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.
Charging, discharging and time constant 16(l)
Question 1
A 20 μF capacitor charges through 0.50 MΩ from a 12 V source. Find τ and the capacitor p.d. after one time constant.
Check the model response
τ = RC = (0.50 × 10⁶)(20 × 10⁻⁶) = 10 s. V = V₀(1 − e⁻¹) = 7.59 V, about 63% of its final value.
repair
2. Repair the common breaks
Use only the correction matching an error, then retry the corresponding diagnostic.
Charging, discharging and time constant 16(l)
Check this idea
Misconception: Charging or discharging is complete after one time constant.
Repair: After one τ, a rise is 63.2% complete and a decay retains 36.8%; the exponential approaches its limit asymptotically.
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.
Charging, discharging and time constant 16(l)
Model 1
A 40 μF capacitor discharges from 12 V through 1.0 MΩ. Find V and I at 20 s.
Check the model response
τ = RC = 40 s. V = 12e⁻⁰⋅⁵ = 7.28 V. Current magnitude is V/R = 7.28 μA and decays with the same exponential; its conventional sign depends on the chosen reference direction.
guided practice
4. Guided practice
Use each hint only to select the correct topology, material model, combination rule or exponential.
Charging, discharging and time constant 16(l)
Question 1
A discharging quantity falls to 0.25x₀. Express t in terms of τ.
Hint: Take the natural logarithm of both sides.
Check the model response
0.25 = e⁻ᵗ⁄τ, so t = −τ ln(0.25) = 1.386τ.
independent practice
5. Independent practice
Solve without repair notes and state graph axes, ideal-meter assumptions, source model and RC initial/final conditions.
Charging, discharging and time constant 16(l)
Question 1
Describe current, charge and capacitor p.d. during charging and discharging, including initial/final values and τ.
Check the model response
During charging, Q and VC rise as x₀(1 − e⁻ᵗ⁄τ), while current falls as I₀e⁻ᵗ⁄τ. During discharge, Q, VC and current magnitude fall as x₀e⁻ᵗ⁄τ. τ = RC; after one τ a rise is 63% complete and a decay retains 37%.
Practice exit check
6. Practice assessment
Use this as extra closed-book practice, then complete the separate recorded assessment in your plan.
Charging, discharging and time constant 16(l)
Question 1
A 25 μF capacitor discharges through 0.80 MΩ from 10 V. Find τ and V after 30 s.
Check the model response
τ = RC = 20 s. V = 10e⁻³⁰⁄²⁰ = 2.23 V.
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.
Charging, discharging and time constant 16(l)
Question 1
After one time constant, state the fractions reached during charging and remaining during discharge.
Check the model response
Charging reaches 1 − e⁻¹ = 0.632 of the final change; discharge retains e⁻¹ = 0.368 of its initial value.