Q70PE - College Physics (Urone)

Question

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?

Step-by-Step Solution

Answer

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

Step 1: Concepts and Principles

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

Step 2: Given Data

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

Step 3: Finding Energy stored

1. 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:

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

 

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

Step 4: Explanation for part (b)

b.

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

Step 5: Explanation for part (c)

c.

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