Cambridge International AS & A Level · 19 · structured
Derive and use capacitor network rules
Derive and use capacitor network rules · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- demanding
- Marks
- 11
- Topics
- 2
- Answer
- Complete
Treat every capacitor as ideal and initially uncharged before the source is connected.
- (a)
Starting with C=Q/V, derive the combined capacitance of capacitors in parallel. State which electrical quantity is common.
3 marks - (b)
Starting with C=Q/V, derive the combined capacitance of capacitors in series. Explain why the charge magnitudes are equal.
4 marks - (c)
A 6.0 μF and 3.0 μF capacitor are in series, and that series branch is parallel with 4.0 μF. The whole network is connected across 12 V. Find total capacitance, source charge and the p.d. across each capacitor in the series branch.
4 marks
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(a)
Starting with C=Q/V, derive the combined capacitance of capacitors in parallel. State which electrical quantity is common.
Identify what the question is testing, organise the response into distinct mark-earning points, and make every conclusion traceable to a physical principle or to the evidence supplied.
- 1
Parallel components share the same two terminals and therefore the same p.d. V.
- 2
Charge supplied is the sum: Q_total=Q₁+Q₂+…=C₁V+C₂V+….
- 3
Divide by V: C_total=C₁+C₂+….
(b)
Starting with C=Q/V, derive the combined capacitance of capacitors in series. Explain why the charge magnitudes are equal.
List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.
- 1
The intermediate conductors are isolated and initially neutral, so equal/opposite charges form on their surfaces, giving the same charge magnitude Q on each capacitor.
- 2
Potential differences add: V_total=V₁+V₂+….
- 3
Substitute V_i=Q/C_i to obtain V_total=Q(1/C₁+1/C₂+…).
- 4
Divide by Q and use V_total/Q=1/C_total.
(c)
A 6.0 μF and 3.0 μF capacitor are in series, and that series branch is parallel with 4.0 μF. The whole network is connected across 12 V. Find total capacitance, source charge and the p.d. across each capacitor in the series branch.
List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.
- 1
C_series=(1/6.0+1/3.0)⁻¹=2.0 μF; adding the parallel branch gives C_total=6.0 μF.
- 2
Q_source=C_totalV=6.0 μF×12 V=72 μC.
- 3
Series-branch charge is 2.0 μF×12 V=24 μC on each capacitor.
- 4
V_6=24/6=4.0 V; V_3=24/3=8.0 V, summing to 12 V.
Complete marking guidance
Use these worked steps alongside the part-by-part answers. They explain the method and mark allocation; they do not add extra marks.
- (a) Parallel components share the same two terminals and therefore the same p.d. V.
- (a) Charge supplied is the sum: Q_total=Q₁+Q₂+…=C₁V+C₂V+….
- (a) Divide by V: C_total=C₁+C₂+….
- (b) The intermediate conductors are isolated and initially neutral, so equal/opposite charges form on their surfaces, giving the same charge magnitude Q on each capacitor.
- (b) Potential differences add: V_total=V₁+V₂+….
- (b) Substitute V_i=Q/C_i to obtain V_total=Q(1/C₁+1/C₂+…).
- (b) Divide by Q and use V_total/Q=1/C_total.
- (c) C_series=(1/6.0+1/3.0)⁻¹=2.0 μF; adding the parallel branch gives C_total=6.0 μF.
- (c) Q_source=C_totalV=6.0 μF×12 V=72 μC.
- (c) Series-branch charge is 2.0 μF×12 V=24 μC on each capacitor.
- (c) V_6=24/6=4.0 V; V_3=24/3=8.0 V, summing to 12 V.
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