It is required to construct a 10 µF capacitor which can be connected across a 200 V battery. Capacitors of capacitance 10 µF are available but they can withstand only 50 V. Design a combination which can yield the desired result.
Let the number of capacitors in series (connected in a row) be x.
The maximum voltage that the capacitors can withstand is 50 V.
The voltage across each row should be equal to 200 V.
x × 50 = 200
x = 4 capacitors
Let there be y such rows.
So, the equivalent capacitance of the combination will be xy.
⇒ xy = 10
⇒ y = 10 x = 4 capacitors
Thus, to yield the required result, the combination of 4 rows, each of 4 capacitors having capacitance 10 µF and breakdown voltage 50 V, is required.