Source: Discussion Questions 5A & 5B
Tags: capacitor networks, series capacitors, parallel capacitors, equivalent capacitance, circuit analysis, P212, PHYS 212, UIUC, University Physics, Electricity and Magnetism
Difficulty: Intermediate Prerequisites: Basic understanding of capacitance (C = Q/ΔV), voltage, and charge. Familiarity with circuit diagrams.
Capacitor network analysis is a core skill in electrostatics and circuit theory. You will use it repeatedly throughout P212 and beyond, whenever a circuit contains more than one capacitor. The method taught here, collapsing a network down to a single equivalent capacitor and then expanding back out, is the same strategy used for resistor networks later in the course. If you can do this confidently with capacitors, the resistor version will feel familiar.
Capacitors in parallel share the same voltage; their capacitances add directly. Capacitors in series share the same charge; their reciprocals add. To solve any capacitor network, collapse it step by step into one equivalent capacitor, find Q and V for that single capacitor, then expand back out, using the sharing rules at each step to find individual charges and voltages.
Equivalent capacitance (C_eq)
The single capacitance that could replace a group of capacitors and draw the same total charge from the battery. In simple terms, it is the "one capacitor to rule them all" that behaves identically to the whole network from the battery's point of view.
Capacitors in parallel
Capacitors whose terminals are both connected to the same two nodes, so they experience the same voltage across them. Think of it as capacitors sitting side by side, each with their own direct path to both battery terminals.
Capacitors in series
Capacitors connected end-to-end in a single chain, so the same charge must appear on each. Think of it as a queue: the charge that flows onto one capacitor's plate must come from the next capacitor's plate, so they all end up holding the same amount.
Collapse and expand method
The systematic procedure for analysing capacitor networks: reduce the circuit to a single equivalent capacitor (collapse), solve for Q and V, then reverse the process (expand), recovering individual voltages and charges at each stage.
Both capacitors sit across the same two nodes, so ΔV₁ = ΔV₂ = V₀
The total charge drawn from the battery splits between them: Q_total = Q₁ + Q₂
Equivalent capacitance:
C_parallel = C₁ + C₂ + C₃ + ...
Because Q = CV and V is shared, the larger capacitor stores more charge
To recover individual charges after collapsing: Q₁ = C₁ · V₀ and Q₂ = C₂ · V₀
Why the voltage is the same: any two components connected between the same pair of nodes must have the same potential difference. There is only one voltage between node A and node B, regardless of which path you take.
Capacitors are daisy-chained so that current flowing onto one plate must flow off the adjacent plate of the next capacitor
All capacitors in the chain carry the same charge: Q₁ = Q₂ = Q_total
The total voltage divides among them: V₀ = ΔV₁ + ΔV₂
Equivalent capacitance:
1/C_series = 1/C₁ + 1/C₂ + 1/C₃ + ...
The equivalent capacitance of a series combination is always smaller than the smallest individual capacitor
To recover individual voltages after collapsing: ΔV₁ = Q/C₁ and ΔV₂ = Q/C₂
Why the charge is the same: the wire between two series capacitors is an isolated conductor. It started with zero net charge and has no connection to the battery, so whatever charge appears on one plate must be drawn from the adjacent plate. Charge is conserved on that island of metal.
This is the universal procedure for any capacitor network:
Phase 1, Collapse:
Identify pairs or groups that are purely in series or purely in parallel
Replace each group with a single equivalent capacitor
Redraw the circuit after each replacement
Repeat until only one capacitor remains across the battery
Phase 2, Solve the single capacitor:
ΔV = V₀ (the battery voltage)
Q = C_eq · V₀
Phase 3, Expand:
Reverse the collapse one step at a time
At each step, apply the appropriate sharing rule:
Parallel: voltage is the same, find each Q from Q = CV
Series: charge is the same, find each V from V = Q/C
Continue until every original capacitor has its own Q and V
Given: V₀ = 12 V, C₁ = 5 µF, C₂ = 8 µF, C₃ = 15 µF, C₄ = 3 µF, C₅ = 10 µF
The circuit has C₁ in series with a parallel combination of (C₂ series C₃) and (C₄ parallel C₅).
