Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Electric eels and electric fish generate large potential differences that are used to stun enemies and prey. These potentials are produced by cells that each can generate 0.10 V. We can plausibly model such cells as charged capacitors. (a) How should these cells be connected (in series or in parallel) to produce a total potential of more than 0.10 V? (b) Using the connection in part (a), how many cells must be connected together to produce the 500-V surge of the electric eel?

Short Answer

Expert verified
Connect cells in series and use 5000 cells for 500 V.

Step by step solution

01

Understanding the Problem

We need to determine how to connect the cells to create a greater potential difference than 0.10 V with each cell modeled as a capacitor producing 0.10 V. We aim for a 500 V total in Part (b).
02

Type of Connection

To increase the total voltage output above 0.10 V, the cells must be connected in series. In a series connection, the voltages add up cumulatively.
03

Calculate Number of Cells Needed

To find the number of cells needed to reach a total of 500 V, divide the desired voltage by the voltage of one cell: \( N = \frac{500 \text{ V}}{0.10 \text{ V}} \).
04

Performing the Division

Calculate the number of cells: \( N = \frac{500 \text{ V}}{0.10 \text{ V}} = 5000 \text{ cells} \).
05

Conclusion

Therefore, 5000 cells need to be connected in series to achieve the 500 V required for an electric eel's surge.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Series Connection
In an electric circuit, a series connection is where components are arranged in a single path. Every part is interconnected, forming a loop. When you connect electrical components, such as capacitors, in series, the electric current must flow through each component one after the other. This results in the voltages across each component adding together.
For example:
  • If you have three capacitors in series, each contributing 0.10 V, the total voltage would be 0.10 V + 0.10 V + 0.10 V = 0.30 V.
  • Series connections are particularly useful when you need to increase the voltage supplied by components like batteries or capacitors.
In our case, connecting the cells in a series allows us to achieve a higher potential difference than each cell can provide individually. Therefore, to achieve the high voltages electric eels generate, cells must be connected in series.
Voltage Calculation
Voltage calculation in a series connection involves adding the voltages of each component. Understanding this principle is key when determining the total voltage achievable with several cells or units.
The formula is simple: - In a series connection, the total voltage, Vtotal, is the sum of the individual voltages: \[ V_{total} = V_1 + V_2 + ext{...} + V_n \] - For our exercise, the calculation involves determining how many 0.10 V cells we need to sum to reach 500 V.
Following the formula, you divide the target voltage by the voltage of one cell:\[ N = \frac{V_{target}}{V_{cell}} = \frac{500\text{ V}}{0.10\text{ V}} = 5000 \text{ cells} \] Therefore, we need 5000 cells to reach the desired total of 500 V, showing how effective a series connection is for achieving high voltage levels.
Capacitor Model
A capacitor is an electrical component that can store and release electrical energy. In our exercise, cells that make up the electric eels are modeled as capacitors. Each capacitor provides a specific voltage, in this case, 0.10 V.
Some key points about capacitors:
  • They store energy in an electric field created between a pair of conductive plates.
  • In this context, capacitors are used to model the potential difference each cell in an electric eel can generate.
  • Capacitors are crucial in circuits where storage and release of energy at certain voltages is required.
When you connect capacitors in series, the total voltage across them becomes the sum of the voltages across each capacitor. Therefore, modeling the eel's cells as capacitors helps us understand and visualize how these cells work to produce such high voltages.
Electric Potential
Electric potential, often referred to as voltage, represents the potential energy per unit charge at a point in an electric field. It's the driving force that pushes charge through a circuit, allowing electrical devices to function.

Some essential aspects include:
  • Electric potential is measured in volts (V).
  • In a circuit, the electric potential determines the amount of work needed to move a charge between two points.
  • The cells in electrical eels generate electric potential to produce powerful surges.
In the context of electric eels, the cells produce a small voltage, 0.10 V, but when connected in a series, they can accumulate to the high electric potentials observed, like the 500 V surge mentioned.
This high electric potential is crucial for the eel's ability to stun its prey, illustrating the practical application of these circuits in nature.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Suppose that the change in V\(_m\) was caused by the entry of Ca\(^{2+}\) instead of Na\(^+\). How many Ca\(^{2+}\) ions would have to enter the cell per unit membrane to produce the change? (a) Half as many as for Na\(^+\); (b) the same as for Na\(^+\); (c) twice as many as for Na\(^+\); (d) cannot say without knowing the inside and outside concentrations of Ca\(^{2+}\).

A capacitor is made from two hollow, coaxial, iron cylinders, one inside the other. The inner cylinder is negatively charged and the outer is positively charged; the magnitude of the charge on each is 10.0 pC. The inner cylinder has radius 0.50 mm, the outer one has radius 5.00 mm, and the length of each cylinder is 18.0 cm. (a) What is the capacitance? (b) What applied potential difference is necessary to produce these charges on the cylinders?

A parallel-plate air capacitor of capacitance 245 pF has a charge of magnitude 0.148 \(\mu\)C on each plate. The plates are 0.328 mm apart. (a) What is the potential difference between the plates? (b) What is the area of each plate? (c) What is the electricfield magnitude between the plates? (d) What is the surface charge density on each plate?

A cylindrical capacitor consists of a solid inner conducting core with radius 0.250 cm, surrounded by an outer hollow conducting tube. The two conductors are separated by air, and the length of the cylinder is 12.0 cm. The capacitance is 36.7 pF. (a) Calculate the inner radius of the hollow tube. (b) When the capacitor is charged to 125 V, \(\textbf{what is the charge per unit length}\) \(\lambda\) on the capacitor?

A parallel-plate air capacitor is made by using two plates 12 cm square, spaced 3.7 mm apart. It is connected to a 12-V battery. (a) What is the capacitance? (b) What is the charge on each plate? (c) What is the electric field between the plates? (d) What is the energy stored in the capacitor? (e) If the battery is disconnected and then the plates are pulled apart to a separation of 7.4 mm, what are the answers to parts (a)-(d)?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free