Problem 41 - College Physics

Question

Current state-of-the-art capacitors are capable of storing many times the energy of older ones. Such a capacitor, with a capacitance of \(1.0 \mathrm{~F}\), is able to light a small 0.50 - \(W\) bulb at steady full power for 5.0 s before it quits. What is the terminal voltage of the battery that charged the capacitor?

Step-by-Step Solution

Answer

The terminal voltage of the battery is approximately 2.24 V.

Step 1: Understanding the Problem

We need to find the terminal voltage of the battery that charged a capacitor, which has a capacitance of \(1.0\, \mathrm{F}\) and can power a \(0.50\, \mathrm{W}\) bulb for \(5.0\, \mathrm{s}\). The power, duration, and capacitance are given to us.

Step 2: Calculating Energy Used by the Bulb

The energy used by the bulb can be calculated using the formula \( E = P \times t \), where \( P \) is power and \( t \) is time. Here, \( P = 0.50\, \mathrm{W} \) and \( t = 5.0\, \mathrm{s} \), so \( E = 0.50 \times 5.0 = 2.5\, \mathrm{J} \).

Step 3: Relating Energy to Capacitance and Voltage

The energy stored in a capacitor is given by \( E = \frac{1}{2} C V^2 \), where \( C \) is the capacitance and \( V \) is the voltage. We already know \( E = 2.5\, \mathrm{J} \) and \( C = 1.0\, \mathrm{F} \).

Step 4: Solving for Voltage

Rearranging the formula \( E = \frac{1}{2} C V^2 \) to solve for \( V \), we get \( V = \sqrt{\frac{2E}{C}} \). Substituting \( E = 2.5\, \mathrm{J} \) and \( C = 1.0\, \mathrm{F} \) gives \( V = \sqrt{\frac{2 \times 2.5}{1.0}} = \sqrt{5.0} \approx 2.24\, \mathrm{V} \).

Key Concepts

Capacitance

Capacitance is a fundamental concept in electronics that represents a capacitor's ability to store charge. A capacitor is essentially a device that holds electrical energy in an electric field. The capacitance, denoted by the symbol \( C \), is measured in farads (F). When a capacitor is charged, it can release the stored energy whenever needed, like lighting up a bulb.

• Capacitance is calculated as the charge stored per unit voltage: \( C = \frac{Q}{V} \), where \( Q \) is the charge in coulombs, and \( V \) is the voltage in volts.
• This ability to store energy makes capacitors useful in various applications, from maintaining power in electronic circuits to energy storage systems.

The exercise in question involves a capacitor with a capacitance of \(1.0\, \mathrm{F}\). A large capacitance indicates that the capacitor can hold a significant amount of charge, which explains why modern capacitors can be much more efficient than older models.

Energy Storage

Energy storage in capacitors is a key feature that allows them to release energy on demand. Unlike batteries, which store energy chemically, capacitors store electrical energy in the form of an electrostatic field between their plates. The amount of energy a capacitor can store is tied to both its capacitance and the voltage applied across it.The energy \( E \) stored in a capacitor is calculated with the formula:\[ E = \frac{1}{2} C V^2 \]Here, \( C \) is the capacitance, and \( V \) is the voltage across the capacitor.For the problem at hand, the capacitor stores \(2.5\, \mathrm{J}\) of energy, enabling it to power a small 0.50-watt bulb for 5 seconds.

• Higher capacitance and voltage encapsulate more energy.
• This stored energy can be strategic in powering devices during brief outages or starting engines.

Electric Power

Electric power is the rate at which electrical energy is consumed or transferred. It is measured in watts (W) and is calculated using the formula \( P = VI \), where \( V \) is the voltage and \( I \) is the current. Another common calculation uses \( P = \frac{E}{t} \) where \( E \) is energy, and \( t \) is time.In this scenario, the bulb uses \( 0.50 \) watts of power over \( 5 \) seconds. This total energy consumption helps us understand how much energy the capacitor must release.

• Power is crucial because it determines how quickly energy is being used.
• High-power devices consume more energy in short durations compared to low-power ones.

This concept is crucial when designing circuits that need a precise amount of energy over a specific time frame.

Terminal Voltage

Terminal voltage is the voltage output that a power source provides to the external circuit under load conditions. It differs from the open-circuit voltage, which is the maximum voltage when the circuit is not connected to any load.In the context of capacitors, understanding terminal voltage is important because it influences how much energy the capacitor can store and release.For this problem:

• We calculate terminal voltage using \( V = \sqrt{\frac{2E}{C}} \).
• Given that the energy \( E \) used is \( 2.5\, \mathrm{J} \) and capacitance \( C \) is \( 1.0\, \mathrm{F} \), we compute \( V \) as \( \sqrt{5.0} \approx 2.24\, \mathrm{V} \).

Terminal voltage lets us know how efficiently the capacitor can discharge energy to its attached load, like the light bulb in this exercise.