Practice assessment
Use this as extra closed-book practice, then complete the separate recorded assessment in your plan.
Resistance, resistivity, I–V characteristics and temperature 16(c)–(f)
Question 1
A 2.0 m wire of radius 0.15 mm has ρ = 4.0 × 10⁻⁷ Ω m. Find R. Then contrast its heated I–V curve with an NTC thermistor's.
Check the model response
A = π(1.5 × 10⁻⁴)² = 7.07 × 10⁻⁸ m², so R = ρl/A = 11.3 Ω. Metal heating reduces drift velocity and raises resistance; NTC heating increases carrier number density and lowers resistance.
Re-test practice
7. Delayed re-test practice
Return after at least three days and solve these fresh contexts without reopening earlier responses. The recorded plan enforces the delay and uses a separate re-test family for selected-response skill-group evidence.
Resistance, resistivity, I–V characteristics and temperature 16(c)–(f)
Question 1
At constant ρ, length triples and area doubles. State the resistance factor.
Check the model response
R = ρl/A, so the factor is 3/2.