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Chapter 19: Problem 58

A \(12.4 \mu \mathrm{F}\) capacitor is connected through a \(0.895 \mathrm{M} \Omega\) resistor to a constant potential difference of \(60.0 \mathrm{~V}\). (a) Compute the charge on the capacitor at the following times after the connections are made: \(0,5.0 \mathrm{~s}, 10.0 \mathrm{~s}, 20.0 \mathrm{~s},\) and \(100.0 \mathrm{~s} .\) (b) Compute the charging currents at the same instants. (c) Graph the results of parts (a) and (b) for \(t\) between 0 and \(20 \mathrm{~s}\)

Short Answer

Expert verified
The charge at 0, 5, 10, 20, and 100 seconds is approximately 0, 4.42x10^-4, 7.6x10^-4, 1.10x10^-3, and 7.44x10^-4 C, respectively. The current at these times is approximately 6.71x10^-5, 4.45x10^-5, 2.95x10^-5, 1.29x10^-5, and 0 A. Graph shows charging curve and exponential current decay.

Step by step solution

01

Identify Known Quantities

For the given capacitor-resistor circuit, the following values are known: capacitance \( C = 12.4 \mu F = 12.4 \times 10^{-6} F \), resistance \( R = 0.895 \times 10^6 \Omega \), and voltage \( V = 60.0 \mathrm{~V} \).
02

Calculate Time Constant

The time constant \( \tau \) for an RC circuit is calculated using \( \tau = R \cdot C \). Substitute the known values: \[ \tau = 0.895 \times 10^6 \cdot 12.4 \times 10^{-6} \approx 11.098 \;s. \]
03

Determine Charge at Each Time

The charge \( q(t) \) on the capacitor as a function of time \( t \) is given by \( q(t) = C \cdot V \cdot (1 - e^{-t/\tau}) \). Calculate \( q(t) \) for each time. - \( q(0) = 0 \)- \( q(5) = 12.4 \times 10^{-6} \cdot 60 \cdot (1 - e^{-5/11.098}) \approx 4.42 \times 10^{-4} \; C \)- \( q(10) = 12.4 \times 10^{-6} \cdot 60 \cdot (1 - e^{-10/11.098}) \approx 7.6 \times 10^{-4} \; C \)- \( q(20) = 12.4 \times 10^{-6} \cdot 60 \cdot (1 - e^{-20/11.098}) \approx 1.10 \times 10^{-3} \; C \)- \( q(100) = 12.4 \times 10^{-6} \cdot 60 \approx 7.44 \times 10^{-4} \; C \) (almost fully charged as \( t \to \infty \))
04

Calculate Current at Each Time

The current \( i(t) \) is given by \( i(t) = \frac{V}{R} \cdot e^{-t/\tau} \). Calculate \( i(t) \) for each time. - \( i(0) = \frac{60}{0.895 \times 10^6} \cdot e^{-0/11.098} \approx 6.71 \times 10^{-5} \; A \)- \( i(5) = \frac{60}{0.895 \times 10^6} \cdot e^{-5/11.098} \approx 4.45 \times 10^{-5} \; A \)- \( i(10) = \frac{60}{0.895 \times 10^6} \cdot e^{-10/11.098} \approx 2.95 \times 10^{-5} \; A \)- \( i(20) = \frac{60}{0.895 \times 10^6} \cdot e^{-20/11.098} \approx 1.29 \times 10^{-5} \; A \)- \( i(100) = \frac{60}{0.895 \times 10^6} \cdot e^{-100/11.098} \approx 0 \; A \) (current approaches zero as \( t \to \infty \))
05

Graph Charge and Current

Create a graph for the charge \( q(t) \) and current \( i(t) \) over time \( t \) from 0 to 20 seconds. - The charge \( q(t) \) would increase over time and will asymptotically approach the final charge value of \( 7.44 \times 10^{-4} \; C \).- The current \( i(t) \) will exponentially decrease, approaching zero as time increases.

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Key Concepts

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

Capacitor Charging
In RC circuits, when a capacitor is charged, it accumulates an electrical charge over time due to the applied voltage. Initially, the capacitor is uncharged. When connected to a voltage source, electrons start moving, and the capacitor begins to store charge. This process is not instantaneous. Instead, it follows an exponential pattern, meaning it charges rapidly at first, then more slowly as it approaches full charge.
The equation governing this process is:
  • \( q(t) = C \cdot V \cdot (1 - e^{-t/\tau}) \)
where \( q(t) \) is the charge at time \( t \), \( C \) is the capacitance, \( V \) is the voltage, and \( \tau \) is the time constant. Understanding the charging process helps visualize how capacitors behave in circuits, especially how they manage the flow of electrical energy over time.
Time Constant Calculation
The time constant \( \tau \) is crucial in determining how quickly a capacitor charges or discharges in an RC circuit. It is defined by the product of resistance \( R \) and capacitance \( C \) in the circuit:
  • \( \tau = R \cdot C \)
This constant (measured in seconds) dictates the rate of charging and discharging. In our example, a time constant of approximately 11.098 seconds was calculated.

This means that after about one time constant, the capacitor charges to about 63% of its full charge. After five time constants, the capacitor is almost fully charged (over 99% of its capacity).
  • Time constant is a key parameter for predicting how long circuits take to stabilize and can guide decisions in circuit design.
Current in Circuits
Current plays a pivotal role in charging a capacitor in an RC circuit. Initially, when the capacitor is uncharged, the current is at its maximum, limited only by the resistor in the circuit. As the capacitor charges, the current exponentially decreases because the potential difference between the capacitor plates lessens.

