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Chapter 27: Problem 65

An electric car is designed to run off a bank of \(12.0-\mathrm{V}\) batteries with total energy storage of \(2.00 \times 10^{7} \mathrm{J} .\) (a) If the electric motor draws \(8.00 \mathrm{kW},\) what is the current delivered to the motor? (b) If the electric motor draws \(8.00 \mathrm{kW}\) as the car moves at a steady speed of \(20.0 \mathrm{m} / \mathrm{s}\), how far will the car travel before it is "out of juice"?

Short Answer

Expert verified
The current delivered to the motor is \(666.67 \, \mathrm{A}\) , and the car can travel a distance of \(50000 \, \mathrm{m}\) or \(50 \, \mathrm{km}\) before it runs out of energy.

Step by step solution

01

Find Current

To find the current delivered to the motor, we use the formula for power: Power = Voltage x Current. We rearrange the formula to find current: Current = Power / Voltage. Substituting the given values in, Current = \(8000 \, \mathrm{W} / 12 \, \mathrm{V}\).
02

Compute Current

After substituting the provided values from the exercise, we can now compute the current. This will give us a current of approximately \(666.67 \, \mathrm{A}\). This is the current being fed to the motor.
03

Calculate the Time

For calculating how long the car can use the energy stored in the battery, we use the formula for energy: Energy = Power x Time. We can rearrange this formula to solve for time: Time = Energy / Power. Substituting the given values, we get: Time = \((2.00 \times 10^{7}) \mathrm{J} / 8000 \, \mathrm{W}\).
04

Compute Time

After substituting the values provided, we can now find the time. This simplifies to a time of approximately \(2500 \, \mathrm{s}\).
05

Calculate the Distance

Now, to find the distance covered in this amount of time, we multiply the time for which the car can run by the speed of the car. Distance = Speed x Time. Substituting in the given values, we get: Distance = \(20.0 \, \mathrm{m/s} \times 2500 \, \mathrm{s}\).
06

Compute Distance

Finally, after substituting the values we calculated earlier, we find the distance, which equals to \(50000 \, \mathrm{m}\) or \(50 \, \mathrm{km}\). This is how far the car will travel before it runs out of energy.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Current Calculation
Understanding how to calculate the current drawn by an electric motor is essential in analyzing electric car performance. To do this, we use the basic relationship between power (P), voltage (V), and current (I). The formula that ties these together is given by:
  • Power = Voltage 脳 Current
To find the current, we rearrange the formula:
  • Current = Power / Voltage
Given that the electric motor draws 8.00 kW, we can substitute this power value alongside the voltage of the battery bank, which is 12.0 V, into the formula:
  • Current = 8000 W / 12 V
After performing the division, we find that the current delivered to the motor is approximately 666.67 A. This large current value is typical for motors that require significant power to operate, like those in electric vehicles.
Energy Storage
Energy storage is a crucial factor for electric cars. It determines how long the car can operate before needing a recharge. The total energy stored in the car's battery bank is given as 2.00 脳 10鈦� J (joules). To find out how long this energy can last, we use the relationship between energy (E), power (P), and time (T):
  • Energy = Power 脳 Time
Solving for time gives us:
  • Time = Energy / Power
Plugging in the values from the problem, we have:
  • Time = 2.00 脳 10鈦� J / 8000 W
Completing this calculation yields a time of approximately 2500 seconds (or about 41.67 minutes). This is the maximum duration the car can run continuously at the specified power draw before depleting the energy stored in the batteries. Knowing this time helps in planning trips and estimating when the next recharge is needed.
Travel Distance Calculation
To find out how far an electric car can travel before exhausting its energy storage, we need to understand the relationship between speed, time, and distance. The formula for distance is:
  • Distance = Speed 脳 Time
In this scenario, we are given a speed of 20.0 m/s for the car. From the previous calculations, we determined that the car can run for 2500 seconds. Using the distance formula, we calculate:
  • Distance = 20.0 m/s 脳 2500 s
Performing the multiplication, we find that the car can travel 50000 meters, which converts to 50 kilometers. This calculation indicates how far the car will travel while drawing the specified power from its fully charged batteries. Understanding travel distance is vital for route planning and ensuring the vehicle stays within the range of its energy reserves.

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Most popular questions from this chapter

A teapot with a surface area of \(700 \mathrm{cm}^{2}\) is to be silver plated. It is attached to the negative electrode of an electrolytic cell containing silver nitrate \(\left(\mathrm{Ag}^{+} \mathrm{NO}_{3}^{-}\right) .\) If the cell is powered by a \(12.0-\mathrm{V}\) battery and has a resistance of \(1.80 \Omega,\) how long does it take for a 0.133 -mm layer of silver to build up on the teapot? (The density of silver is \(\left.10.5 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3} .\right)\)

In a hydroelectric installation, a turbine delivers \(1500 \mathrm{hp}\) to a generator, which in turn transfers \(80.0 \%\) of the mechanical energy out by electrical transmission. Under these conditions, what current does the generator deliver at a terminal potential difference of \(2000 \mathrm{V} ?\)

An aluminum wire having a cross-sectional area of \(4.00 \times 10^{-6} \mathrm{m}^{2}\) carries a current of \(5.00 \mathrm{A}\). Find the drift speed of the electrons in the wire. The density of aluminum is \(2.70 \mathrm{g} / \mathrm{cm}^{3} .\) Assume that one conduction electron is supplicd by each atom.

A straight cylindrical wire lying along the \(x\) axis has a length \(L\) and a diameter \(d\). It is made of a material that obeys Ohm's law with a resistivity \(\rho .\) Assume that potential \(V\) is maintained at \(x=0,\) and that the potential is zero at \(x=L .\) In terms of \(L, d, V, \rho,\) and physical constants, derive expressions for (a) the electric field in the wire, (b) the resistance of the wire, (c) the electric current in the wire, and (d) the current density in the wire. Express vectors in vector notation. (e) Prove that \(\mathbf{E}=\rho \mathbf{J}\).

A close analogy exists between the flow of energy by heat because of a temperature difference (see Section 20.7 ) and the flow of electric charge because of a potential difference. The energy \(d Q\) and the electric charge \(d q\) can both be transported by free electrons in the conducting material. Consequently, a good electrical conductor is usually a good thermal conductor as well. Consider a thin conducting slab of thickness \(d x,\) area \(A,\) and electrical conductivity \(\sigma,\) with a potential difference \(d V\) between opposite faces. Show that the current \(I=d q / d t\) is given by the equation on the left below: Charge Conduction\(\frac{d q}{d t}=\sigma A\left|\frac{d V}{d x}\right|\) Thermal Conduction \((E q, 20.14)$$\frac{d Q}{d t}=k A\left|\frac{d T}{d x}\right|\) In the analogous thermal conduction equation on the right, the rate of energy flow dQ/de (in SI units of joules per second) is due to a temperature gradient \(d T / d x,\) in a material of thermal conductivity \(k .\) State analogous rules relating the direction of the electric current to the change in potential, and relating the direction of energy flow to the change in temperature.
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