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Chapter 26: Problem 44

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

Expert verified
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

01

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: \(R = \dfrac{(12.0\,\mathrm{V})^2}{1.21 \times 10^9\,\mathrm{W}}\)
02

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.
03

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: \(t = -(1.19\times 10^{-7}\,\Omega) (1.00\,\mathrm{F}) \ln(1-0.9)\)
04

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.

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Most popular questions from this chapter

An RC circuit has a time constant of 3.1 s. At \(t=0\), the process of charging the capacitor begins. At what time will the energy stored in the capacitor reach half of its maximum value?

You want to make an ohmmeter to measure the resistance of unknown resistors. You have a battery with voltage \(\mathrm{V}_{\mathrm{emf}}=9.00 \mathrm{~V}\), a variable resistor, \(R,\) and an ammeter that measures current on a linear scale from 0 to \(10.0 \mathrm{~mA}\) a) What resistance should the variable resistor have so that the ammeter gives its full-scale (maximum) reading when the ohmmeter is shorted? b) Using the resistance from part (a), what is the unknown resistance if the ammeter reads \(\frac{1}{4}\) of its full scale?

How long would it take, in multiples of the time constant, \(\tau,\) for the capacitor in an \(\mathrm{RC}\) circuit to be \(98 \%\) charged? a) \(9 \tau\) c) \(90 \tau\) e) \(0.98 \tau\) b) \(0.9 \tau\) d) \(4 \tau\)

The ammeter your physics instructor uses for in-class demonstrations has internal resistance \(R_{\mathrm{i}}=75 \Omega\) and measures a maximum current of \(1.5 \mathrm{~mA}\). The same ammeter can be used to measure currents of much greater magnitudes by wiring a shunt resistor of relatively small resistance, \(R_{\text {shunt }}\), in parallel with the ammeter. (a) Sketch the circuit diagram, and explain why the shunt resistor connected in parallel with the ammeter allows it to measure larger currents. (b) Calculate the resistance the shunt resistor has to have to allow the ammeter to measure a maximum current of 15 A.

A circuit consists of a source of emf, a resistor, and a capacitor, all connected in series. The capacitor is fully charged. How much current is flowing through it? a) \(i=V / R\) b) zero c) neither (a) nor (b)
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