Collapse step 1: Identify sub-groups.
C₂ and C₃ are in series: 1/C₂₃ = 1/8 + 1/15 = 23/120, so C₂₃ = 120/23 ≈ 5.22 µF
C₄ and C₅ are in parallel: C₄₅ = 3 + 10 = 13 µF
Collapse step 2: C₂₃ and C₄₅ are in parallel:
C₂₃₄₅ = 5.22 + 13 = 18.22 µF (more precisely 540/23 µF ≈ 23.48 µF... let me redo this)
Actually, looking at the circuit diagram more carefully: C₂ and C₃ are in series (top-to-bottom between nodes A and B), and C₄ and C₅ are in parallel with each other (both between nodes A and B). So C₂₃ is in parallel with C₄₅.
C₂₃ = 120/23 ≈ 5.217 µF
C₄₅ = 13 µF
C₂₃₄₅ = C₂₃ + C₄₅ = 120/23 + 13 = (120 + 299)/23 = 419/23 ≈ 18.217 µF
Collapse step 3: C₁ is in series with C₂₃₄₅:
1/C_eq = 1/5 + 23/419 = (419 + 115)/(5 × 419) = 534/2095
C_eq = 2095/534 ≈ 3.924 µF
Solve the single capacitor:
Q_total = C_eq · V₀ = 3.924 × 12 ≈ 47.08 µC
Expand step 1: C₁ and C₂₃₄₅ are in series, so Q₁ = Q₂₃₄₅ = 47.08 µC
ΔV₁ = Q₁/C₁ = 47.08/5 ≈ 9.42 V
ΔV₂₃₄₅ = Q₂₃₄₅/C₂₃₄₅ = 47.08/18.217 ≈ 2.58 V
V_A = V₀ − ΔV₁ = 12 − 9.42 ≈ 2.58 V (this is the voltage at point A)
Expand step 2: C₂₃ and C₄₅ are in parallel, so ΔV₂₃ = ΔV₄₅ = 2.58 V
Q₂₃ = C₂₃ · 2.58 = 5.217 × 2.58 ≈ 13.46 µC
Q₄₅ = C₄₅ · 2.58 = 13 × 2.58 ≈ 33.55 µC
Expand step 3: Break apart the sub-groups.
C₂ and C₃ in series: Q₂ = Q₃ = Q₂₃ ≈ 13.46 µC
ΔV₂ = Q₂/C₂ = 13.46/8 ≈ 1.68 V
ΔV₃ = Q₃/C₃ = 13.46/15 ≈ 0.90 V
V_B = V_A − ΔV₂ ≈ 2.58 − 1.68 ≈ 0.90 V
C₄ and C₅ in parallel: ΔV₄ = ΔV₅ = 2.58 V
Q₅ = C₅ · ΔV₅ = 10 × 2.58 ≈ 25.8 µC
Energy stored on C₁:
U₁ = ½ C₁ (ΔV₁)² = ½ × 5 × 10⁻⁶ × (9.42)² ≈ 222 µJ
Parallel:
C_eq = C₁ + C₂ + ... (capacitances add)
ΔV₁ = ΔV₂ = V₀
Series:
1/C_eq = 1/C₁ + 1/C₂ + ... (reciprocals add)
Q₁ = Q₂ = Q_total
For two capacitors in series (shortcut):
C_eq = (C₁ · C₂) / (C₁ + C₂)
Energy stored in a capacitor:
U = ½ CV² = ½ Q²/C = ½ QV
Capacitor networks appear in power supply filtering, signal coupling, and timing circuits. Engineers combine capacitors in parallel to increase total capacitance (common in power supplies where a single component with the required value may not exist or may be too expensive), and in series to increase the voltage rating of the overall combination.
