It is required to construct a \(10\text{ \mu F}\) capacitor which can be connected across a \(200\text{ V}\) battery. Capacitors of capacitance \(10\text{ \mu F}\) are available but they can withstand only \(50\text{ V}\). Design a combination which can yield the desired result.
Four capacitors in parallel and four such combination in series.
Four capacitors in series and four such combination in parallel.
Eight capacitors in parallel and four such combination in series.
Eight capacitors in series and four such combination in parallel.
Solution:
To withstand \(200\text{ V}\) using \(50\text{ V}\) capacitors, \(4\) capacitors must be in series (\(4 \times 50 = 200\text{ V}\)). Each series combination has \(C_{eq} = 10/4 = 2.5\text{ \mu F}\). To get \(10\text{ \mu F}\) total, \(10/2.5 = 4\) such series combinations must be in parallel. Thus, 4 capacitors in series, and 4 such combinations in parallel.
Leave a Reply