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A capacitor \(C\) is initially uncharged. At time \(t=0,\) the capacitor is attached through a resistor \(R\) to a battery. The energy stored in the capacitor increases, eventually reaching a value \(U\) as \(t \rightarrow \infty\), After a time equal to the time constant \(\tau=R C\), the energy stored in the capacitor is given by a) \(U / e\). c) \(U(1-1 / e)^{2}\) b) \(U / e^{2}\) d) \(U(1-1 / e)\).

Short Answer

Expert verified
Answer: d) \(U(1-1 / e)\).

Step by step solution

01

Calculate the charge on the capacitor as a function of time

To find the charge on the capacitor as a function of time, we can use the formula: \(Q(t) = Q_{max}(1-e^{-\frac{t}{RC}})\) where \(Q(t)\) is the charge on the capacitor at time \(t\), \(Q_{max}\) is the maximum charge that can be stored in the capacitor, \(R\) is the resistor's resistance, and \(C\) is the capacitance of the capacitor.
02

Calculate the voltage across the capacitor at time \(t\)

To find the voltage across the capacitor at time \(t\), we can use the relationship: \(V(t) = \frac{Q(t)}{C}\) So, the voltage will be: \(V(t) = \frac{Q_{max}}{C}(1-e^{-\frac{t}{RC}})\)
03

Calculate the energy stored in the capacitor at any given time

The energy stored in the capacitor at any given time (\(t\)) can be calculated using the formula: \(U(t) = \frac{1}{2}CV^2(t)\) Plugging the expression for voltage from Step 2, we get: \(U(t) = \frac{1}{2}C\left[\frac{Q_{max}}{C}(1-e^{-\frac{t}{RC}})\right]^2\)
04

Calculate the energy stored in the capacitor when \(t = RC = \tau\)

Now that we have the expression for the energy stored in the capacitor, we can find the energy at time \(t=RC=\tau\) by substituting the value of \(t\) into the expression: \(U(\tau) = \frac{1}{2}C\left[\frac{Q_{max}}{C}(1-e^{-\frac{RC}{RC}})\right]^2\) Simplifying and calculating, we get: \(U(\tau) = \frac{1}{2}C\left[\frac{Q_{max}}{C}(1-e^{-1})\right]^2\)
05

Express the energy stored in terms of the maximum energy \(U\)

The maximum energy stored in the capacitor can be calculated when the capacitor is fully charged: \(U = \frac{1}{2}CV_{max}^2\) Since \(V_{max} = \frac{Q_{max}}{C},\) we can rewrite the maximum energy as: \(U = \frac{1}{2}C\left(\frac{Q_{max}}{C}\right)^2\) Now we can express the energy stored at time \(t=RC=\tau\) in terms of the maximum energy \(U\): \(U(\tau) = \frac{1}{2}C\left[\frac{Q_{max}}{C}(1-e^{-1})\right]^2 \div \frac{1}{2}C\left(\frac{Q_{max}}{C}\right)^2\) Simplifying, we find that the energy stored in the capacitor at time \(\tau\) is: \(U(\tau) = U(1-e^{-1})^2\) Comparing with the given options, the answer is: d) \(U(1-1 / e)\).

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Key Concepts

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

Capacitor Charging
The process of capacitor charging is like filling a bucket with water. A capacitor stores energy in the form of an electric field between its plates, much like a bucket stores water. When connected to a battery through a resistor, the charge starts flowing into the capacitor, and it begins to charge. As the charge accumulates, the voltage across the capacitor also increases, until it equals the battery voltage. This charging process can be modeled mathematically by the equation
\(Q(t) = Q_{max}(1 - e^{-\frac{t}{RC}})\),
where \(Q(t)\) is the charge at time \(t\), \(Q_{max}\) is the maximum charge, \(R\) is the resistance, and \(C\) is the capacitance. Understanding this equation provides insight into how the charge on the capacitor evolves over time during the charging process.