Current at any time \( t \) when charging can be calculated using:
  • \( i(t) = \frac{V}{R} \cdot e^{-t/\tau} \)
where \( i(t) \) is the current at time \( t \), \( V \) is the potential difference, and \( R \) is the resistance.

It's important to note that the current decreases to nearly zero as the capacitor reaches its full charge. This behavior highlights the dependency of current on both the voltage across and the resistance within a circuit.
Exponential Decay in Physics
In the context of RC circuits, exponential decay describes how quantities like current and voltage decrease over time. This is seen when examining how current decreases as a capacitor charges.
  • The decay can be described mathematically by an exponential function: \( e^{-t/\tau} \)
This function indicates that everything decreases rapidly at first and then slows as it approaches a certain value (in this case, zero for current and initial voltage stored in the capacitor).

Exponential decay is not exclusive to circuits; it appears in various physical processes, such as radioactive decay, population decline, cooling of objects, and more.
Understanding this behavior in RC circuits provides insight into energy transfer dynamics and time-dependent behaviors of other systems.

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Most popular questions from this chapter

Some types of spiders build webs that consist of threads made of dry silk coated with a solution of a variety of compounds. This coating leaves the threads, which are used to capture prey, hygroscopic-that is, they attract water from the atmosphere. It has been hypothesized that this aqueous coating makes the threads good electrical conductors. To test the electrical properties of coated thread, researchers placed a \(5 \mathrm{~mm}\) length of thread between two electrical contacts. The researchers stretched the thread in \(1 \mathrm{~mm}\) increments to more than twice its original length, and then allowed it to return to its original length, again in \(1 \mathrm{~mm}\) increments. Some of the resistance measurements are given in the table: $$\begin{array}{l|cccccccc}\begin{array}{l}\text { Resistance of } \\\\\text { thread }\left(10^{9} \Omega\right)\end{array} & 9 & 19 & 41 & 63 & 102 & 76 & 50 & 24 \\ \hline \begin{array}{l}\text { Length of } \\\\\text { thread }(\mathrm{mm})\end{array} & 5 & 7 & 9 & 11 & 13 & 9 & 7 & 5\end{array}$$ What is the best explanation for the behavior exhibited in the data? A. Longer threads can carry more current than shorter threads and so make better electrical conductors. B. The thread stops being a conductor when it is stretched to 13 \(\mathrm{mm}\) due to breaks that occur in the thin coating. C. As the thread is stretched, the coating thins and its resistance increases; as the thread is relaxed, the coating returns nearly to its original state. D. The resistance of the thread increases with distance from the end of the thread.

Some types of spiders build webs that consist of threads made of dry silk coated with a solution of a variety of compounds. This coating leaves the threads, which are used to capture prey, hygroscopic-that is, they attract water from the atmosphere. It has been hypothesized that this aqueous coating makes the threads good electrical conductors. To test the electrical properties of coated thread, researchers placed a \(5 \mathrm{~mm}\) length of thread between two electrical contacts. The researchers stretched the thread in \(1 \mathrm{~mm}\) increments to more than twice its original length, and then allowed it to return to its original length, again in \(1 \mathrm{~mm}\) increments. Some of the resistance measurements are given in the table: $$\begin{array}{l|cccccccc}\begin{array}{l}\text { Resistance of } \\\\\text { thread }\left(10^{9} \Omega\right)\end{array} & 9 & 19 & 41 & 63 & 102 & 76 & 50 & 24 \\ \hline \begin{array}{l}\text { Length of } \\\\\text { thread }(\mathrm{mm})\end{array} & 5 & 7 & 9 & 11 & 13 & 9 & 7 & 5\end{array}$$ In another experiment, a piece of the web is suspended so that it can move freely. When either a positively charged object or a negatively charged object is brought near the web, the thread is observed to move toward the charged object. What is the best interpretation of this observation? The web is A. a negatively charged conductor. B. a positively charged conductor. C. either a positively or negatively charged conductor. D. an electrically neutral conductor.

With a \(1500 \mathrm{M} \Omega\) resistor across its terminals, the terminal voltage of a certain battery is \(2.50 \mathrm{~V}\). With only a \(5.00 \Omega\) resistor across its terminals, the terminal voltage is \(1.75 \mathrm{~V}\). (a) Find the internal emf and the internal resistance of this battery. (b) What would be the terminal voltage if the \(5.00 \Omega\) resistor were replaced by a \(7.00 \Omega\) resistor?

A \(1.50 \mathrm{~m}\) cylindrical rod of diameter \(0.500 \mathrm{~cm}\) is connected to a power supply that maintains a constant potential difference of \(15.0 \mathrm{~V}\) across its ends, while an ammeter measures the current through it. You observe that at room temperature \(\left(20.0^{\circ} \mathrm{C}\right)\) the ammeter reads \(18.5 \mathrm{~A},\) while at \(92.0^{\circ} \mathrm{C}\) it reads \(17.2 \mathrm{~A} .\) You can ignore any thermal expansion of the rod. Find (a) the resistivity and (b) the temperature coefficient of resistivity at \(20^{\circ} \mathrm{C}\) for the material of the rod.

In an ionic solution, a current consists of \(\mathrm{Ca}^{2}+\) ions (of charge \(+2 e\) ) and \(\mathrm{Cl}^{-}\) ions (of charge \(-e\) ) traveling in opposite directions. If \(5.11 \times 10^{18} \mathrm{Cl}^{-}\) ions go from \(A\) to \(B\) every \(0.50 \mathrm{~min},\) while \(3.24 \times 10^{18} \mathrm{Ca}^{2+}\) ions move from \(B\) to \(A,\) what is the current (in mA) through this solution, and in which direction (from \(A\) to \(B\) or from \(B\) to \(A\) ) is it going?
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