Question

The battery of an electric car stores 60.51 MJ of energy. If 6990 supercapacitors, each with capacitance \(C=3.423 \mathrm{kF}\), are required to supply this amount of energy, what is the potential difference across each supercapacitor?

Short answer

Answer: The potential difference across each supercapacitor is approximately 71.11 Volts.

Step-by-step solution

Convert energy to Joules

First, convert the energy from Mega Joules to Joules: \(60.51\, MJ = 60.51 \times 10^6\, J\)

Compute energy stored in one supercapacitor

Divide the total energy by the number of supercapacitors to find the energy stored in one supercapacitor: \(E_{1} = \frac{60.51\times 10^6\, J}{6990} = 8657.94\, J\)

Convert capacitance to Farads

Since the capacitance is given in kilo-Farads, convert it to Farads: \(C = 3.423\, kF = 3.423\times 10^3\, F\)

Rearrange the energy equation

Rearrange the energy equation to get the potential difference \(V\): \(V^2 =\frac{2E_{1}}{C}\)

Solve for potential difference

Now, substitute the energy and capacitance values into the rearranged equation and solve for the potential difference: \(V^2 = \frac{2 \times 8657.94\, J}{3.423\times 10^3\, F} = 5057.24\) \(V = \sqrt{5057.24} = 71.11\, V\) The potential difference across each supercapacitor is approximately 71.11 Volts.