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What is the potential difference between the plates of a 3.3-F capacitor that stores sufficient energy to operate a \(75-\) W light bulb for one minute?

Short Answer

Expert verified
The potential difference is approximately 52.23 V.

Step by step solution

01

Understanding the Problem

We are given a capacitor, with capacitance of 3.3 F, and a light bulb that requires 75 W of power. We need to find the potential difference (voltage) across the capacitor plates that allows it to store enough energy to operate the light bulb for one minute.
02

Calculating Energy Requirement

The light bulb requires 75 W for 60 seconds, hence it needs a total energy of:\[E = ext{Power} \times ext{Time} = 75 ext{ W} \times 60 ext{ s} = 4500 ext{ J}\]This energy must be stored in the capacitor.
03

Using the Energy Stored in a Capacitor Formula

The formula for the energy stored in a capacitor is:\[E = \frac{1}{2} C V^2\]where \(E\) is energy, \(C\) is capacitance, and \(V\) is the potential difference (voltage). We know \(E = 4500 \text{ J}\) and \(C = 3.3 \text{ F}\).
04

Solving for Voltage

Rearrange the formula to solve for \(V\):\[V^2 = \frac{2E}{C} = \frac{2 \times 4500}{3.3} = 2727.27\]Taking the square root gives:\[V = \sqrt{2727.27} \approx 52.23 \text{ V}\]
05

Conclusion

The potential difference required across the capacitor to store enough energy to power the light bulb for one minute is approximately 52.23 volts.

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Key Concepts

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

Capacitor
A capacitor is a device that stores electrical energy in an electric field. It consists of two conductive plates separated by an insulating material called a dielectric. Capacitors are important in electronic circuits for several reasons:

  • They store and release energy quickly, making them useful for supplying power during short interruptions.
  • They help smooth out fluctuations in voltage, improving the stability of power supplies.
  • They can block direct current (DC) while allowing alternating current (AC) to pass, which is useful in filtering applications.
Capacitors come in various types and sizes, suited to different applications. In our exercise, a 3.3 farad (F) capacitor is used, which is relatively large and capable of storing a considerable amount of energy. Understanding how a capacitor operates is key to appreciating how it can power a device like a light bulb when needed.
Energy Storage
Energy storage is a critical concept in electronics and electrical engineering. It refers to the capture of energy for later use. Capacitors store energy by holding charge on their plates, which creates an electric field. This stored energy can be released when needed, providing power to various devices.

In the exercise, energy storage allows the capacitor to operate a 75-watt light bulb for one minute. This means that the capacitor must hold at least 4500 joules of energy. The ability to store and release this amount of energy efficiently depends on the type and size of the capacitor:

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  • Efficient energy storage requires minimal loss, which is achieved via materials with high dielectric strength.
  • The form factor of capacitors (i.e., large, small, compact) affects how much energy can be stored. Larger capacitors usually have higher energy storage capacity.
  • The concept of energy storage in capacitors is important in many applications, including power conditioning, signal processing, and backup power supply systems.
    Capacitance
    Capacitance is a measure of a capacitor's ability to store electrical energy. It is defined as the charge stored per unit voltage across the capacitor's plates, and is measured in farads (F). The higher the capacitance, the more charge the capacitor can store.

    The formula for capacitance is:\[C = \frac{Q}{V}\]where \(C\) is the capacitance, \(Q\) is the charge in coulombs, and \(V\) is the potential difference in volts. In the given exercise, a 3.3 F capacitor is used to store energy, which indicates a high storage capacity.

    • Capacitance depends on the surface area of the plates, the distance between them, and the permittivity of the dielectric material between the plates.
    • Increasing the area of the plates or using a dielectric with a higher permittivity improves capacitance.
    Understanding capacitance is crucial when designing circuits that need precise energy storage and discharge characteristics.
    Electrical Power
    Electrical power is the rate at which electrical energy is transferred by a circuit. It is measured in watts (W), and is given by the product of voltage and current:\[P = V \times I\]where \(P\) is power, \(V\) is voltage, and \(I\) is current. In the exercise, the light bulb uses 75 watts of power, which means it needs a continuous supply of energy to operate.

    An understanding of electrical power is essential for determining how much energy a device needs. In this scenario:

    • The bulb requires 75 W for 60 seconds, leading to an energy consumption of 4500 joules.
    • To supply this, the capacitor must release stored energy at the required power level.
    Being able to calculate and understand electrical power is fundamental in designing and evaluating electronic circuits and devices.

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    Most popular questions from this chapter

    Two identical point charges \(\left(q=+7.20 \times 10^{-6} \mathrm{C}\right)\) are fixed at diagonally opposite corners of a square with sides of length \(0.480 \mathrm{m}\). A test charge \(\left(q_{0}=-2.40 \times 10^{-8} \mathrm{C}\right),\) with a mass of \(6.60 \times 10^{-8} \mathrm{kg},\) is released from rest at one of the empty corners of the square. Determine the speed of the test charge when it reaches the center of the square.

    At a distance of \(1.60 \mathrm{m}\) from a point charge of \(+2.00 \mu \mathrm{C},\) there is an equipotential surface. At greater distances there are additional equipotential surfaces. The potential difference between any two successive surfaces is \(1.00 \times 10^{3} \mathrm{V} .\) Starting at a distance of \(1.60 \mathrm{m}\) and moving radially outward, how many of the additional equipotential surfaces are crossed by the time the electric field has shrunk to one-half of its initial value? Do not include thenstarting surface.

    Refer to Multiple-Concept Example 3 to review the concepts that are needed here. A cordless electric shaver uses energy at a rate of 4.0 W from a rechargeable 1.5-V battery. Each of the charged particles that the battery delivers to the shaver carries a charge that has a magnitude of \(1.6 \times\) \(10^{-19} \mathrm{C} .\) A fully charged battery allows the shaver to be used for its maximum operation time, during which \(3.0 \times 10^{22}\) of the charged particles pass between the terminals of the battery as the shaver operates. What is the shaver's maximum operation time?

    Identical \(+1.8 \mu \mathrm{C}\) charges are fixed to adjacent corners of a square. What charge (magnitude and algebraic sign) should be fixed to one of the empty corners, so that the total electric potential at the remaining empty corner is \(0 \mathrm{V} ?\)

    Suppose that the electric potential outside a living cell is higher than that inside the cell by 0.070 V. How much work is done by the electric force when a sodium ion (charge \(=+e\) ) moves from the outside to the inside?

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