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.

  • H2 Physics 9478 · 2027
  • Internally reviewed by MiniEducation Team
  • Recorded selected-response study loop available

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.

Series, parallel and potential-divider networks 16(h)–(j)

Question 1

A 12 V source supplies 3.0 kΩ in series with a parallel pair of 6.0 kΩ and 2.0 kΩ. Find total resistance, source current and p.d. across the parallel branch.

Check the model response

The parallel pair is 1.5 kΩ; total resistance is 4.5 kΩ. I = 12/4500 = 2.67 mA and branch p.d. = I(1.5 kΩ) = 4.0 V.

repair

2. Repair the common breaks

Use only the correction matching an error, then retry the corresponding diagnostic.

Series, parallel and potential-divider networks 16(h)–(j)

Check this idea

Misconception: A divider ratio is unchanged when a load is connected.

Repair: A finite load changes the effective resistance of the arm it parallels and therefore changes Vout.

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.

Series, parallel and potential-divider networks 16(h)–(j)

Model 1

A 10 V divider uses 8.0 kΩ above 4.0 kΩ; a 6.0 kΩ load is placed across the lower resistor. Calculate the loaded output.

Check the model response

The lower branch becomes 4.0 || 6.0 = 2.4 kΩ. It is in series with 8.0 kΩ, so Vout = 10[2.4/(8.0 + 2.4)] = 2.31 V, below the unloaded 3.33 V.

guided practice

4. Guided practice

Use each hint only to select the correct topology, material model, combination rule or exponential.

Series, parallel and potential-divider networks 16(h)–(j)

Question 1

Find the equivalent resistance of 4.0 Ω and 12 Ω in parallel, then in series with 2.0 Ω.

Hint: Reduce one network block at a time.

Check the model response

Parallel resistance is (4 × 12)/(4 + 12) = 3.0 Ω; total is 5.0 Ω.

independent practice

5. Independent practice

Solve without repair notes and state graph axes, ideal-meter assumptions, source model and RC initial/final conditions.

Series, parallel and potential-divider networks 16(h)–(j)

Question 1

Derive the series and parallel resistor rules, then explain how an NTC thermistor changes divider output when heated.

Check the model response

Series components share current and their p.d.s add, giving Rtotal = ΣR. Parallel branches share p.d. and their currents add, giving 1/Rtotal = Σ(1/R). Heating lowers an NTC resistance; whether Vout rises or falls depends on whether output is across the NTC or the other arm.

Practice exit check

6. Practice assessment

Use this as extra closed-book practice, then complete the separate recorded assessment in your plan.

Series, parallel and potential-divider networks 16(h)–(j)

Question 1

A 9.0 V divider has 3.0 kΩ above an NTC of 6.0 kΩ. Find Vout across the NTC, then its value when heating reduces it to 2.0 kΩ.

Check the model response

Initially Vout = 9[6/(3 + 6)] = 6.0 V. Heated, Vout = 9[2/(3 + 2)] = 3.6 V. Output across the NTC falls as its resistance falls.

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.

Series, parallel and potential-divider networks 16(h)–(j)

Question 1

Find the resistance of 6.0 Ω in series with two 8.0 Ω resistors in parallel.

Check the model response

The parallel pair is 4.0 Ω, giving 10.0 Ω total.

Continue with established practice

Use the established six-question structured set after the delayed re-test. It directly samples resistance, materials, sources, networks, capacitance and RC transients; symbols and I–V sketching remain assessed in this chain and the quiz.

Open D.C. Circuits structured practice