Students often assume that capacitors in series add directly, the way resistors in series do. They do not. Series capacitors add by reciprocals. The analogy flips: capacitors in parallel add directly (like resistors in series), and capacitors in series add by reciprocals (like resistors in parallel).
Confusing which quantity is shared. In parallel, voltage is shared. In series, charge is shared. A useful mnemonic: "parallel plates see the same potential" (voltage), while "series means single file" (same charge passes through all).
Forgetting to redraw the circuit after each collapse step. Skipping the redraw is the single biggest source of errors in multi-step networks. Always sketch the simplified circuit before moving to the next collapse.
Treating the expand phase as optional. Students sometimes find C_eq and Q_total and stop, but the exam almost always asks for the voltage or charge on a specific individual capacitor, which requires expanding back out.
⚠️ The collapse-and-expand method is the backbone of every capacitor (and later, resistor) network problem. Expect at least one multi-step network on every exam.
⚠️ "Find the voltage at point X" means voltage relative to ground (the negative terminal of the battery), not relative to some other node. Watch the reference point.
⚠️ Energy questions (U = ½CV²) require you to first find V across the specific capacitor, which means going through the full expand procedure.
⚠️ Sanity check: the sum of voltage drops across series elements must equal the source voltage. The sum of charges on parallel branches must equal the total charge. Use these to catch arithmetic errors.
True or False: The equivalent capacitance of capacitors in series is always less than the smallest individual capacitor.
Fill in the blank: For capacitors in parallel, the ______ across each capacitor is the same.
True or False: If two capacitors are in series across a 10 V battery, and one has twice the capacitance of the other, the larger capacitor has the larger voltage drop.
Fill in the blank: The first step in any capacitor network problem is to ______ the network down to one equivalent capacitor.
True or False: Energy stored in a capacitor is proportional to the square of the charge on it.
Answers: 1. True. 2. Voltage. 3. False (the smaller capacitor gets the larger voltage drop, since V = Q/C and Q is the same). 4. Collapse. 5. True (U = Q²/2C).
Q: Two capacitors, 4 µF and 6 µF, are connected in parallel across a 9 V battery. What is the charge on each?
A: Both see 9 V. Q₁ = 4 × 9 = 36 µC. Q₂ = 6 × 9 = 54 µC.
Q: The same two capacitors (4 µF and 6 µF) are now connected in series across the same 9 V battery. What is the voltage across each?
A: C_eq = (4 × 6)/(4 + 6) = 2.4 µF. Q = 2.4 × 9 = 21.6 µC. ΔV₁ = 21.6/4 = 5.4 V. ΔV₂ = 21.6/6 = 3.6 V.
Q: In a series combination, why must the charge on each capacitor be the same?
A: The conductor between two series capacitors is electrically isolated. It began with zero net charge and has no external connection, so by conservation of charge, whatever positive charge accumulates on one side must be drawn from the other side. The magnitudes must match.
Q: You collapse a five-capacitor network and find C_eq = 4 µF across a 12 V battery. What is the total charge, and what is the total energy stored?
A: Q = 4 × 12 = 48 µC. U = ½ × 4 × 10⁻⁶ × 144 = 288 µJ.
Q: After expanding back to individual capacitors, you find ΔV₁ = 7.5 V and ΔV₂ = 4.5 V for two series capacitors. How can you check your work?
A: The voltages must sum to V₀. 7.5 + 4.5 = 12 V, which matches a 12 V battery. If they did not sum correctly, there is an arithmetic error somewhere.
This connects to resistor network analysis (coming later in P212), which uses the same collapse-and-expand logic but with the series/parallel rules swapped. It also connects to energy storage in electric fields, since U = ½CV² ties capacitor analysis directly to the energy density of the electric field between the plates. If you continue to AC circuits, the impedance of capacitors in series and parallel follows identical combination rules.
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