Study guide
1.7 Capacitor and Inductor Combination Calculations
ACMA Amateur Radio Standard Recognition Certificate - Theory Exam
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1 Chapter 1: Electrical Principles and Components
1.1 Mathematics, Units and Metric Prefixes 1.2 DC Circuits, Voltage, Current, and Ohm's Law 1.3 Power, Energy, and Resistance Calculations 1.4 Resistor Colour Code and Component Symbol Identification 1.5 Capacitors: Construction, Ratings and Safety 1.6 Inductors: Construction, Factors Affecting Inductance and Combinations 1.7 Capacitor and Inductor Combination Calculations 1.8 AC Circuits, Frequency, Wavelength, and RMS Values 1.9 Reactance and Impedance in AC Circuits 1.10 Resonance, Q Factor and Tuned Circuits 1.11 Transformers: Turns Ratio, Impedance Matching and Eddy Currents 1.12 Rectifiers, PIV and Power Supply Smoothing 1.13 Active Components: Semiconductors, Diodes, and Transistors 1.14 Transistor and FET Symbols, Configurations and External Behaviour
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6 Chapter 6: Interference and Electromagnetic Compatibility (EMC)
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1.7 Capacitor and Inductor Combination Calculations
Key Takeaways Explain This Topic AI
Inductors combine exactly like resistors while capacitors combine the opposite way, which settles most combination questions before any formula is used.
The ACMA syllabus restricts these calculations to equal-value components, so four shortcuts cover the paper: C/n and nC for capacitors, nL and L/n for inductors.
A series capacitor string always totals less than its smallest member, and a parallel inductor group always totals less than its smallest member.
Four 10 uF capacitors as two parallel pairs joined in series total 10 uF, with double the voltage rating of a single part.
Series capacitors share the applied voltage, so four 63 V parts stand about 252 V, while a parallel bank is limited to the lowest rating present.
Last updated: July 2026
1.7 Capacitor and Inductor Combination Calculations
ACMA Exam Focus: Syllabus items 4.12 and 4.15 — understand and apply the formulae for capacitors and for inductors connected in series, in parallel and in series-parallel. Both items carry the note that calculations will only involve components of the same value , so the equal-value shortcuts are all the paper actually demands.
Two opposite rules, one pair of shortcuts
Resistors set the pattern most candidates already know: in series they add, and in parallel the total falls below the smallest. Inductors follow exactly the same pattern as resistors. Capacitors follow the opposite one. Fix that single sentence in your head and half the combination questions answer themselves before you reach for a formula.
- Connection Resistors Inductors Capacitors
- Series add add reciprocal — total is less than the smallest
- Parallel reciprocal — total is less than the smallest reciprocal — total is less than the smallest add
The physical reason is worth thirty seconds of your revision time. Wiring capacitors in parallel effectively enlarges the plate area, and capacitance rises with area, so the values add. Wiring them in series effectively increases the plate separation, and capacitance falls as separation grows, so the total drops below the smallest member. Inductors need no such special pleading: putting coils in series simply puts more turns into the same current path.
The general formulae
Capacitors in parallel: C T = C 1 + C 2 + C 3 + … C_T = C_1 + C_2 + C_3 + \dots C T = C 1 + C 2 + C 3 + …
Capacitors in series: 1 C T = 1 C 1 + 1 C 2 + 1 C 3 + … \frac{1}{C_T} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \dots C T 1 = C 1 1 + C 2 1 + C 3 1 + …
and, for exactly two, the product-over-sum form C T = C 1 C 2 C 1 + C 2 C_T = \dfrac{C_1 C_2}{C_1 + C_2} C T = C 1 + C 2 C 1 C 2 .
Inductors in series (assuming no magnetic coupling between them): L T = L 1 + L 2 + L 3 + … L_T = L_1 + L_2 + L_3 + \dots L T = L 1 + L 2 + L 3 + …
Inductors in parallel: 1 L T = 1 L 1 + 1 L 2 + 1 L 3 + … \frac{1}{L_T} = \frac{1}{L_1} + \frac{1}{L_2} + \frac{1}{L_3} + \dots L T 1 = L 1 1 + L 2 1 + L 3 1 + …
with the two-component form L T = L 1 L 2 L 1 + L 2 L_T = \dfrac{L_1 L_2}{L_1 + L_2} L T = L 1 + L 2 L 1 L 2 .
The equal-value shortcuts the paper needs
Because the syllabus restricts calculations to components of the same value, every examinable question collapses to one of four one-line results. For n n n identical components:
- n n n capacitors of value C C C in series : C T = C / n C_T = C / n C T = C / n
- n n n capacitors of value C C C in parallel : C T = n × C C_T = n \times C C T = n × C
- n n n inductors of value L L L in series : L T = n × L L_T = n \times L L T = n × L
- n n n inductors of value L L L in parallel : L T = L / n L_T = L / n L T = L / n
Memorise those four lines. In the examination you will multiply or divide by 2, 3 or 4 and nothing more — no reciprocals and no calculator gymnastics.
