Source: High school physics (Chinese)
Problem
A collection of identical capacitors is available; each has capacitance $2.0\ \mu\mathrm{F}$ and voltage rating $200$ V. They are to be combined to form (a) a capacitor of $C_1 = 0.40\ \mu\mathrm{F}$ rated $1000$ V, and (b) a capacitor of $C_2 = 1.2\ \mu\mathrm{F}$ rated $1000$ V.
- How many capacitors are needed and how should they be connected for $C_1 = 0.40\ \mu\mathrm{F}$, $1000$ V?
- How many capacitors are needed and how should they be connected for $C_2 = 1.2\ \mu\mathrm{F}$, $1000$ V?
(1) $5$ capacitors in series. (2) $15$ capacitors arranged as $3$ parallel branches, each branch containing $5$ in series.
To withstand $1000$ V using $200$ V units, the voltage must be divided across at least $1000/200 = 5$ capacitors in series. A series chain of $5$ identical capacitors has equivalent capacitance $2.0/5 = 0.40\ \mu\mathrm{F}$ and is rated $5 \times 200 = 1000$ V --- exactly matching $C_1$.
(1) For $C_1 = 0.40\ \mu\mathrm{F}$, $1000$ V: one chain of $5$ capacitors in series. \textbf{Total: 5 capacitors.}
(2) For $C_2 = 1.2\ \mu\mathrm{F}$, $1000$ V: each series chain delivers $0.40\ \mu\mathrm{F}$. Need $1.2/0.40 = 3$ such chains in parallel. Each chain is still rated $1000$ V, so the parallel combination is also rated $1000$ V. \textbf{Total: $3 \times 5 = 15$ capacitors arranged as $3$ parallel branches of $5$ in series.}