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(II) Compact "ultracapacitors" with capacitance values up to several thousand farads are now commercially available. One application for ultracapacitors is in providing power for electrical circuits when other sources (such as a battery) are turned off. To get an idea of how much charge can be stored in such a component, assume a \(1200-\mathrm{F}\) ultracapacitor is initially charged to 12.0 \(\mathrm{V}\) by a battery and is then disconnected from the battery. If charge is then drawn off the plates of this capacitor at a rate of \(1.0 \mathrm{mC} / \mathrm{s},\) say, to power the backup memory of some electrical gadget, how long (in days) will it take for the potential difference across this capacitor to drop to 6.0 \(\mathrm{V}\) ?

Short Answer

Expert verified
It will take approximately 83.33 days for the voltage to drop to 6.0V.

Step by step solution

01

Understand the Problem

We need to determine how long it takes for the voltage across a discharged ultracapacitor to drop from 12.0V to 6.0V when charge is withdrawn at a constant rate. We will use the relationship between capacitance, charge, and voltage.
02

Calculate Initial Charge

The initial charge on the capacitor when it is charged to 12.0V can be calculated using the formula: \[ Q = C \times V \]where \( C = 1200 \text{F} \) and \( V = 12.0 \text{V} \). Plug in the values: \[ Q = 1200 \times 12 = 14400 \text{C} \]
03

Calculate Final Charge

To find the final charge when the voltage drops to 6.0V, again use the formula: \[ Q = C \times V \]where \( V = 6.0 \text{V} \). Plug in the values: \[ Q = 1200 \times 6 = 7200 \text{C} \]
04

Determine Change in Charge

Calculate the change in charge \( \Delta Q \) as the difference between initial and final charge: \[ \Delta Q = Q_{initial} - Q_{final} = 14400 \text{C} - 7200 \text{C} = 7200 \text{C} \]
05

Calculate Time to Discharge

Since the charge is drawn off at a rate of 1.0 mC/s (which is \( 0.001 \text{C/s} \)), the time \( t \) required can be calculated using:\[ t = \frac{\Delta Q}{\text{rate}} = \frac{7200}{0.001} \] Compute the result: \[ t = 7,200,000 \text{s} \]
06

Convert Time to Days

Convert the time from seconds to days; there are 86400 seconds in a day: \[ t = \frac{7,200,000}{86400} \approx 83.33 \text{days} \]

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Key Concepts

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

Capacitance
Capacitance is a key property of capacitors and ultracapacitors, which are components used in electrical circuits to store charge. It is the ability of a system to store an electric charge. The unit of capacitance is the farad, named after Michael Faraday. A farad is defined as one coulomb of electric charge stored per one volt of electric potential difference.
Capacitance can be thought of as the potential of a capacitor to hold charge and is given by the equation: \[C = \frac{Q}{V}\]where \(C\) is the capacitance, \(Q\) is the charge stored, and \(V\) is the potential difference across the capacitor.
In the context of ultracapacitors, which have capacitance values up to thousands of farads, they are extremely capable of storing large charges compared to regular capacitors. This attribute makes them exceptionally useful for applications requiring quick energy release and recharge cycles.
Electrical Circuits
Electrical circuits are networks consisting of interconnected electrical components such as resistors, capacitors, and power sources. In a circuit, current flows through the connectivity path when there is a potential difference across its terminals.
Ultracapacitors, due to their high capacitance, can provide a steady source of power, similar to batteries, but with the advantage of longer cycle lives and rapid charge-discharge capabilities. This makes them valuable in backup power applications where batteries might fall short.
When integrated into a circuit, ultracapacitors can help maintain functionality during power transitions, such as when a primary power source is temporarily turned off. They release their stored energy to keep the circuit operational, making them a powerful component in maintaining the stability of electrical circuits.
Charge Storage
Charge storage is a fundamental function of capacitors, where energy is stored in the form of electric charge. This is crucial for devices that need to store and release energy quickly. The quantity of charge \( Q \) stored in a capacitor is directly proportional to its capacitance \( C \) and the voltage \( V \) applied across its terminals, as given by: \[Q = C \times V\]In our exercise, an ultracapacitor with a capacitance of 1200 farads stores charge when connected across a 12V battery. The high capacitance allows it to store a significant amount of charge, calculated as 14400 coulombs in this scenario.
As charge storage takes place, the capacitor retains energy until it is needed. When disconnected, it can discharge and supply energy at a controlled rate to other devices, like providing consistent power to memory components in electronic gadgets.
Potential Difference
Potential difference, often referred to as voltage, is the measure of electrical tension between two points. It is the force that pushes electric charge through a circuit. Potential difference is measured in volts and represents the work needed to move a unit charge between two points.
In our exercise, the potential difference across the ultracapacitor initially charged by the battery is 12 volts. As charge is drawn from the capacitor, this potential difference reduces, reflecting the energy released into the circuit.
When the potential difference drops to 6 volts, it indicates how much energy remains stored in the capacitor. Measuring potential difference is critical for understanding how long a capacitor can continue to supply energy before it needs to be recharged, which is precisely what the given exercise calculates through time to discharge.

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Most popular questions from this chapter

(I) There is an electric field near the Earth's surface whose intensity is about 150 \(\mathrm{V} / \mathrm{m}\) . How much energy is stored per cubic meter in this field?

(1I) Given three capacitors, \(C_{1}=2.0 \mu \mathrm{F}, C_{2}=1.5 \mu \mathrm{F},\) and \(C_{3}=3.0 \mu \mathrm{F}, \quad\) what arrangement of parallel and series connections with a \(12-\mathrm{V}\) battery will give the minimum voltage drop across the \(2.0-\mu \mathrm{F}\) capacitor? What is the minimum voltage drop?

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