Answer to Question #261503 in Electric Circuits for eissa

Question

Suppose many capacitors, each with 𝐶 =90.0 μ𝐹, are connected in parallel across a battery

with a potential difference of ∆𝑉 =160.0 𝑉. How many capacitors are needed to store 95.6 J

of energy?

Answer 1

$C=90\mu F\\∆V=160volt\\U=95.6J$

$U=\frac{1}{2}C'∆V^2$

Assume n capacitor connected in parallel combination

C'=nC

$U=\frac{1}{2}nC∆V^2$

$$n=\frac{2U}{C∆V^2}=\frac{2\times95.6\times}{90\times10^{-6}\times160\times160}=82$$

n=82