Question

In the movie Back to the Future, time travel is made possible by a flux capacitor, which generates 1.21 GW of power. Assuming that a 1.00 - \(F\) capacitor is charged to its maximum capacity with a \(12.0-\mathrm{V}\) car battery and is discharged through a resistor, what resistance is necessary to produce a peak power output of \(1.21 \mathrm{GW}\) in the resistor? How long would it take for a \(12.0-\mathrm{V}\) car battery to charge the capacitor to \(90.0 \%\) of its maximum capacity through this resistor?

Short answer

Answer: The necessary resistance to produce a peak power output of 1.21 GW in the resistor is approximately \(1.20 \times 10^{-7}\,\Omega\), and it would take approximately \(3.07 \times 10^{-6}\,\text{s}\) for a 12.0 V car battery to charge the capacitor to 90% of its maximum capacity through this resistor.

Step-by-step solution

Calculate the peak voltage across the capacitor

When the capacitor is fully charged, the maximum voltage across it will be 12.0 V. We can use this value to calculate the peak power output.

Apply the power-resistance relationship

The power output (P) in the resistor can be calculated using the equation \( P = \frac{V^2}{R}\), where V is the voltage across the resistor and R is the resistance. Rearranging the equation, we get \( R = \frac{V^2}{P}\).

Calculate the necessary resistance

Now, we can plug the values into the equation: \(R = \frac{(12.0\,\text{V})^2}{1.21 \times 10^9\,\text{W}} = 1.20 \times 10^{-7}\,\Omega\) Now, let's find the time required to charge the capacitor to 90% of its maximum capacity through this resistor using a 12.0 V car battery.

Calculate the RC time constant

The RC time constant (τ) can be calculated using the equation \( \tau = R \times C\), where R is the resistance and C is the capacitance. Substituting the values, we get \(\tau = (1.20 \times 10^{-7}\,\Omega) \times (1.00\,\text{F}) = 1.20 \times 10^{-7}\,\text{s}\)

Use the RC time constant to find the charge time

The voltage across the charging capacitor (V_c) as a function of time (t) is given by the equation \(V_c = V \times (1 - e^{-\frac{t}{\tau}})\), where V is the battery voltage and τ is the RC time constant. We need to find the time t when the capacitor has charged to 90% of its maximum capacity, i.e., the voltage across the capacitor is 90% of the maximum voltage, \(0.90\times12.0\,\text{V} = 10.8\,\text{V}\). We can rearrange the equation to find t: \(t = -\tau \times \ln (1-\frac{V_c}{V})\)

Calculate the required charging time

Now, we can plug the values into the equation: \(t = -(1.20 \times 10^{-7}\,\text{s}) \times \ln (1-\frac{10.8\,\text{V}}{12.0\,\text{V}}) = 3.07 \times 10^{-6}\,\text{s}\) So, the required resistance to produce a peak power output of 1.21 GW in the resistor is approximately \(1.20 \times 10^{-7}\,\Omega\), and it would take approximately \(3.07 \times 10^{-6}\,\text{s}\) for a 12.0 V car battery to charge the capacitor to 90% of its maximum capacity through this resistor.