Worked example 1 — capacitors in parallel
Four 100 nF capacitors are connected in parallel. C T = 4 × 100 = 400 nF = 0.4 μ F C_T = 4 \times 100 = 400\ \text{nF} = 0.4\ \mu\text{F} C T = 4 × 100 = 400 nF = 0.4 μ F
Worked example 2 — the same capacitors in series C T = 100 4 = 25 nF C_T = \frac{100}{4} = 25\ \text{nF} C T = 4 100 = 25 nF
The answer, 25 nF, is smaller than any single member — check it against the general formula if you doubt the shortcut: 1 / C T 1/C_T 1/ C T equals four lots of 1 / 100 1/100 1/100 nF, which returns 100/4 = 25 nF. Shortcut and long form agree.
Worked example 3 — inductors in series
Three 60 µH radio-frequency chokes are wired in series and mounted so that their magnetic fields do not interact. L T = 3 × 60 = 180 μ H L_T = 3 \times 60 = 180\ \mu\text{H} L T = 3 × 60 = 180 μ H
Worked example 4 — the same inductors in parallel L T = 60 3 = 20 μ H L_T = \frac{60}{3} = 20\ \mu\text{H} L T = 3 60 = 20 μ H
Inductors behave here exactly as resistors would, which is the fastest way to sanity-check an answer under exam pressure.
Worked example 5 — series-parallel capacitors
Four 10 µF capacitors are wired as two parallel pairs, and the two pairs are then joined in series. Always work from the inside out.
Each parallel pair: 2 × 10 = 20 μ F 2 \times 10 = 20\ \mu\text{F} 2 × 10 = 20 μ F .
The two 20 µF banks in series: 20 / 2 = 10 μ F 20 / 2 = 10\ \mu\text{F} 20/2 = 10 μ F .
The network behaves as a single 10 µF capacitor, but with twice the voltage rating and better sharing of ripple current, which is exactly why power-supply designers build reservoir banks this way.
Worked example 6 — series-parallel inductors
Four 8 µH coils are wired as two series pairs, and the two pairs are then paralleled.
Each series pair: 2 × 8 = 16 μ H 2 \times 8 = 16\ \mu\text{H} 2 × 8 = 16 μ H .
The two 16 µH branches in parallel: 16 / 2 = 8 μ H 16 / 2 = 8\ \mu\text{H} 16/2 = 8 μ H .
The result is the value of a single coil. That is the general rule for four identical components arranged two-by-two: series-then-parallel and parallel-then-series both bring you back to the value you started with.
Ratings change as well as values
Combination questions are not only about the number of farads or henries.
Capacitors in series share the applied voltage. A string of four 100 nF capacitors each rated 63 V will stand roughly 4 × 63 = 252 4 \times 63 = 252 4 × 63 = 252 V. High-voltage stacks in valve amplifiers are built this way, usually with equalising resistors across each capacitor to force the sharing.
Capacitors in parallel share no voltage at all. The whole bank is limited to the lowest rating present: one 25 V part among 450 V capacitors sets the bank at 25 V.
Inductors in parallel share the current , so paralleled chokes carry more total current before the core saturates than any one of them could alone.
Beyond the syllabus: unequal values
You will not be asked this, but working it once makes the shortcuts safer to trust.
A 470 pF capacitor in series with a 220 pF capacitor: C T = 470 × 220 470 + 220 = 103 400 690 ≈ 150 pF C_T = \frac{470 \times 220}{470 + 220} = \frac{103\,400}{690} \approx 150\ \text{pF} C T = 470 + 220 470 × 220 = 690 103 400 ≈ 150 pF
The result sits below 220 pF, precisely as the series rule promises.
A 10 µH inductor in parallel with a 15 µH inductor: L T = 10 × 15 10 + 15 = 150 25 = 6 μ H L_T = \frac{10 \times 15}{10 + 15} = \frac{150}{25} = 6\ \mu\text{H} L T = 10 + 15 10 × 15 = 25 150 = 6 μ H
Again below the smaller of the two, exactly as parallel resistors behave.
One caution about real coils. The inductor formulae above assume no mutual coupling — that the magnetic field of one coil does not link the turns of another. Two toroids, or two screened chokes, satisfy that assumption. Two air-wound coils mounted end to end on the same axis do not: depending on the winding sense their fields aid or oppose, and the measured total comes out higher or lower than the calculation. For the examination, assume no coupling unless a question tells you otherwise.