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Found in: Page 699

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# A ${\mathbf{165}}{\mathbf{}}{\mathbf{\mu F}}$capacitor is used in conjunction with a motor. How much energy is stored in it when 119 V is applied?

The energy stored in the capacitor is 1.17 J.

See the step by step solution

## Step 1: Definition of Capacitor

The energy stored in a capacitance C capacitor charged to a potential difference ${\mathbf{\Delta V}}$ is as follows:

${{\mathbf{U}}}_{{\mathbf{E}}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{C}}\mathbf{\left(}\mathbf{\Delta V}{\mathbf{\right)}}^{\mathbf{2}}$

## Step 2: Given information

The capacitor's capacitance is: $\begin{array}{rcl}{\mathbf{C}}& {\mathbf{=}}& {\mathbf{\left(}}{\mathbf{165}}{\mathbf{}}{\mathbf{\mu F}}{\mathbf{\right)}}\left(\frac{1F}{{10}^{6}\mathrm{\mu F}}\right)\\ & {\mathbf{=}}& {\mathbf{165}}{\mathbf{×}}{{\mathbf{10}}}^{\mathbf{-}\mathbf{6}}{\mathbf{}}{\mathbf{}}{\mathbf{F}}\end{array}$

The capacitance's potential difference is: ${\mathbf{\Delta V}}{\mathbf{=}}{\mathbf{119}}{\mathbf{}}{\mathbf{}}{\mathbf{V}}$

## Step 3: Finding the energy  stored in the capacitor

The amount of energy stored in the capacitor can be found from the equation:

${{\mathbf{U}}}_{{\mathbf{E}}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{C}}\mathbf{\left(}\mathbf{\Delta V}{\mathbf{\right)}}^{\mathbf{2}}$

Now, substitute the values of C and $∆V$

We get,

$\begin{array}{rcl}{{\mathbf{U}}}_{{\mathbf{E}}}& {\mathbf{=}}& \frac{\mathbf{1}}{\mathbf{2}}\left(165×{10}^{-6}F\right)\mathbf{\left(}\mathbf{119}\mathbf{}\mathbf{}\mathbf{V}{\mathbf{\right)}}^{\mathbf{2}}\\ & {\mathbf{=}}& {\mathbf{1}}{\mathbf{.}}{\mathbf{17}}{\mathbf{ }}{\mathbf{}}{\mathbf{J}}\end{array}$

Therefore, the energy stored in the capacitor is 1.17 J.