Factors Influencing Capacitor Charging

  • Resistance (R): A higher resistance slows down the charging process, similar to a narrow pipe restricting water flow into the bucket.
  • Capacitance (C): A larger capacitance means a greater storage capacity, like a bigger bucket taking longer to fill.
  • Battery Voltage: The higher the battery voltage, the more charge can be stored, up to the limit described by \(Q_{max}\).
RC Circuits
RC circuits, which combine a resistor (R) and a capacitor (C) in series, are fundamental in understanding electrical circuits. They serve as simple models for analyzing charging and discharging behavior in complex systems. When connected to a power source, the capacitor in an RC circuit charges up through the resistor, and when disconnected, it discharges through that same resistor.
These circuits are also pivotal in creating various electronic filters, timers, and signal-processing systems. The resistor in the circuit controls the rate at which the capacitor charges or discharges by limiting the flow of current, in the same way a faucet controls the flow of water into a bucket.

Applications of RC Circuits

  • Timing elements in electronic devices.
  • Providing phase shifts in audio equipment.
  • Smoothing voltage in power supplies.
Understanding the behavior of RC circuits is crucial for designing and analyzing electronic systems, as they can affect the time it takes for a system to respond to changes in input signals.
Time Constant
The time constant, represented by the Greek letter \(\tau\), is a significant parameter in RC circuits. It symbolizes the time required for the voltage across the capacitor to reach approximately 63% of its maximum value, or conversely, to drop to about 37% of its maximum value when discharging.
Mathematically, \(\tau = RC\), where \(R\) is the resistance and \(C\) is the capacitance. The time constant determines how quickly an RC circuit reacts to a change in voltage. In the context of capacitor charging, after one time constant has passed, the stored energy within the capacitor is:
\(U(\tau) = U(1-1/e)^2\),
indicating that it is not fully charged but getting closer. This parameter is crucial when one needs to predict how long a capacitor will take to charge or discharge in applications that require precise timing.

Understanding Time Constant

  • Larger Capacitance (C): Leads to a longer time constant, indicating a slower charging/discharging process.
  • Higher Resistance (R): Also results in a greater time constant, further slowing the rate of charge or discharge.
  • Environmental Factors: Temperature and other environmental conditions can affect the resistance and thus the time constant.
Exponential Decay
Exponential decay is a pattern of reduction over time that can be applied to various physical and mathematical phenomena, including the discharge of a capacitor in an RC circuit. The term 'decay' describes how the quantity decreases at a rate proportional to its current value.
In RC circuits, the charge and voltage during discharge exhibit exponential decay, described by equations like:
\(V(t) = V_{max}e^{-\frac{t}{RC}}\).
This equation illustrates how the voltage across the capacitor decreases rapidly at first and then gradually reaches zero. Exponential decay is not limited to just electrical systems; it also describes processes such as radioactive decay and cooling of objects.

Characteristics of Exponential Decay

  • Initial Rapid Decrease: The quantity drops quickly at the beginning.
  • Gradual Leveling Off: Over time, the rate of decrease slows down significantly.
  • Never Zero: Theoretically, the decaying quantity never reaches zero, constantly approaching but not touching it.
Comprehending exponential decay is key to predicting how fast a system loses its energy or how a signal fades over time, impacting the design of electrical circuits and the understanding of natural processes.

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Most popular questions from this chapter

A battery, a resistor, and a capacitor are connected in series in an RC circuit. What happens to the current through a resistor after a long time? Explain using Kirchhoff's rules.

A capacitor bank is designed to discharge 5.0 J of energy through a \(10.0-\mathrm{k} \Omega\) resistor array in under \(2.0 \mathrm{~ms}\) To what potential difference must the bank be charged, and what must the capacitance of the bank be?

A circuit consists of two \(1.00-\mathrm{k} \Omega\) resistors in series with an ideal \(12.0-\mathrm{V}\) battery. a) Calculate the current flowing through each resistor. b) A student trying to measure the current flowing through one of the resistors inadvertently connects an ammeter in parallel with that resistor rather than in series with it. How much current will flow through the ammeter, assuming that it has an internal resistance of \(1.0 \Omega ?\)

Kirchhoff's Junction Rule states that a) the algebraic sum of the currents at any junction in a circuit must be zero. b) the algebraic sum of the potential changes around any closed loop in a circuit must be zero. c) the current in a circuit with a resistor and a capacitor varies exponentially with time. d) the current at a junction is given by the product of the resistance and the capacitance. e) the time for the current development at a junction is given by the product of the resistance and the capacitance.

Two capacitors in series are charged through a resistor. Identical capacitors are instead connected in parallel and charged through the same resistor. How do the times required to fully charge the two sets of capacitors compare?

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