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Q70PE

Expert-verifiedFound in: Page 699

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**Question: Unreasonable Results**

**(a) On a particular day, it takes **\(9.60 \times {10^3}\;J\)** of electric energy to start a truck's engine. Calculate the capacitance of a capacitor that could store that amount of energy at **\(12.0\;V\)**.**

**(b) What is unreasonable about this result?**

**(c) Which assumptions are responsible?**

- The capacitance of a capacitor that could store that amount of energy at \(12.0\;V\) is \(133\;\;F\).
- A capacitor of capacitance \(133\;\;F\) might be too big in phrases of dimensions to be carried at the truck.
- It is unreasonable to expect that the capacitor can keep that quantity of electrical energy.

**The following is the amount of energy held in a capacitor with capacitance **\(C\)** that is charged to a potential difference **\(\Delta V\)**: **\({U_E} = \frac{1}{2}C{(\Delta V)^2}...(1)\)**.**

- The electric energy required to start the engine is: \({U_E} = 9.60 \times {10^3}\;J\).
- The potential difference across the capacitor is: \(\Delta V = 12.0\;V\).

- The energy stored in the capacitor is found from Equation \((1)\):

\({U_E} = \frac{1}{2}C{(\Delta V)^2}\)

Solve for \(C\):

\(C = \frac{{2{U_E}}}{{{{(\Delta V)}^2}}}\)

Entering the values for \({U_E}\) and \(\Delta V\), we obtain:

\(\begin{aligned}{}C = \frac{{2\left( {9.60 \times {{10}^3}\;J} \right)}}{{12.0\;V}}\\ = 133\;\;F\end{aligned}\)

Therefore, the capacitance value is obtained as \(133\;\;F\).

b.

A capacitor of capacitance \(133\;\;F\)might be too big in phrases of dimensions to be carried at the truck.

c.

It is unreasonable to expect that the capacitor can keep that quantity of electrical energy

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