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 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 required resistance is approximately \(1.19 \times 10^{-7}\,\Omega\) to produce a peak power output of 1.21 GW. It would take approximately \(2.57 \times 10^{-7}\,\mathrm{s}\) (257 nanoseconds) to charge the capacitor to 90% of its maximum capacity.

Step-by-step solution

Find the Resistance

First, we will use the power formula to find the resistance (R) required to produce a peak power output of 1.21 GW (1.21 x 10^9 W). The power formula can be given as: \(P = \dfrac{V^2}{R}\) Where P is the power, V is the voltage, and R is the resistance. Now, we can find the resistance (R) by rearranging the formula: \(R = \dfrac{V^2}{P}\) Let's plug in the voltage and power values: \(R = \dfrac{(12.0\,\mathrm{V})^2}{1.21 \times 10^9\,\mathrm{W}}\)

Calculate the Resistance

Now, let's calculate the resistance: \( R = \dfrac{144\,\mathrm{V^2}}{1.21 \times 10^9\,\mathrm{W}} = 1.19 \times 10^{-7}\,\Omega\) Therefore, the required resistance is approximately \(1.19 \times 10^{-7}\,\Omega\) to produce a peak power output of 1.21 GW in the resistor.

Find the Time to Charge the Capacitor

Next, we need to find the time it takes for the capacitor to charge to 90% of its maximum capacity through this resistor. To do this, we can use the charging equation for a capacitor: \(t = -RC\ln(1-\dfrac{Q}{Q_{max}})\) Where \(t\) is the time, \(R\) is the resistance, \(C\) is the capacitance, \(Q\) is the charge at time \(t\), and \(Q_{max}\) is the maximum charge. In this case, we want the charge to be 90% of the maximum charge, so: \(\dfrac{Q}{Q_{max}} = 0.9\) We can now plug in the values already known: \(t = -(1.19\times 10^{-7}\,\Omega) (1.00\,\mathrm{F}) \ln(1-0.9)\)

Calculate the Time to Charge the Capacitor

Now, let's calculate the time it takes for the capacitor to charge to 90% of its maximum capacity: \(t \approx -(1.19\times 10^{-7}\,\Omega) (1.00\,\mathrm{F}) \ln(0.1)\) \(t \approx 2.57 \times 10^{-7}\,\mathrm{s}\) Therefore, it would take approximately \(2.57 \times 10^{-7}\,\mathrm{s}\) (257 nanoseconds) for a \(12.0\,\mathrm{V}\) car battery to charge the capacitor to 90% of its maximum capacity through this resistor.

Key concepts

Capacitor

A capacitor is like an electrical storage device that temporarily holds electric charge. Imagine it as a small rechargeable energy tank. Its main job is to store energy when a voltage is applied, and then release it when needed. The capacitance of a capacitor is measured in farads (F). This is essentially how much charge the capacitor can hold per unit voltage.
For example, in our exercise, we are dealing with a 1.00-F capacitor. This means it can store a lot of charge. In many circuits, capacitors are used to manage power surges or to smooth out electrical flow. They make sure that the electric devices they are part of can operate smoothly without interruptions.

Resistance

Resistance is a measure of how much a material opposes the flow of electric current. Imagine resistance as a narrow road for electricity to travel through. The narrower the road, the harder it is for the electricity to pass and the higher the resistance. Resistance is measured in ohms (Ω).
In our exercise, we found the resistance needed to produce a peak power output of 1.21 GW, which turned out to be very small, approximately 1.19 x 10^{-7} Ω. Having such a low resistance allows for a large amount of power to be delivered, just like racing down a very wide highway would allow for high-speed travel.

Power Output

Power is the rate at which energy is used or produced. In the world of electric circuits, power is calculated using the formula: \[ P = \frac{V^2}{R} \] where \(P\) is power, \(V\) is voltage, and \(R\) is resistance. The power output tells us how much energy is being transferred at any given second.
In our scenario, the goal was to achieve a power output of 1.21 GW, which is indeed a lot of power — enough to power significant machinery or, as the Back to the Future movie suggests, a time machine! To put it in perspective, 1 gigawatt (GW) is a billion watts and is roughly equivalent to the output of a medium-sized power plant.

Charging Time

The charging time of a capacitor is how long it takes to fill up with energy. It's crucial because it tells us how quickly a capacitor can be ready to do its job. The formula to calculate the charging time of a capacitor is: \[ t = -RC \ln(1-\frac{Q}{Q_{max}}) \] where \(t\) is the time, \(R\) is the resistance, \(C\) is the capacitance, and \(Q/Q_{max}\) is the ratio of the charge to the maximum charge.
In this exercise, we wanted the capacitor to reach 90% of its capacity, which gives a charging time of approximately 257 nanoseconds with our setup. This is incredibly fast and shows that with a suitable resistor, we can charge a capacitor almost instantaneously for practical purposes.