P26_063.PDF

63. (a) Put ﬁve such capacitors in series. Then, the equivalent capacitance is 2 . 0 µ F / 5=0 . 40 µ F. With

each capacitor taking a 200-V potential diﬀerence, the equivalent capacitor can withstand 1000V.

(b) As one possibility, you can take three identical arrays of capacitors, each array being a ﬁve-capacitor

combination described in part (a) above, and hookup the arrays in parallel. The equivalent

capacitance is now C eq =3(0 . 40 µ F) = 1 . 2 µ F. With each capacitor taking a 200-V potential

diﬀerence the equivalent capacitor can withstand 1000V.