- First capacitance C1 (F)
- 2F
- Second capacitance C2 (F)
- 3F
1.200000F
Open with these values1.200000F
Result: 1.200000 FCapacitors in series follow product over sum: 2 F and 3 F give 6 ÷ 5 = 1.2 F, below either part. That is the same arithmetic as resistors in parallel, and the reverse of capacitors in parallel, which simply add. In series the voltage splits, so the pair withstands more than one alone.
1.200000F
Open with these values2.000000F
Open with these values2.000000F
Open with these valuesC = (C1 × C2) ÷ (C1 + C2)
| C1 (F), C2 (F) | What it is | Total (F) |
|---|---|---|
| 1, 1 | half of one part | 0.5 |
| 2, 3 | the default pair | 1.2 |
| 4, 4 | two equal parts | 2 |
| 6, 3 | unequal, same result | 2 |
| 10, 10 | two equal parts | 5 |
By the product-over-sum rule: C = (C1 × C2) ÷ (C1 + C2). For 2 F and 3 F that is 6 ÷ 5 = 1.2 F. Capacitors sit in series when the same charge flows through both, and they then combine exactly like resistors in parallel.
Wiring capacitors end to end effectively increases the plate separation, and capacitance falls as the plates move apart. The pair therefore stores less charge per volt than either part alone. Two 2 F capacitors in series give 1 F.
The total is exactly half of one of them, because C times C over C plus C is C ÷ 2. Two 4 F capacitors give 2 F and two 10 F capacitors give 5 F.
It is the opposite. In parallel you add the values and the total grows above either part; in series the total falls below the smaller one. For capacitors, series behaves like parallel resistors and parallel behaves like series resistors.
Yes. In series the applied voltage splits across the capacitors, so each one sees only part of the total. That is why series connections are used to exceed the rating of a single part, at the cost of a smaller total capacitance.
Information, not professional advice.
Diese Seite gibt es auch auf Deutsch.
Zu Deutsch wechseln