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