Question

The battery of an electric car stores 67.39 MJ of energy. If 6845 supercapacitors, each with capacitance \(C\) and charged to a potential difference of \(2.377 \mathrm{~V}\), can supply this amount of energy, what is the value of \(C\) for each supercapacitor?

Short answer

Answer: The capacitance of each supercapacitor is approximately \(1738.76\,\text{F}\).

Step-by-step solution

Given parameters

The given parameters are: Total energy in the battery: $ E_b = 67.39\,\text{MJ} = 67.39 \times 10^6\,\text{J}$ Number of supercapacitors: \( n = 6845\) Potential difference across each supercapacitor: \( V = 2.377\,\text{V}\) We'll need to find the capacitance of each supercapacitor, denoted by \(C\).

Calculate the energy per supercapacitor

As the energy is uniformly spread between the supercapacitors, to find the energy stored in each supercapacitor (\(E_c\)), we will divide the total energy in the battery (\(E_b\)) by the number of supercapacitors (\(n\)): $$ E_c = \frac{E_b}{n} = \frac{67.39 \times 10^6\,\text{J}}{6845} = 9850.54\,\text{J} $$

Apply the energy formula for capacitors and solve for \(C\)

The energy stored in a capacitor can be calculated using the formula: $$ E_c = \frac{1}{2}CV^2 $$ We know the energy per supercapacitor (\(E_c\)) and the potential difference (\(V\)), so we can solve for the capacitance (\(C\)): $$ C = \frac{2E_c}{V^2} = \frac{2 \times 9850.54\,\text{J}}{(2.377\,\text{V})^2} = 1738.76\,\text{F} $$

Final answer

The capacitance of each supercapacitor is approximately \(1738.76\,\text{F}\).