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Unformatted text preview: 63. (a) Put ﬁve such capacitors in series. Then, the equivalent capacitance is 2.0 µF/5 = 0.40 µF. With
each capacitor taking a 200-V potential diﬀerence, the equivalent capacitor can withstand 1000 V.
(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 hook up the arrays in parallel. The equivalent
capacitance is now Ceq = 3(0.40 µF) = 1.2 µF. With each capacitor taking a 200-V potential
diﬀerence the equivalent capacitor can withstand 1000 V. ...
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