Question

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

Short answer

The energy stored in the capacitor is 1.17 J.

Step-by-step solution

Definition of Capacitor

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

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

Given information

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

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

Finding the energy  stored in the capacitor

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

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

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

We get,

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

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